I have an input string like this
$(xx.xx.xx)abcde$(yyy.yyy.yyy)fghijk$(zzz.zz.zz.zzz)
I want to be able to pull out each subset of strings matching $(anything inside here), so for the example above I would like to get 3 substrings.
the characters in between the brackets do not necessarily always match the same pattern.
I have tried using the following regex
(\$\([a-z]+.*\))
but this matches whole string, due to the fact it starts with '$', anything in middle, and ends with ')'
Hopefully this makes sense.
I should also note that I have very limited experience using regex.
Thanks
(\$\([a-z]+.*?\))
Use ? to make your search non greedy.* is greedy and consumes the max it can.adding ? to * makes it non greedy and it will stop at the first instance of ).
See demo.
http://regex101.com/r/sU3fA2/28
try the below
\((.*?)\)\g
for the given string $(xx.xx.xx)abcde$(yyy.yyy.yyy)fghijk$(zzz.zz.zz.zzz) it returns the three substring..
MATCH 1
1. [2-10] `xx.xx.xx`
MATCH 2
1. [18-29] `yyy.yyy.yyy`
MATCH 3
1. [38-51] `zzz.zz.zz.zzz`
http://regex101.com/r/bX7qR2/1
Related
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
I have a regex that I thought was working correctly until now. I need to match on an optional character. It may be there or it may not.
Here are two strings. The top string is matched while the lower is not. The absence of a single letter in the lower string is what is making it fail.
I'd like to get the single letter after the starting 5 digits if it's there and if not, continue getting the rest of the string. This letter can be A-Z.
If I remove ([A-Z]{1}) +.*? + from the regex, it will match everything I need except the letter but it's kind of important.
20000 K Q511195DREWBT E00078748521
30000 K601220PLOPOH Z00054878524
Here is the regex I'm using.
/^([0-9]{5})+.*? ([A-Z]{1}) +.*? +([A-Z]{1})([0-9]{3})([0-9]{3})([A-Z]{3})([A-Z]{3}) +([A-Z])[0-9]{3}([0-9]{4})([0-9]{2})([0-9]{2})/
Use
[A-Z]?
to make the letter optional. {1} is redundant. (Of course you could also write [A-Z]{0,1} which would mean the same, but that's what the ? is there for.)
You could improve your regex to
^([0-9]{5})+\s+([A-Z]?)\s+([A-Z])([0-9]{3})([0-9]{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])[0-9]{3}([0-9]{4})([0-9]{2})([0-9]{2})
And, since in most regex dialects, \d is the same as [0-9]:
^(\d{5})+\s+([A-Z]?)\s+([A-Z])(\d{3})(\d{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])\d{3}(\d{4})(\d{2})(\d{2})
But: do you really need 11 separate capturing groups? And if so, why don't you capture the fourth-to-last group of digits?
You can make the single letter optional by adding a ? after it as:
([A-Z]{1}?)
The quantifier {1} is redundant so you can drop it.
You have to mark the single letter as optional too:
([A-Z]{1})? +.*? +
or make the whole part optional
(([A-Z]{1}) +.*? +)?
You also could use simpler regex designed for your case like (.*)\/(([^\?\n\r])*) where $2 match what you want.
here is the regex for password which will require a minimum of 8 characters including a number and lower and upper case letter and optional sepecial charactor
/((?=.\d)(?=.[a-z])(?=.*[A-Z])(?![~##$%^&*_-+=`|{}:;!.?"()[]]).{8,25})/
/((?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?![~##\$%\^&\*_\-\+=`|{}:;!\.\?\"()\[\]]).{8,25})/
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}$")
I solved by using trial and error:
Regex pattern = new Regex(#"^(?!.*(.)\1\1)(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,20}$");
I am trying to use regular expressions to parse a method in the following format from a text:
mvAddSell[value, type1, reference(Moving, 60)]
so using the regular expressions, I am doing the following
tokensizedStrs = Regex.Split(target, "([A-Za-z ]+[\\[ ][A-Za-z0-9 ]+[ ,][A-Za-z0-9 ]+[ ,][A-Za-z0-9 ]+[\\( ][A-Za-z0-9 ]+[, ].+[\\) ][\\] ])");
It is working, but the problem is that it always gives me an empty array at the beginning if the string started with a method in the given format and the same happens if it comes at the end. Also if two methods appeared in the string, it catches only the first one! why is that ?
I think what is causing the parser not to catch two methods is the existance of ".+" in my patern, what I wanted to do is that I want to tell it that there will be a number of a date in that location, so I tell it that there will be a sequence of any chars, is that wrong ?
it woooorked with ,e =D ... I replaced ".+" by ".+?" which meant as few as possible of any number of chars ;)
Your goal is quite unclear to me. What do you want as result? If you split on that method pattern, you will get the part before your pattern and the part after your pattern in an array, but not the method itself.
Answer to your question
To answer your concrete question: your .+ is greedy, that means it will match anything till the last )] (in the same line, . does not match newline characters by default).
You can change this behaviour by adding a ? after the quantifier to make it lazy, then it matches only till the first )].
tokensizedStrs = Regex.Split(target, "([A-Za-z ]+[\\[ ][A-Za-z0-9 ]+[ ,][A-Za-z0-9 ]+[ ,][A-Za-z0-9 ]+[\\( ][A-Za-z0-9 ]+[, ].+?[\\) ][\\] ])");
Problems in your regex
There are several other problems in your regex.
I think you misunderstood character classes, when you write e.g. [\\[ ]. this construct will match either a [ or a space. If you want to allow optional space after the [ (would be logical to me), do it this way: \\[\\s*
Use a verbatim string (with a leading #) to define your regex to avoid excessive escaping.
tokensizedStrs = Regex.Split(target, #"([A-Za-z ]+\[\s*[A-Za-z0-9 ]+\s*,\s*[A-Za-z0-9 ]+\s*,\s*[A-Za-z0-9 ]+\(\s*[A-Za-z0-9 ]+\s*,\s*.+?\)s*\]\s*)");
You can simplify your regex, by avoiding repeating parts
tokensizedStrs = Regex.Split(target, #"([A-Za-z ]+\[\s*[A-Za-z0-9 ]+(?:\s*,\s*[A-Za-z0-9 ]+){2}\(\s*[A-Za-z0-9 ]+\s*,\s*.+?\)s*\]\s*)");
This is an non capturing group (?:\s*,\s*[A-Za-z0-9 ]+){2} repeated two times.
I'm having a hard time understanding why the following expression \\[B.+\\] and code returns a Matches count of 1:
string r = "\\[B.+\\]";
return Regex.Matches(Markup, sRegEx);
I want to find all the instances (let's call them 'tags') (in a variable length HTML string Markup that contains no line breaks) that are prefixed by B and are enclosed in square brackets.
If the markup contains [BName], I get one match - good.
If the markup contains [BName] [BAddress], I get one match - why?
If the markup contains [BName][BAddress], I also only get one match.
On some web-based regex testers, I've noticed that if the text contains a CR character, I'll get a match per line - but I need some way to specify that I want matches returned independent of line breaks.
I've also poked around in the Groups and Captures collections of the MatchCollection, but to no avail - always just one result.
You are getting only one match because, by default, .NET regular expressions are "greedy"; they try to match as much as possible with a single match.
So if your value is [BName][BAddress] you will have one match - which will match the entire string; so it will match from the [B at the beginning all the way to the last ] - instead of the first one. If you want two matches, use this pattern instead: \\[B.+?\\]
The ? after the + tells the matching engine to match as little as possible... leaving the second group to be its own match.
Slaks also noted an excellent option; specifying specifically that you do not wish to match the ending ] as part of the content, like so: \\[B[^\\]]+\\] That keeps your match 'greedy', which might be useful in some other case. In this specific instance, there may not be much difference - but it's an important thing to keep in mind depending on what data/patterns you might be dealing with specifically.
On a side note, I recommend using the C# "literal string" specifier # for regular expression patterns, so that you do not need to double-escape things in regex patterns; So I would set the pattern like so:
string pattern = #"\[B.+?\]";
This makes it much easier to figure out regular expressions that are more complex
Try the regex string \\[B.+?\\] instead. .+ on it's own (same is pretty much true for .*) will match against as many characters as possible, whereas .+? (or .*?) will match against the bare minimum number of characters whilst still satisfying the rest of the expression.
.+ is a greedy match; it will match as much as possible.
In your second example, it matches BName] [BAddress.
You should write \[B[^\]]+\].
[^\]] matches every character except ], so it is forced to stop before the first ].