I currently have a regex that pulls up a 16 digit number from a file e.g.:
Regex:
Regex.Match(l, #"\d{16}")
This would work well for a number as follows:
1234567891234567
Although how could I also include numbers in the regex such as:
1234 5678 9123 4567
and
1234-5678-9123-4567
If all groups are always 4 digit long:
\b\d{4}[ -]?\d{4}[ -]?\d{4}[ -]?\d{4}\b
to be sure the delimiter is the same between groups:
\b\d{4}(| |-)\d{4}\1\d{4}\1\d{4}\b
If it's always all together or groups of fours, then one way to do this with a single regex is something like:
Regex.Match(l, #"\d{16}|\d{4}[- ]\d{4}[- ]\d{4}[- ]\d{4}")
You could try something like:
^([0-9]{4}[\s-]?){3}([0-9]{4})$
That should do the trick.
Please note:
This also allows
1234-5678 9123 4567
It's not strict on only dashes or only spaces.
Another option is to just use the regex you currently have, and strip all offending characters out of the string before you run the regex:
var input = fileValue.Replace("-",string.Empty).Replace(" ",string.Empty);
Regex.Match(input, #"\d{16}");
Here is a pattern which will get all the numbers and strip out the dashes or spaces. Note it also checks to validate that there is only 16 numbers. The ignore option is so the pattern is commented, it doesn't affect the match processing.
string value = "1234-5678-9123-4567";
string pattern = #"
^ # Beginning of line
( # Place into capture groups for 1 match
(?<Number>\d{4}) # Place into named group capture
(?:[\s-]?) # Allow for a space or dash optional
){4} # Get 4 groups
(?!\d) # 17th number, do not match! abort
$ # End constraint to keep int in 16 digits
";
var result = Regex.Match(value, pattern, RegexOptions.IgnorePatternWhitespace)
.Groups["Number"].Captures
.OfType<Capture>()
.Aggregate (string.Empty, (seed, current) => seed + current);
Console.WriteLine ( result ); // 1234567891234567
// Shows False due to 17 numbers!
Console.WriteLine ( Regex.IsMatch("1234-5678-9123-45678", pattern, RegexOptions.IgnorePatternWhitespace));
Related
I'm using Replace(#"[^a-zA-Z]+", "");
leave only letters, but I have a set of numbers or characters that I want to keep as well, ex: 122456 and 112466. But I'm having trouble leaving it only if it's this sequence:
ex input:
abc 1239 asm122456000
I want to:
abscasm122456
tried this: ([^a-zA-Z])+|(?!122456)
My answer doesn't applying Replace(), but achieves a similar result:
(?:[a-zA-Z]+|\d{6})
which captures the group (non-capturing group) with the alphabetic character(s) or a set of digits with 6 occurrences.
Regex 101 & Test Result
Join all the matching values into a single string.
using System.Linq;
Regex regex = new Regex("(?:[a-zA-Z]+|\\d{6})");
string input = "abc 1239 asm12245600";
string output = "";
var matches = regex.Matches(input);
if (matches.Count > 0)
output = String.Join("", matches.Select(x => x.Value));
Sample .NET Fiddle
Alternate way,
using .Split() and .All(),
string input = "abc 1239 asm122456000";
string output = string.Join("", input.Split().Where(x => !x.All(char.IsDigit)));
.NET Fiddle
It is very simple: you need to match and capture what you need to keep, and just match what you need to remove, and then utilize a backreference to the captured group value in the replacement pattern to put it back into the resulting string.
Here is the regex:
(122456|112466)|[^a-zA-Z]
See the regex demo. Details:
(122456|112466) - Capturing group with ID 1: either of the two alternatives
| - or
[^a-zA-Z] - a char other than an ASCII letter (use \P{L} if you need to match any char other than any Unicode letter).
Note the removed + quantifier as [^A-Za-z] also matches digits.
You need to use $1 in the replacement:
var result = Regex.Replace(text, #"(122456|112466)|[^a-zA-Z]", "$1");
I need to check if a string begins with 2 specific letters and then is followed by any 4 numbers.
the 2 letters are "BR" so BR1234 would be valid so would BR7412 for example.
what bit of code do I need to check that the string is a match with the Regex in C#?
the regex I have written is below, there is probably a more efficient way of writing this (I'm new to RegEx)
[B][R][0-9][0-9][0-9][0-9]
You can use this:
Regex regex = new Regex(#"^BR\d{4}");
^ defines the start of the string (so there should be no other characters before BR)
BR matches - well - BR
\d is a digit (0-9)
{4} says there must be exactly 4 of the previously mentioned group (\d)
You did not specify what is allowed to follow the four digits. If this should be the end of the string, add a $.
Usage in C#:
string matching = "BR1234";
string notMatching = "someOther";
Regex regex = new Regex(#"^BR\d{4}");
bool doesMatch = regex.IsMatch(matching); // true
doesMatch = regex.IsMatch(notMatching); // false;
BR\d{4}
Some text to make answer at least 30 characters long :)
I'm trying to create a regex which will extract a complete phone number from a string (which is the only thing in the string) but leaving out any cruft like decorative brackets, etc.
The pattern I have mostly appears to work, but returns a list of matches - whereas I want it to return the phone number with the characters removed. Unfortunately, it completely fails if I add the start and end of line matchers...
^(?!\(\d+\)\s*){1}(?:[\+\d\s]*)$
Without the ^ and $ this matches the following numbers:
12345-678-901 returns three groups: 12345 678 901
+44-123-4567-8901 returns four groups: +44 123 4567 8901
(+48) 123 456 7890 returns four groups: +48 123 456 7890
How can I get the groups to be returned as a single, joined up whole?
Other than that, the only change I would like to include is to return nothing if there are any non-numeric, non-bracket, non-+ characters anywhere. So, this should fail:
(+48) 123 burger 7890
I'd keep it simple, makes it more readable and maintainable:
public string CleanPhoneNumber(string messynumber){
if(Regex.IsMatch(messynumber, "[a-z]"))
return "";
else
return Regex.Replace(messynumber, "[^0-9+]", "");
}
If any alphameric characters are present (extend this range if you wish) return blank else replace every char that is not 0-9 or +, with nothing. This produces output like 0123456789 and +481234567 with all the brackets, spaces and hyphens etc removed too. If you want to keep those in the output, add them to the Regex
Side note: It's not immediately clear or me what you think is "cruft" that should be stripped (non a-z?) and what you think is "cruft" that should cause blank (a-z?). I struggled with this because you said (paraphrase) "non digit, non bracket, non plus should cause blank" but earlier in your examples your processing permitted numbers that had hyphens and also spaces - being strictly demanding of spec hyphens/spaces would be "cruft that causes the whole thing to return blank" too
I've assumed that it's lowercase chars from the "burger" example but as noted you can extend the range in the IF part should you need to include other chars that return blank
If you have a lot of them to do maybe pre compile a regex as a class level variable and use it in the method:
private Regex _strip = new Regex( "[^0-9+]", RegexOptions.Compiled);
public string CleanPhoneNumber(string messynumber){
if(Regex.IsMatch(messynumber, "[a-z]"))
return "";
else
return _strip.Replace(messynumber, "");
}
...
for(int x = 0; x < millionStrArray.Length; x++)
millionStrArray[x] = CleanPhoneNumber(millionStrArray[x], "");
I don't think you'll gain much from compiling the IsMatch one but you could try it in a similar pattern
Other options exist if you're avoiding regex, you cold even do it using LINQ, or looping on char arrays, stringbuilders etc. Regex is probably the easiest in terms of short maintainable code
The strategy here is to use a look ahead and kick out (fail) a match if word characters are found.
Then when there are no characters, it then captures the + and all numbers into a match group named "Phone". We then extract that from the match's "Phone" capture group and combine as such:
string pattern = #"
^
(?=[\W\d+\s]+\Z) # Only allows Non Words, decimals and spaces; stop match if letters found
(?<Phone>\+?) # If a plus found at the beginning; allow it
( # Group begin
(?:\W*) # Match but don't *capture* any non numbers
(?<Phone>[\d]+) # Put the numbers in.
)+ # 1 to many numbers.
";
var number = "+44-123-33-8901";
var phoneNumber =
string.Join(string.Empty,
Regex.Match(number,
pattern,
RegexOptions.IgnorePatternWhitespace // Allows us to comment the pattern
).Groups["Phone"]
.Captures
.OfType<Capture>()
.Select(cp => cp.Value));
// phoneNumber is `+44123338901`
If one looks a the match structure, the data it houses is this:
Match #0
[0]: +44-123-33-8901
["1"] → [1]: -8901
→1 Captures: 44, -123, -33, -8901
["Phone"] → [2]: 8901
→2 Captures: +, 44, 123, 33, 8901
As you can see match[0] contains the whole match, but we only need the captures under the "Phone" group. With those captures { +, 44, 123, 33, 8901 } we now can bring them all back together by the string.Join.
I have a string with 3 dates in it like this:
XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx
I want to select the 2nd date in the string, the 20180208 one.
Is there away to do this purely in the regex, with have to resort to pulling out the 2 match in code. I'm using C# if that matters.
Thanks for any help.
You could use
^(?:[^_]+_){2}(\d+)
And take the first group, see a demo on regex101.com.
Broken down, this says
^ # start of the string
(?:[^_]+_){2} # not _ + _, twice
(\d+) # capture digits
C# demo:
var pattern = #"^(?:[^_]+_){2}(\d+)";
var text = "XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx";
var result = Regex.Match(text, pattern)?.Groups[1].Value;
Console.WriteLine(result); // => 20180208
Try this one
MatchCollection matches = Regex.Matches(sInputLine, #"\d{8}");
string sSecond = matches[1].ToString();
You could use the regular expression
^(?:.*?\d{8}_){1}.*?(\d{8})
to save the 2nd date to capture group 1.
Demo
Naturally, for n > 2, replace {1} with {n-1} to obtain the nth date. To obtain the 1st date use
^(?:.*?\d{8}_){0}.*?(\d{8})
Demo
The C#'s regex engine performs the following operations.
^ # match the beginning of a line
(?: # begin a non-capture group
.*? # match 0+ chars lazily
\d{8} # match 8 digits
_ # match '_'
) # end non-capture group
{n} # execute non-capture group n (n >= 0) times
.*? # match 0+ chars lazily
(\d{8}) # match 8 digits in capture group 1
The important thing to note is that the first instance of .*?, followed by \d{8}, because it is lazy, will gobble up as many characters as it can until the next 8 characters are digits (and are not preceded or followed by a digit. For example, in the string
_1234abcd_efghi_123456789_12345678_ABC
capture group 1 in (.*?)_\d{8}_ will contain "_1234abcd_efghi_123456789".
You can use System.Text.RegularExpressions.Regex
See the following example
Regex regex = new Regex(#"^(?:[^_]+_){2}(\d+)"); //Expression from Jan's answer just showing how to use C# to achieve your goal
GroupCollection groups = regex.Match("XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx").Groups;
if (groups.Count > 1)
{
Console.WriteLine(groups[1].Value);
}
In C# I'm trying to validate a string that looks like:
I#paramname='test'
or
O#paramname=2827
Here is my code:
string t1 = "I#parameter='test'";
string r = #"^([Ii]|[Oo])#\w=\w";
var re = new Regex(r);
If I take the "=\w" off the end or variable r I get True. If I add an "=\w" after the \w it's False. I want the characters between # and = to be able to be any alphanumeric value. Anything after the = sign can have alphanumeric and ' (single quotes). What am I doing wrong here. I very rarely have used regular expressions and normally can find example, this is custom format though and even with cheatsheets I'm having issues.
^([Ii]|[Oo])#\w+=(?<q>'?)[\w\d]+\k<q>$
Regular expression:
^ start of line
([Ii]|[Oo]) either (I or i) or (O or o)
\w+ 1 or more word characters
= equals sign
(?<q>'?) capture 0 or 1 quotes in named group q
[\w\d]+ 1 or more word or digit characters
\k<q> repeat of what was captured in named group q
$ end of line
use \w+ instead of \w to one character or more. Or \w* to get zero or more:
Try this: Live demo
^([Ii]|[Oo])#\w+=\'*\w+\'*
If you are being a bit more strict with using paramname:
^([Ii]|[Oo])#paramname=[']?[\w]+[']?
Here is a demo
You could try something like this:
Regex rx = new Regex( #"^([IO])#(\w+)=(.*)$" , RegexOptions.IgnoreCase ) ;
Match group 1 will give you the value of I or O (the parameter direction?)
Match group 2 will give you the name of the parameter
Match group 3 will give you the value of the parameter
You could be stricter about the 3rd group and match it as
(([^']+)|('(('')|([^']+))*'))
The first alternative matches 1 or more non quoted character; the second alternative match a quoted string literal with any internal (embedded) quotes escape by doubling them, so it would match things like
'' (the empty string
'foo bar'
'That''s All, Folks!'