I am looking for regular expression that evaluates below.
0-9 a-Z A-Z - / '
The C# version of this pattern is:
#"[0-9a-zA-Z/'-]"
Used in code:
var regex = new Regex(#"[0-9a-zA-Z/'-]");
or
var regex = new Regex(#"[0-9a-z/'-]", RegexOptions.IgnoreCase);
Note that the - is at the very end of the character class (the part in the brackets). For - to mean a literal hyphen inside a character class, it must be at the beginning or end of the class (i.e. [-blah] or [blah-]), or escaped with a backslash: [ab\-c] will match a, b, c, or -.
Note also the # at the beginning of the quoted string. This isn't important for this pattern, but it's a good habit to get into with C# regex. Regular expressions often contain backslashes, and the #"..." form will allow you to use backslashes in your pattern without having to escape them.
Use bellow code to validate(Regex patterns) Alphabetic and numbers:
String name="123ABCabc";
if(System.Text.RegularExpressions.Regex.Match(name, #"[0-9a-zA-Z_]") == true)
{
return true;
}
else
{
return false;
}
In the case you want to match digits, lower- and higher-case latin characters, "-", "/" and "'" then I would suggest the following:
[0-9a-zA-Z-\/\']
Related
Regex.IsMatch method returns the wrong result while checking the following condition,
string text = "$0.00";
Regex compareValue = new Regex(text);
bool result = compareValue.IsMatch(text);
The above code returns as "False". Please let me know if i missed anything.
The Regex class has a special method for escaping characters in a pattern: Regex.Escape()
Change your code like this:
string text = "$0.00";
Regex compareValue = new Regex(Regex.Escape(text)); // Escape characters in text
bool result = compareValue.IsMatch(text);
"$" is a special character in C# regex. Escape it first.
Regex compareValue = new Regex(#"\$0\.00");
bool result = compareValue.IsMatch("$0.00");
Regex expressions: https://msdn.microsoft.com/en-us/library/az24scfc(v=vs.110).aspx
Both '.' and '$' are special characters and thus you need to escape them if you want to match the character itself. '.' matches any character and '$' matches the end of a string
see: https://regex101.com/r/pK2uY6/1
You have to escape $ since it is a special (reserved) character which means "end of string". In case . means just dot (say, decimal separator) you have to escape it as well (when not escaped, . means "any symbol"):
string pattern = #"\$0\.00";
bool result = RegEx.IsMatch(text, pattern);
As for your original pattern, it has no chance to match any string, since $0.00 means
$ end of string, followed by
0 zero
. any character
0 zero
0 zero
but end of string can't be followed by...
I have a list of special characters that includes ^ $ ( ) % . [ ] * + - ?. I want put % in front of this special characters in a string value.
I need this to generate a Lua script to use in Redis.
For example Test$String? must be change to Test%$String%?.
Is there any way to do this with regular expressions in C#?
In C#, you just need a Regex.Replace:
var LuaEscapedString = Regex.Replace(input, #"[][$^()%.*+?-]", "%$&");
See the regex demo
The [][$^()%.*+?-] character class will match a single character, either a ], [, $, ^, (, ), %, ., *, +, ? or - and will reinsert it back with the $& backreference in the replacement pattern pre-pending with a % character.
A lookahead is just a redundant overhead here (or a show-off trick for your boss).
You can use lookaheads and replace with %
/(?=[]*+$?)[(.-])/
Regex Demo
(?=[]*+$?)[(.-]) Postive lookahead, checks if the character following any one from the altenation []. If yes, substitutes with %
You can use this regex: ([\\^$()%.\\[\\]*+\\-?])
It will match and capture characters inside the character class. Then you can use $1 to reference the captured character and insert % before it, like so: %$1.
Here is an example code and demo:
string input = "Test$String?";
string pattern = "([\\^$()%.\\[\\]*+\\-?])";
string replacement = "%$1";
Regex rgx = new Regex(pattern);
string result = rgx.Replace(input, replacement);
Console.WriteLine("Original String: {0}", input);
Console.WriteLine("Replacement String: {0}", result);
You can use (?=[\^\$()\%.[]*+-\?]) regex replaced String as "%"
I have a CSV file that has rows resembling this:
1, 4, 2, "PUBLIC, JOHN Q" ,ACTIVE , 1332
I am looking for a regular expression replacement that will match against these rows and spit out something resembling this:
1,4,2,"PUBLIC, JOHN Q",ACTIVE,1332
I thought this would be rather easy: I made the expression ([ \t]+,) and replaced it with ,. I made a complement expression (,[ \t]+) with a replacement of , and I thought I had achieved a good means of right-trimming and left-trimming strings.
...but then I noticed that my "PUBLIC, JOHN Q" was now "PUBLIC,JOHN Q" which isn't what I wanted. (Note the space following the comma is now gone).
What would be the appropriate expression to trim the white space before and after a comma, but leave quoted text untouched?
UPDATE
To clarify, I am using an application to handle the file. This application allows me to define multiple regular expression replacements; it does not provide a parsing capability. While this may not be the ideal mechanism for this, it would sure beat making another application for this one file.
If the engine used by your tool is the C# regular expression engine, then you can try the following expression:
(?<!,\s*"(?:[^\\"]|\\")*)\s+(?!(?:[^\\"]|\\")*"\s*,)
replace with empty string.
The guys answers assumed the quotes are balanced and used counting to determine if the space is part of a quoted value or not.
My expression looks for all spaces that are not part of a quoted value.
RegexHero Demo
Something like this might do the job:
(?<!(^[^"]*"[^"]*(("[^"]*){2})*))[\t ]*,[ \t]*
Which matches [\t ]*,[ \t]*, only when not preceded by an odd number of quotes.
Going with some CSV library or parsing the file yourself would be much more easier, and IMO should be preferable option here.
But if you really insist on a regex, you can use this one:
"\s+(?=([^\"]*\"[^\"]*\")*[^\"]*$)"
And replace it with empty string - ""
This regex matches one or more whitespaces, followed by an even number of quotes. This will of course work only if you have balanced quote.
(?x) # Ignore Whitespace
\s+ # One or more whitespace characters
(?= # Followed by
( # A group - This group captures even number of quotes
[^\"]* # Zero or more non-quote characters
\" # A quote
[^\"]* # Zero or more non-quote characters
\" # A quote
)* # Zero or more repetition of previous group
[^\"]* # Zero or more non-quote characters
$ # Till the end
) # Look-ahead end
string format(string val)
{
if (val.StartsWith("\"")) val = " " + val;
string[] vals = val.Split('\"');
for (int i = 0; i < vals.Length; i += 2) vals[i] = vals[i].Replace(" ", "").Replace("\t", "");
return string.Join("\t", vals);
}
This will work if you have properly closed quoted strings in between
Forget the regex (See Bart's comment on the question, regular expressions aren't suitable for CSV).
public static string ReduceSpaces( string input )
{
char[] a = input.ToCharArray();
int placeComma = 0, placeOther = 0;
bool inQuotes = false;
bool followedComma = true;
foreach( char c in a ) {
inQuotes ^= (c == '\"');
if (c == ' ') {
if (!followedComma)
a[placeOther++] = c;
}
else if (c == ',') {
a[placeComma++] = c;
placeOther = placeComma;
followedComma = true;
}
else {
a[placeOther++] = c;
placeComma = placeOther;
followedComma = false;
}
}
return new String(a, 0, placeComma);
}
Demo: http://ideone.com/NEKm09
I'm trying to create a regex expression what will accept a certain format of command. The pattern is as follows:
Can start with a $ and have two following value 0-9,A-F,a-f (ie: $00 - $FF)
or
Can be any value except for "&<>'/"
*if the value start with $ the next two values after need to be a valid hex value from 00-ff
So far I have this
Regex correctValue = new Regex("($[0-9a-fA-F][0-9a-fA-F])");
Any help will be greatly appreciated!
You just need to add "\" symbol before your "$" and it works:
string input = "$00";
Match m = Regex.Match(input, #"^\$[0-9a-fA-F][0-9a-fA-F]$");
if (m.Success)
{
foreach (Group g in m.Groups)
Console.WriteLine(g.Value);
}
else
Console.WriteLine("Didn't match");
If I'm following you correctly, the net result you're looking for is any value that is not in the list "&<>'/", since any combination of $ and two alphanumeric characters would also not be in that list. Thus you could make your expression:
Regex correctValue = new Regex("[^&<>'/]");
Update: But just in case you do need to know how to properly match the $00 - $FF, this would do the trick:
Regex correctValue = new Regex("\$[0-9A-Fa-f]{2}");
In Regular Expression $ use for Anchor assertion, and means:
The match must occur at the end of the string or before \n at the end of the line or string.
try using [$] (Character Class for single character) or \$ (Character Escape) instead.
An input string:
string datar = "aag, afg, agg, arg";
I am trying to get matches: "aag" and "arg", but following won't work:
string regr = "a[a-z&&[^fg]]g";
string regr = "a[a-z[^fg]]g";
What is the correct way of ignoring regex matches in C#?
The obvious way is to use a[a-eh-z]g, but you could also try with a negative lookbehind like this :
string regr = "a[a-z](?<!f|g)g"
Explanation :
a Match the character "a"
[a-z] Match a single character in the range between "a" and "z"
(?<!XXX) Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind)
f|g Match the character "f" or match the character "g"
g Match the character "g"
Character classes aren't quite that fancy. The simple solution is:
a[a-eh-z]g
If you really want to explicitly list out the letters that don't belong, you could try something like:
a[^\W\d_A-Zfg]g
This character class matches everything except:
\W excludes non-word characters, i.e. punctuation, whitespace, and other special characters. What's left are letters, digits, and the underscore _.
\d removes digits so now we have letters and the underscore _.
_ removes the underscore so now we only match letters.
A-Z removes uppercase letters so now we only match lowercase letters.
Finally at this point we can list the individual lowercase letters we don't want to match.
All in all way more complicated than we'd likely ever want. That's regular expressions for ya!
What you're using is Java's set intersection syntax:
a[a-z&&[^fg]]g
..meaning the intersection of the two sets ('a' THROUGH 'z') and (ANYTHING EXCEPT 'f' OR 'g'). No other regex flavor that I know of uses that notation. The .NET flavor uses the simpler set subtraction syntax:
a[a-z-[fg]]g
...that is, the set ('a' THROUGH 'z') minus the set ('f', 'g').
Java demo:
String s = "aag, afg, agg, arg, a%g";
Matcher m = Pattern.compile("a[a-z&&[^fg]]g").matcher(s);
while (m.find())
{
System.out.println(m.group());
}
C# demo:
string s = #"aag, afg, agg, arg, a%g";
foreach (Match m in Regex.Matches(s, #"a[a-z-[fg]]g"))
{
Console.WriteLine(m.Value);
}
Output of both is
aag
arg
Try this if you want match arg and aag:
a[ar]g
If you want to match everything except afg and agg, you need this regex:
a[^fg]g
It seems like you're trying to match any three alphabetic characters, with the condition that the second character cannot be f or g. If this is the case, why not use the following regular expression:
string regr = "a[a-eh-z]g";
Regex: a[a-eh-z]g.
Then use Regex.Matches to get the matched substrings.