My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.
You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).
location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c
How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/
Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.
Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?
The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.
Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/
import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary
Related
What are these two terms in an understandable way?
Greedy will consume as much as possible. From http://www.regular-expressions.info/repeat.html we see the example of trying to match HTML tags with <.+>. Suppose you have the following:
<em>Hello World</em>
You may think that <.+> (. means any non newline character and + means one or more) would only match the <em> and the </em>, when in reality it will be very greedy, and go from the first < to the last >. This means it will match <em>Hello World</em> instead of what you wanted.
Making it lazy (<.+?>) will prevent this. By adding the ? after the +, we tell it to repeat as few times as possible, so the first > it comes across, is where we want to stop the matching.
I'd encourage you to download RegExr, a great tool that will help you explore Regular Expressions - I use it all the time.
'Greedy' means match longest possible string.
'Lazy' means match shortest possible string.
For example, the greedy h.+l matches 'hell' in 'hello' but the lazy h.+?l matches 'hel'.
Greedy quantifier
Lazy quantifier
Description
*
*?
Star Quantifier: 0 or more
+
+?
Plus Quantifier: 1 or more
?
??
Optional Quantifier: 0 or 1
{n}
{n}?
Quantifier: exactly n
{n,}
{n,}?
Quantifier: n or more
{n,m}
{n,m}?
Quantifier: between n and m
Add a ? to a quantifier to make it ungreedy i.e lazy.
Example:
test string : stackoverflow
greedy reg expression : s.*o output: stackoverflow
lazy reg expression : s.*?o output: stackoverflow
Greedy means your expression will match as large a group as possible, lazy means it will match the smallest group possible. For this string:
abcdefghijklmc
and this expression:
a.*c
A greedy match will match the whole string, and a lazy match will match just the first abc.
As far as I know, most regex engine is greedy by default. Add a question mark at the end of quantifier will enable lazy match.
As #Andre S mentioned in comment.
Greedy: Keep searching until condition is not satisfied.
Lazy: Stop searching once condition is satisfied.
Refer to the example below for what is greedy and what is lazy.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String args[]){
String money = "100000000999";
String greedyRegex = "100(0*)";
Pattern pattern = Pattern.compile(greedyRegex);
Matcher matcher = pattern.matcher(money);
while(matcher.find()){
System.out.println("I'm greedy and I want " + matcher.group() + " dollars. This is the most I can get.");
}
String lazyRegex = "100(0*?)";
pattern = Pattern.compile(lazyRegex);
matcher = pattern.matcher(money);
while(matcher.find()){
System.out.println("I'm too lazy to get so much money, only " + matcher.group() + " dollars is enough for me");
}
}
}
The result is:
I'm greedy and I want 100000000 dollars. This is the most I can get.
I'm too lazy to get so much money, only 100 dollars is enough for me
Taken From www.regular-expressions.info
Greediness: Greedy quantifiers first tries to repeat the token as many times
as possible, and gradually gives up matches as the engine backtracks to find
an overall match.
Laziness: Lazy quantifier first repeats the token as few times as required, and
gradually expands the match as the engine backtracks through the regex to
find an overall match.
From Regular expression
The standard quantifiers in regular
expressions are greedy, meaning they
match as much as they can, only giving
back as necessary to match the
remainder of the regex.
By using a lazy quantifier, the
expression tries the minimal match
first.
Greedy matching. The default behavior of regular expressions is to be greedy. That means it tries to extract as much as possible until it conforms to a pattern even when a smaller part would have been syntactically sufficient.
Example:
import re
text = "<body>Regex Greedy Matching Example </body>"
re.findall('<.*>', text)
#> ['<body>Regex Greedy Matching Example </body>']
Instead of matching till the first occurrence of ‘>’, it extracted the whole string. This is the default greedy or ‘take it all’ behavior of regex.
Lazy matching, on the other hand, ‘takes as little as possible’. This can be effected by adding a ? at the end of the pattern.
Example:
re.findall('<.*?>', text)
#> ['<body>', '</body>']
If you want only the first match to be retrieved, use the search method instead.
re.search('<.*?>', text).group()
#> '<body>'
Source: Python Regex Examples
Greedy Quantifiers are like the IRS
They’ll take as much as they can. e.g. matches with this regex: .*
$50,000
Bye-bye bank balance.
See here for an example: Greedy-example
Non-greedy quantifiers - they take as little as they can
Ask for a tax refund: the IRS sudden becomes non-greedy - and return as little as possible: i.e. they use this quantifier:
(.{2,5}?)([0-9]*) against this input: $50,000
The first group is non-needy and only matches $5 – so I get a $5 refund against the $50,000 input.
See here: Non-greedy-example.
Why do we need greedy vs non-greedy?
It becomes important if you are trying to match certain parts of an expression. Sometimes you don't want to match everything - as little as possible. Sometimes you want to match as much as possible. Nothing more to it.
You can play around with the examples in the links posted above.
(Analogy used to help you remember).
Greedy means it will consume your pattern until there are none of them left and it can look no further.
Lazy will stop as soon as it will encounter the first pattern you requested.
One common example that I often encounter is \s*-\s*? of a regex ([0-9]{2}\s*-\s*?[0-9]{7})
The first \s* is classified as greedy because of * and will look as many white spaces as possible after the digits are encountered and then look for a dash character "-". Where as the second \s*? is lazy because of the present of *? which means that it will look the first white space character and stop right there.
Best shown by example. String. 192.168.1.1 and a greedy regex \b.+\b
You might think this would give you the 1st octet but is actually matches against the whole string. Why? Because the.+ is greedy and a greedy match matches every character in 192.168.1.1 until it reaches the end of the string. This is the important bit! Now it starts to backtrack one character at a time until it finds a match for the 3rd token (\b).
If the string a 4GB text file and 192.168.1.1 was at the start you could easily see how this backtracking would cause an issue.
To make a regex non greedy (lazy) put a question mark after your greedy search e.g
*?
??
+?
What happens now is token 2 (+?) finds a match, regex moves along a character and then tries the next token (\b) rather than token 2 (+?). So it creeps along gingerly.
To give extra clarification on Laziness, here is one example which is maybe not intuitive on first look but explains idea of "gradually expands the match" from Suganthan Madhavan Pillai answer.
input -> some.email#domain.com#
regex -> ^.*?#$
Regex for this input will have a match. At first glance somebody could say LAZY match(".*?#") will stop at first # after which it will check that input string ends("$"). Following this logic someone would conclude there is no match because input string doesn't end after first #.
But as you can see this is not the case, regex will go forward even though we are using non-greedy(lazy mode) search until it hits second # and have a MINIMAL match.
try to understand the following behavior:
var input = "0014.2";
Regex r1 = new Regex("\\d+.{0,1}\\d+");
Regex r2 = new Regex("\\d*.{0,1}\\d*");
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // "0014.2"
input = " 0014.2";
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // " 0014"
input = " 0014.2";
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // ""
I have this regular expression :
string[] values = Regex
.Matches(mystring4, #"([\w-[\d]][\w\s-[\d]]+)|([0-9]+)")
.OfType<Match>()
.Select(match => match.Value.Trim())
.ToArray();
This regular expression turns this string :
MY LIMITED COMPANY (52100000 / 58447000)";
To these strings :
MY LIMITED COMPANY - 52100000 - 58447000
This also works on non-English characters.
But there is one problem, when I have this string : MY. LIMITED. COMPANY. , it splits that too. I don't want that. I don't want that regular expression to work on dots. How can I do that? Thanks.
You may add the dot after each \w in your pattern, and I also suggest removing unnecessary ( and ):
string[] values = Regex
.Matches("MY. LIMITED. COMPANY. (52100000 / 58447000)", #"[\w.-[\d]][\w.\s-[\d]]+|[0-9]+")
.OfType<Match>()
.Select(match => match.Value.Trim())
.ToArray();
foreach (var s in values)
Console.WriteLine(s);
See the C# demo
Pattern:
[\w.-[\d]] - one Unicode letter or underscore ([\w-[\d]]) or a dot (.)
[\w.\s-[\d]]+ - 1 or more (due to + quantifier at the end) characters that are either Unicode letters or underscore, ., or whitespace (\s)
| - or
[0-9]+ - one or more ASCII-only digits
I'd simplify the expression. What if the names in the front include numbers? Not that my solution doesn't exactly mimic the original expression. It will allow numbers in the name part.
Let's start from the beginning:
To match words all you need is a sequence of word characters:
\w+
This will match any alphanumerical characters including underscores (_).
Considering you want the possibility of the word ending with a dot, you can add it and make it optional (one or zero matches):
\w+\.?
Note the escape to make it an actual character rather than a character class "any character".
To match another potential word following, we now simply duplicate this match, add a white space before, and once again make it optional using the * quantifier:
\w+\.?(?:\w+\.?)*
In case you haven't seen a group starting with ?: is a non-matching group. In essence this works like a usual group, but won't save a matching group in your results.
And that's it already. This pattern will split your demo string as expected. Of course there could be other possible characters not being covered by this.
You can see the results of this matching online here and also play around with it.
To test your regular expressions (and to learn them), I'd really recommend you using a tool such as http://regex101.com
It has an input mask allowing you to provide your pattern and your target string. On the right hand side it will first explain the pattern to you (to see if it's indeed what you had in mind) and below it will show all the groups matched. Just keep in mind it actually uses slightly different flavors of regular expressions, but this shouldn't matter for such simple patterns. (I'm not affiliated with that site, just consider it really useful.)
As an alternative, to directly use C#'s regex parser, you can also try this Regex Tester. This works in a similar way, although doesn't include any explanations, which might be not as ideal for someone just getting started.
I have an app where users can specify regular expressions in a number of places. These are used while running the app to check if text (e.g. URLs and HTML) matches the regexes. Often the users want to be able to say where the text matches ABC and does not match XYZ. To make it easy for them to do this I am thinking of extending regular expression syntax within my app with a way to say 'and does not contain pattern'. Any suggestions on a good way to do this?
My app is written in C# .NET 3.5.
My plan (before I got the awesome answers to this question...)
Currently I'm thinking of using the ¬ character: anything before the ¬ character is a normal regular expression, anything after the ¬ character is a regular expression that can not match in the text to be tested.
So I might use some regexes like this (contrived) example:
on (this|that|these) day(s)?¬(every|all) day(s) ?
Which for example would match 'on this day the man said...' but would not match 'on this day and every day after there will be ...'.
In my code that processes the regex I'll simply split out the two parts of the regex and process them separately, e.g.:
public bool IsMatchExtended(string textToTest, string extendedRegex)
{
int notPosition = extendedRegex.IndexOf('¬');
// Just a normal regex:
if (notPosition==-1)
return Regex.IsMatch(textToTest, extendedRegex);
// Use a positive (normal) regex and a negative one
string positiveRegex = extendedRegex.Substring(0, notPosition);
string negativeRegex = extendedRegex.Substring(notPosition + 1, extendedRegex.Length - notPosition - 1);
return Regex.IsMatch(textToTest, positiveRegex) && !Regex.IsMatch(textToTest, negativeRegex);
}
Any suggestions on a better way to implement such an extension? I'd need to be slightly cleverer about splitting the string on the ¬ character to allow for it to be escaped, so wouldn't just use the simple Substring() splitting above. Anything else to consider?
Alternative plan
In writing this question I also came across this answer which suggests using something like this:
^(?=(?:(?!negative pattern).)*$).*?positive pattern
So I could just advise people to use a pattern like, instead of my original plan, when they want to NOT match certain text.
Would that do the equivalent of my original plan? I think it's quite an expensive way to do it peformance-wise, and since I'm sometimes parsing large html documents this might be an issue, whereas I suppose my original plan would be more performant. Any thoughts (besides the obvious: 'try both and measure them!')?
Possibly pertinent for performance: sometimes there will be several 'words' or a more complex regex that can not be in the text, like (every|all) in my example above but with a few more variations.
Why!?
I know my original approach seems weird, e.g. why not just have two regexes!? But in my particular application administrators provide the regular expressions and it would be rather difficult to give them the ability to provide two regular expressions everywhere they can currently provide one. Much easier in this case to have a syntax for NOT - just trust me on that point.
I have an app that lets administrators define regular expressions at various configuration points. The regular expressions are just used to check if text or URLs match a certain pattern; replacements aren't made and capture groups aren't used. However, often they would like to specify a pattern that says 'where ABC is not in the text'. It's notoriously difficult to do NOT matching in regular expressions, so the usual way is to have two regular expressions: one to specify a pattern that must be matched and one to specify a pattern that must not be matched. If the first is matched and the second is not then the text does match. In my application it would be a lot of work to add the ability to have a second regular expression at each place users can provide one now, so I would like to extend regular expression syntax with a way to say 'and does not contain
pattern'.
You don't need to introduce a new symbol. There already is support for what you need in most regex engines. It's just a matter of learning it and applying it.
You have concerns about performance, but have you tested it? Have you measured and demonstrated those performance problems? It will probably be just fine.
Regex works for many many people, in many many different scenarios. It probably fits your requirements, too.
Also, the complicated regex you found on the other SO question, can be simplified. There are simple expressions for negative and positive lookaheads and lookbehinds.
?! ?<! ?= ?<=
Some examples
Suppose the sample text is <tr valign='top'><td>Albatross</td></tr>
Given the following regex's, these are the results you will see:
tr - match
td - match
^td - no match
^tr - no match
^<tr - match
^<tr>.*</tr> - no match
^<tr.*>.*</tr> - match
^<tr.*>.*</tr>(?<tr>) - match
^<tr.*>.*</tr>(?<!tr>) - no match
^<tr.*>.*</tr>(?<!Albatross) - match
^<tr.*>.*</tr>(?<!.*Albatross.*) - no match
^(?!.*Albatross.*)<tr.*>.*</tr> - no match
Explanations
The first two match because the regex can apply anywhere in the sample (or test) string. The second two do not match, because the ^ says "start at the beginning", and the test string does not begin with td or tr - it starts with a left angle bracket.
The fifth example matches because the test string starts with <tr.
The sixth does not, because it wants the sample string to begin with <tr>, with a closing angle bracket immediately following the tr, but in the actual test string, the opening tr includes the valign attribute, so what follows tr is a space. The 7th regex shows how to allow the space and the attribute with wildcards.
The 8th regex applies a positive lookbehind assertion to the end of the regex, using ?<. It says, match the entire regex only if what immediately precedes the cursor in the test string, matches what's in the parens, following the ?<. In this case, what follows that is tr>. After evaluating ``^.*, the cursor in the test string is positioned at the end of the test string. Therefore, thetr>` is matched against the end of the test string, which evaluates to TRUE. Therefore the positive lookbehind evaluates to true, therefore the overall regex matches.
The ninth example shows how to insert a negative lookbehind assertion, using ?<! . Basically it says "allow the regex to match if what's right behind the cursor at this point, does not match what follows ?<! in the parens, which in this case is tr>. The bit of regex preceding the assertion, ^<tr.*>.*</tr> matches up to and including the end of the string. Because the pattern tr> does match the end of the string. But this is a negative assertion, therefore it evaluates to FALSE, which means the 9th example is NOT a match.
The tenth example uses another negative lookbehind assertion. Basically it says "allow the regex to match if what's right behind the cursor at this point, does not match what's in the parens, in this case Albatross. The bit of regex preceding the assertion, ^<tr.*>.*</tr> matches up to and including the end of the string. Checking "Albatross" against the end of the string yields a negative match, because the test string ends in </tr>. Because the pattern inside the parens of the negative lookbehind does NOT match, that means the negative lookbehind evaluates to TRUE, which means the 10th example is a match.
The 11th example extends the negative lookbehind to include wildcards; in english the result of the negative lookbehind is "only match if the preceding string does not include the word Albatross". In this case the test string DOES include the word, the negative lookbehind evaluates to FALSE, and the 11th regex does not match.
The 12th example uses a negative lookahead assertion. Like lookbehinds, lookaheads are zero-width - they do not move the cursor within the test string for the purposes of string matching. The lookahead in this case, rejects the string right away, because .*Albatross.* matches; because it is a negative lookahead, it evaluates to FALSE, which mean the overall regex fails to match, which means evaluation of the regex against the test string stops there.
example 12 always evaluates to the same boolean value as example 11, but it behaves differently at runtime. In ex 12, the negative check is performed first, at stops immediately. In ex 11, the full regex is applied, and evaluates to TRUE, before the lookbehind assertion is checked. So you can see that there may be performance differences when comparing lookaheads and lookbehinds. Which one is right for you depends on what you are matching on, and the relative complexity of the "positive match" pattern and the "negative match" pattern.
For more on this stuff, read up at http://www.regular-expressions.info/
Or get a regex evaluator tool and try out some tests.
like this tool:
source and binary
You can easily accomplish your objectives using a single regex. Here is an example which demonstrates one way to do it. This regex matches a string containing "cat" AND "lion" AND "tiger", but does NOT contain "dog" OR "wolf" OR "hyena":
if (Regex.IsMatch(text, #"
# Match string containing all of one set of words but none of another.
^ # anchor to start of string.
# Positive look ahead assertions for required substrings.
(?=.*? cat ) # Assert string has: 'cat'.
(?=.*? lion ) # Assert string has: 'lion'.
(?=.*? tiger ) # Assert string has: 'tiger'.
# Negative look ahead assertions for not-allowed substrings.
(?!.*? dog ) # Assert string does not have: 'dog'.
(?!.*? wolf ) # Assert string does not have: 'wolf'.
(?!.*? hyena ) # Assert string does not have: 'hyena'.
",
RegexOptions.Singleline | RegexOptions.IgnoreCase |
RegexOptions.IgnorePatternWhitespace)) {
// Successful match
} else {
// Match attempt failed
}
You can see the needed pattern. When assembling the regex, be sure to run each of the user provided sub-strings through the Regex.escape() method to escape any metacharacters it may contain (i.e. (, ), | etc). Also, the above regex is written in free-spacing mode for readability. Your production regex should NOT use this mode, otherwise whitespace within the user substrings would be ignored.
You may want to add \b word boundaries before and after each "word" in each assertion if the substrings consist of only real words.
Note also that the negative assertion can be made a bit more efficient using the following alternative syntax:
(?!.*?(?:dog|wolf|hyena))
Greetings.
I've been tasked with debugging part of an application that involves a Regex -- but, I have never dealt with Regex before. Two questions:
1) I know that the regexes are supposed to be testing whether or not two strings are equivalent, but what specifically do the two regex statements, below, mean in plain English?
2) Does anyone have a recommendation on websites / sources where I can learn more about Regexes? (preferably in C#)
if (Regex.IsMatch(testString, #"^(\s*?)(" + tag + #")(\s*?),", RegexOptions.IgnoreCase))
{
result = true;
}
else if (Regex.IsMatch(testString, #",(\s*?)(" + tag + #")(\s*?),", RegexOptions.IgnoreCase))
{
result = true;
}
It's going to be difficult to tell what that regex means, without knowing what's in tag. In fact, it looks like that regex is broken (or, at least, doesn't properly escape inputs).
Roughly speaking, for the first regex:
The ^ says to match at the beginning of the string.
The (...) sets up a capturing group (which is available, although this example apparently doesn't use it).
The \s matches any white space characters (spaces, tabs, etc.)
The *? matches zero or more of the previous character (in this case, whitespace), and because it has a question-mark, it matches the minimum number of characters needed to make the rest of the expression work.
The (" + tag + #") inserts the contents of the tag into the regex. As I mention, that's dangerous, without escaping.
The (\s*?) matches the same as the before (the minimum number of whitespace characters)
The , matches a trailing comma.
The second regex is very similar, but looks for a starting comma (rather than the beginning of the string).
I like the Python documentation for Regular Expressions, but it looks like this site
has a pretty good, basic introduction, with C# examples.
One word - Cribsheet (or is that two?) :)
I'm not c# savvy but I can recommend an awesome guide to regular expressions that I use for Bash and Java programming. It applies to pretty much all languages:
http://www.amazon.com/Mastering-Regular-Expressions-Jeffrey-Friedl/dp/0596528124/ref=tmm_pap_title_0
It is totally worth $30 to own this book. It is VERY thorough and helped my fundamental understanding of Regex a lot.
-Ryan
Since you specifically tagged C#, I recommend the Regex Hero as a tool you can use to play around with them since it's running on .NET. It also lets you toggle the different RegexOptions flags as you would pass them into the constructor when creating a new Regex.
Also, if you're using a version of Visual Studio 2010 that supports extensions, I would take a look at the Regex Editor extension... it will popup whenever you type new Regex( and offer you some guidance and autocomplete for your regex pattern.
Using The Regex Coach
The regular expression is a sequence consisting of the expression '(\s*?)', the expression '(tag)', the expression '(\s*?)', and the character ','.
where (\s*?) is defined as The regular expression is a repetition which matches a whitespace character as often as necessary.
the second one matches a , at the start too
As for good learning websites, I like www.regular-expressions.info/
Super simple version:
At the start of a string 0 or more spaces, whatever Tag is, 0 or More spaces, a comma.
the second one is
a comma, 0 or more spaces, whatever Tag is, 0 or More spaces, a comma.
Once you have the very basic idea about regex (it's full of resources over there) I recommend you to use Expresso for creating your regular expressions.
Expresso editor is equally suitable as a teaching tool for the beginning user of regular expressions or as a full-featured development environment for the experienced programmer or web designer with an extensive knowledge of regular expressions.
Your premise is not correct. Regular expressions are not used to tell if two strings are equivalent, but rather if the input string matches a certain pattern.
The first test above looks for any text that does not contain "zero or more whitespace charaters" searching "non-greedy". Then matches the text of the variable "tag" in the middle, then "zero or more whitespace characters, non greedy" again.
The second one is very similar, except that it allows for beginning whitespace as long as it follows a comma.
It is hard to explain "non-greedy" in this context, especially involving whitespace characters, so look here for more information.
A regular expression is a way to describe a set of strings that have some particular characteristics.
They don't merely need just to compare two strings.. what you usually do it to test if a string matches a particular regular expression. They can also be used to do simple parsing of a string in tokens that respect some patterns..
The good thing about regexps is that they allow you to express certain constraints inside a string keeping it general and able to match a group of strings that respect those constraints.. then they follow a formal specification that doesn't leave ambiguities around..
Here you can find a comparison table of various regular expression languages in many different programming languages and a specific guide for C# if you follow its link.
Usually the implementations for the various languages are quite similar since the syntax is somewhat standardized from the theoretical topics regexps come from, so any tutorial about regexp will be fine, then you'll just need to get into C# API.
1) The first regex is trying to do a case-insensitive match starting at the beginning of the test string. It then matches optional whitespace, followed by whatever is in tag, followed by optional whitespace then finally a comma.
The second matches a string containing a comma, followed by optional whitespace, followed by whatever is in tag, followed by optional whitespace then finally a comma.
Thought it's for C# I recommend picking up the Perl Pocket Reference which has a great Regex syntax reference. It helped my out a lot when I was learning regexes 14 years ago.
http://www.myregextester.com/ is a decent regular expression tester that also has an explain option for C# regexps - For Instance check out this example:
The regular expression:
(?-imsx:^(\s*?)(tagtext)(\s*?),)
matches as follows:
NODE EXPLANATION
----------------------------------------------------------------------
(?-imsx: group, but do not capture (case-sensitive)
(with ^ and $ matching normally) (with . not
matching \n) (matching whitespace and #
normally):
----------------------------------------------------------------------
^ the beginning of the string
----------------------------------------------------------------------
( group and capture to \1:
----------------------------------------------------------------------
\s*? whitespace (\n, \r, \t, \f, and " ") (0
or more times (matching the least amount
possible))
----------------------------------------------------------------------
) end of \1
----------------------------------------------------------------------
( group and capture to \2:
----------------------------------------------------------------------
tagtext 'tagtext'
----------------------------------------------------------------------
) end of \2
----------------------------------------------------------------------
( group and capture to \3:
----------------------------------------------------------------------
\s*? whitespace (\n, \r, \t, \f, and " ") (0
or more times (matching the least amount
possible))
----------------------------------------------------------------------
) end of \3
----------------------------------------------------------------------
, ','
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------
A regular expression does not tell you if two strings match, but rather if a given string matches a pattern.
This site is my favorite for learning and testing regular expressions:
http://gskinner.com/RegExr/
It allows you to interactively test regular expressions as you write them, and provides a built-in tutorial.
Although it doesn't use C#, Rejex is a simple tool for testing and learning about regular expressions which includes a quick reference for the special characters
It looks like that they are trying to match some kind of list of words delimited by colons (UPDATE: commas).
The first one is probably matching first item and the second one some item after the first one excluding the last one. I hope you will understand :).
A good source of information about regular expressions is at http://www.regular-expressions.info/
also a great site to test your regular expressions with extra info: http://regex101.com/
i'm having a hard time finding a solution to this and am pretty sure that regex supports it. i just can't recall the name of the concept in the world of regex.
i need to search and replace a string for a specific pattern but the patterns can be different and the replacement needs to "remember" what it's replacing.
For example, say i have an arbitrary string: 134kshflskj9809hkj
and i want to surround the numbers with parentheses,
so the result would be: (134)kshflskj(9809)hkj
Finding numbers is simple enough, but how to surround them?
Can anyone provide a sample or point me in the right direction?
In some various langauges:
// C#:
string result = Regex.Replace(input, #"(\d+)", "($1)");
// JavaScript:
thestring.replace(/(\d+)/g, '($1)');
// Perl:
s/(\d+)/($1)/g;
// PHP:
$result = preg_replace("/(\d+)/", '($1)', $input);
The parentheses around (\d+) make it a "group" specifically the first (and only in this case) group which can be backreferenced in the replacement string. The g flag is required in some implementations to make it match multiple times in a single string). The replacement string is fairly similar although some languages will use \1 instead of $1 and some will allow both.
Most regex replacement functions allow you to reference capture groups specified in the regex (a.k.a. backreferences), when defining your replacement string. For instance, using preg_replace() from PHP:
$var = "134kshflskj9809hkj";
$result = preg_replace('/(\d+)/', '(\1)', $var);
// $result now equals "(134)kshflskj(9809)hkj"
where \1 means "the first capture group in the regex".
Another somewhat generic solution is this:
search : /([\d]+)([^\d]*)/g
replace: ($1)$2
([\d]+): match a set of one or more digits and retain them in a group
([^\d]*): match a set of non-digits, and retain them as well. \D could work here, too.
g: indicate this is a global expression, to work multiple times on the input.
($1): in the replace block, parens have no special meaning, so output the first group, surrounding it with parens.
$2: output the second group
I used a pretty good online regex tool to test out my expression. The next step would be to apply it to the language that you are using, as each has its own implemention nuance.
Backreferences (grouping) are not necessary if you're just looking to search for numbers and replace with the found regex surrounded by parens. It is simpler to use the whole regex match in the replacement string.
e.g for perl
$text =~ s/\d+/($&)/g;
This searches for 1 or more digits and replaces with parens surrounding the match (specified by $&), with trailing g to find and replace all occurrences.
see http://www.regular-expressions.info/refreplace.html for the correct syntax for your regex language.
Depending on your language, you're looking to match groups.
So typically you'll make a pattern in the form of
([0-9]{1,})|([a-zA-Z]{1,})
Then, you'll iterate over the resulting groups in (specific to your language).