I have the following regular expression:
Defaults(.*)Class=\"(?<class>.*)\"(.*)StorePath=\"(?<storePath>.*)\"
And the following string:
Defaults Class="Class name here" StorePath="Any store path here" SqlTable="SqlTableName"
I'm trying to achieve the following:
class Class name here
storePath Any store path here
But, what I'm getting as a result is:
class Class name here
storePath Any store path here SqlTable="SqlTableName"
How to stop before the Sqltable text?
The language is C# and the regex engine is the built in for .NET framework.
Thanks a lot!
The solution proposed by #ahmed-abdelhameed solves the problem, I forgot the non-greedy.
Defaults(.*)Class=\"(?<class>.*)\"(.*)StorePath=\"(?<storePath>.*?)\"
Thanks!
In the storePath group, you're matching zero or more times of any character (greedy match). What greedy match means is that it will return as many characters as possible, so it keeps matching characters until it reaches the last occurrence of ".
What you need to do is to convert your greedy match into a lazy match by replacing .* with .*?. What lazy match means is that it will return as few characters as possible, so in your case, it'll keep matching character until it reaches the first occurrence of ".
Simply replace your regex with:
Defaults(.*)Class=\"(?<class>.*)\"(.*)StorePath=\"(?<storePath>.*?)\"
References:
Laziness Instead of Greediness.
What do 'lazy' and 'greedy' mean in the context of regular expressions?
Alittle easier to read:
Class="(.+?)".+?StorePath="(.+?)"
The .+? is saying match un-greedy, basically match as little as possible.
That will cause it to capture up to the next "
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
In this url:
http://example.com/SearchResult-Empty.html?caty[]=12345&caty[]=45678
I am trying to use the following regex to grab the first occurence of caty which should be "12345". However, instead, the regex below is giving me the final occurrence 45678. I tried using the "?" limiter to make it non-greedy per other stack overflow questions, but it isn't working. How can I do this?
^SearchResult(?:.*)(caty)(?:.*)\=([0-9]+)\&?$
As far as I can tell, two things are messing you up:
The anchors ^ and $ seem to be forcing the regex to produce bad matches
You are using greedy .* instead of non-greedy .*?
SearchResult(?:.*?)(caty)(?:.*?)\=([0-9]+)\&?
Should do the job
^SearchResult(?:.*)(caty)(?:.*)\=([0-9]+)\&?$
^^
.* is greedy matching, meaning that it will go the the last occurrence of caty rather than the first. You could check that by providing three caty's in the input string and it will then skip the first two.
.*? makes it non-greedy (aka reluctant), which will consume as little as possible to make a match - stopping at the first occurrence of caty.
I'm very new at regex, and to be completely honest it confounds me. I need to grab the string after a certain character is reached in said string. I figured the easiest way to do this would be using regex, however like I said I'm very new to it. Can anyone help me with this or point me in the right direction?
For instance:
I need to check the string "23444:thisstring" and save "thisstring" to a new string.
If this is your string:
I'm very new at regex, and to be completely honest it confounds me
and you want to grab everything after the first "c", then this regular expression will work:
/c(.*)/s
It will return this match in the first matched group:
"ompletely honest it confounds me"
Try it at the regex tester here: regex tester
Explanation:
The c is the character you are looking for
.* (in combination with /s) matches everything left
(.*) captures what .* matched, making it available in $1 and returned in list context.
Regex for deleting characters before a certain character!
You can use lookahead like this
.*(?=x)
where x is a particular character or word or string.{using characters like .,$,^,*,+ have special meaning in regex so don't forget to escape when using it within x}
EDIT
for your sample string it would be
.*(?=thisstring)
.* matches 0 to many characters till thisisstring
Here is a one-line solution for matching everything after "before"
print $1."\n" if "beforeafter" =~ m/before(.*)/;
Edit:
While using lookbehind is possible, it's not required. Grouping provides an easier solution.
To get the string before : in your example, you have to use [^:][^:]*:\(.*\). Notice that you should have at least one [^:] followed by any number of [^:]s followed by an actual :, the character you are searching for.
I have the following text string:
$ABCD(file="somefile.txt")$' />Some more text followed by a dollar like this one)$. Some more random text
I am trying to match the $ABCD(file="somefile.txt")$ part of the string using a regular expression.
I am using this (?=[$]ABCD[(]file=).*(?<=[)][$]) regular expression pattern to make the intended match. It's not working as expected because I am getting a match all the way to the second )$ in the string.
For example, the match will be as follows:
$ABCD(file="somefile.txt")$' />Some more text followed by a dollar like this one)$
How should I modify the pattern to match to the end of the first occurrence of the )$?
Here is a good online regular expression engine tester:
http://derekslager.com/blog/posts/2007/09/a-better-dotnet-regular-expression-tester.ashx
try appending a ? to the greedy *
(?=[$]ABCD[(]file=).*?(?<=[)][$])
Lazy quantification
The standard quantifiers in regular expressions are greedy, meaning
they match as much as they can. Modern regular expression tools allow a quantifier to be specified as lazy (also known as > non-greedy, reluctant, minimal, or ungreedy) by putting a question mark after the quantifier
You could just use this:
\$ABCD\(file="[a-z.]+"\)\$
to get $ABCD(file="somefile.txt")$.
Your problem was the .* bit, it was too general and thus matched everything up to the last $.
I would advance you to use the second quote to define the end of the searched pattern: [^"]* will match to anything except ".
So the pattern for the file name would be: \$ABCD\(file="([^"]*)
I'm searching the pattern (.*)\\1 on the text blabl with regexec(). I get successful but empty matches in regmatch_t structures. What exactly has been matched?
The regex .* can match successfully a string of zero characters, or the nothing that occurs between adjacent characters.
So your pattern is matching zero characters in the parens, and then matching zero characters immediately following that.
So if your regex was /f(.*)\1/ it would match the string "foo" between the 'f' and the first 'o'.
You might try using .+ instead of .*, as that matches one or more instead of zero or more. (Using .+ you should match the 'oo' in 'foo')
\1 is the backreference typically used for replacement later or when trying to further refine your regex by getting a match within a match. You should just use (.*), this will give you the results you want and will automatically be given the backreference number 1. I'm no regex expert but these are my thoughts based on my limited knowledge.
As an aside, I always revert back to RegexBuddy when trying to see what's really happening.
\1 is the "re-match" instruction. The question is, do you want to re-match immediately (e.g., BLABLA)
/(.+)\1/
or later (e.g., BLAahemBLA)
/(.+).*\1/