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The value of regex match groups remain empty [duplicate]

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

Simple phone number regex to match numbers, spaces, etc

I'm trying to modify a fairly basic regex pattern in C# that tests for phone numbers.
The patterns is -
[0-9]+(\.[0-9][0-9]?)?
I have two questions -
1) The existing expression does work (although it is fairly restrictive) but I can't quite understand how it works. Regexps for similar issues seem to look more like this one -
/^[0-9()]+$/
2) How could I extend this pattern to allow brackets, periods and a single space to separate numbers. I tried a few variations to include -
[0-9().+\s?](\.[0-9][0-9]?)?
Although i can't seem to create a valid pattern.
Any help would be much appreciated.
Thanks,

[0-9]+(\.[0-9][0-9]?)?
First of all, I recommend checking out either regexr.com or regex101.com, so you yourself get an understanding of how regex works. Both websites will give you a step-by-step explanation of what each symbol in the regex does.
Now, one of the main things you have to understand is that regex has special characters. This includes, among others, the following: []().-+*?\^$. So, if you want your regex to match a literal ., for example, you would have to escape it, since it's a special character. To do so, either use \. or [.]. Backslashes serve to escape other characters, while [] means "match any one of the characters in this set". Some special characters don't have a special meaning inside these brackets and don't require escaping.
Therefore, the regex above will match any combination of digits of length 1 or more, followed by an optional suffix (foobar)?, which has to be a dot, followed by one or two digits. In fact, this regex seems more like it's supposed to match decimal numbers with up to two digits behind the dot - not phone numbers.
/^[0-9()]+$/
What this does is pretty simple - match any combination of digits or round brackets that has the length 1 or greater.
[0-9().+\s?](\.[0-9][0-9]?)?
What you're matching here is:
one of: a digit, round bracket, dot, plus sign, whitespace or question mark; but exactly once only!
optionally followed by a dot and one or two digits
A suitable regex for your purpose could be:
(\+\d{2})?((\(0\)\d{2,3})|\d{2,3})?\d+
Enter this in one of the websites mentioned above to understand how it works. I modified it a little to also allow, for example +49 123 4567890.
Also, for simplicity, I didn't include spaces - so when using this regex, you have to remove all the spaces in your input first. In C#, that should be possible with yourString.Replace(" ", ""); (simply replacing all spaces with nothing = deleting spaces)

The + after the character set is a quantifier (meaning the preceeding character, character set or group is repeated) at least one, and unlimited number of times and it's greedy (matched the most possible).
Then [0-9().+\s]+ will match any character in set one or more times.

Regular Expression to not allow 3 consecutive characters

I have the following regex:
Regex pattern = new Regex(#"^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,20}/(.)$");
(?=.*\d) //should contain at least one digit
(?=.*[a-z]) //should contain at least one lower case
(?=.*[A-Z]) //should contain at least one upper case
[a-zA-Z0-9]{8,20} //should contain at least 8 characters and maximum of 20
My problem is I also need to check if 3 consecutive characters are identical. Upon searching, I saw this solution:
/(.)\1\1/
However, I can't make it to work if I combined it to my existing regex, still no luck:
Regex(#"^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,20}$/(.)\1\1/");
What did I missed here? Thanks!

The problem is that /(.)\1\1/ includes the surrounding / characters which are used to quote literal regular expressions in some languages (like Perl). But even if you don't use the quoting characters, you can't just add it to a regular expression.
At the beginning of your regex, you have to say "What follows cannot contain a character followed by itself and then itself again", like this: (?!.*(.)\1\1). The (?! starts a zero-width negative lookahead assertion. The "zero-width" part means that it does not consume any characters in the input string, and the "negative lookahead assertions" means that it looks forward in the input string to make sure that the given pattern does not appear anywhere.
All told, you want a regex like this:
new Regex(#"^(?!.*(.)\1\1)(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,20}$")

I solved by using trial and error:
Regex pattern = new Regex(#"^(?!.*(.)\1\1)(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,20}$");

.NET Regular expression which check length and non-alphanumeric characters

I need Regexp to validate string has minimum length 6 and it is contains at least one non-alphanumeric character e.g: "eN%{S$u)", "h9YI!>4j", "{9YI!;4j", "eN%{S$usdf)", "dfh9YI!>4j", "ghffg{9YI!;4j".
This one is working well ^.*(?=.{6,})(?=.*\\d).*$" but in cases when string does not contain any numbers(e.g "eN%{S$u)") it is not working.

^(?=.{6})(.*[^0-9a-zA-Z].*)$
We use positive lookahead to assure there are at least 6 characters. Then we match the pattern that looks for at least one non-alphanumeric character ([^0-9a-zA-Z]). The .*'s match any number of any characters around this one non-alphanumeric character, but by the time we've reached here we've already checked that we're matching at least 6.
^.*(?=.{6,})(?=.*\\d).*$"
is the regex you tried. Here are some suggestions:
You don't need to match more than 6 characters in the lookahead. Matching only 6 here does no restrict the rest of the regular expression from matching more than 6.
\d matches a digit, and (?=.*\\d) is a lookahead for one of them. This is why you are experiencing the problems you mentioned with strings like eN%{S$u).
Even if the point above wasn't incorrect and the regular expression here was correct, you can combine the second lookahead with the .* that follows by just using .*\\d.*.

marcog's answer is pretty good, but I'd do it the other way around so that it's easier to add even more conditions (such as having at least one digit or whatever), and I'd use lazy quantifiers because they are cheaper for certain patterns:
^(?=.*?[^0-9a-zA-Z]).{6}
So if you were to add the mentioned additional condition, it would be like this:
^(?=.*?[^0-9a-zA-Z])(?=.*?[0-9]).{6}
As you can see, this pattern is easily extensible. Note that is is designed to be used for checking matches only, its capture is not useful.

Keep it easy.
// long enough and contains something not digit or a-z
x.Length >= 6 && Regex.IsMatch(x, #"[^\da-zA-Z]")
Happy coding.
Edit, pure "regular expression":
This first asserts there are 6 letters of anything in the look-ahead, and then ensures that within the look-ahead there is something that is not alpha-numeric (it will "throw away" up to the first 5 characters trying to match).
(?=.{6}).{0,5}[^\da-zA-Z]

What about that(fixed): ^(?=.{6})(.*[^\w].*)$
Check this out http://www.ultrapico.com/Expresso.htm it is cool tool which could help you a lot in Regexps learning.

How Can I Check If a C# Regular Expression Is Trying to Match 1-(and-only-1)-Character Strings?

Maybe this is a very rare (or even dumb) question, but I do need it in my app.
How can I check if a C# regular expression is trying to match 1-character strings?
That means, I only allow the users to search 1-character strings. If the user is trying to search multi-character strings, an error message will be displaying to the users.
Did I make myself clear?
Thanks.
Peter
P.S.: I saw an answer about calculating the final matched strings' length, but for some unknown reason, the answer is gone.
I thought it for a while, I think calculating the final matched strings length is okay, though it's gonna be kind of slow.
Yet, the original question is very rare and tedious.

a regexp would be .{1}
This will allow any char though. if you only want alpanumeric then you can use [a-z0-9]{1} or shorthand /w{1}
Another option its to limit the number of chars a user can type in an input field. set a maxlength on it.
Yet another option is to save the forms input field to a char and not a string although you may need some handling around this to prevent errors.
Why not use maxlength and save to a char.

You can look for unescaped *, +, {}, ? etc. and count the number of characters (don't forget to flatten the [] as one character).
Basically you have to parse your regex.

Instead of validating the regular expression, which could be complicated, you could apply it only on single characters instead of the whole string.
If this is not possible, you may want to limit the possibilities of regular expression to some certain features. For instance the user can only enter characters to match or characters to exclude. Then you build up the regex in your code.
eg:
ABC matches [ABC]
^ABC matches [^ABC]
A-Z matches [A-Z]
# matches [0-9]
\w matches \w
AB#x-z matches [AB]|[0-9]|[x-z]|\w
which cases do you need to support?
This would be somewhat easy to parse and validate.