My current regex is only picking up part of my string. It creates a match as soon as one if found, even though I need the longer version of that match to hit. For example, I am creating matches for both:
SSS111
and
SSS111-L
The first SSS111 matches fine with my current regex, but the SSS111-L is only getting matched to the SSS111, leaving the -L out.
How can I create a greedy regex to read the whole line before matching? I am currently using
[-A-Z0-9]{3,12}
to capture the numbers and letters, but have not had any luck outside of this.
Regex are allways greedy. This ist mostly the Problem.
Here i think you have only to escape the '-'
#"[-A-Z]{3-12}"
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 need to extract the first element ("adidas-originals") after "designer" in the following URL using regular expressions.
xxx/en-ca/men/designers/adidas-originals/shorts
This needs to be done in Google Big Query API (standard SQL). To this end, I have tried several ways to get the desired valued without any success. Below is the best solution that I have found so far which obviously is not the right one as it returns "/adidas-originals/shorts".
REGEXP_EXTRACT(hits.page.pagePath, r'designers([^\n]*)')
Thanks!
The [^\n]* matches 0 or more chars other than a newline, LF, so no wonder it matches too much.
You need a pattern to match up to the next /, so you may use
designers/([^/]+)
Or a more precise:
(?:^|/)designers/([^/]+)
See the regex demo
Details
(?:^|/) - either start of a string or / (you may just use / if designers is always preceded with /)
designers/ a designers/ substring
([^/]+) - Capturing group 1 (just what will be returned with the REGEXP_EXTRACT function): one or more chars other than /.
This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*
I have been using strapplyc in R to select different portions of a string that match one particular set of criteria. These have worked successfully until I found a portion of the string where the required portion could be defined one of two ways.
Here is an example of the string which is liberally sprinkled with \t:
\t\t\tsome words here\t\t\tDefect: some more words here Action: more words
I can write the strapply statement to capture the text between Defect: and the start of Action:
strapplyc(record[i], "Defect:(.*?)Action")
This works and selects the chosen text between Defect: and Action. In some cases there is no action section to the string and I've used the following code to capture these cases.
strapplyc(record[i], "Defect:(.*?)$")
What I have been trying to do is capture the text that either ends with Action, or with the end of the string (using $).
This is the bit that keeps failing. It returns nothing for either option. Here is my failing code:
strapplyc(record[i], "Defect:(.*?)Action|$")
Any idea where I'm going wrong, or a better solution would be much appreciated.
If you are up for a more efficient solution, you could drop the .*? matching and unroll your pattern like:
Defect:((?:[^A]+|A(?!ction))*)
This matches Defect: followed by any amount of characters that are not an A or are an A and not followed by ction. This avoids the expanding that is needed for the lazy dot matching. It will work for both ways, as it does stop matching when it hits Action or the end of your string.
As suggested by Wiktor, you can also use
Defect:([^A]*(?:A(?!ction)[^A]*)*)
Which is a little bit faster when there are many As in the string.
You might want to consider to use A(?!ction:) or A(?!ction\s*:), to avoid false early matches.
The alternation operator | is the regex operator with the lowest precedence. That means the regex Defect:(.*?)Action|$ is actually a combination of Defect:(.*?)Action and $ - since an empty string is a valid match for $, your regex returns the empty string.
To solve that, you should combine the regexes Defect:(.*?)Action and Defect:(.*?)$ with an OR:
Defect:(.*?)Action|Defect:(.*?)$
Or you can enclose Action|$ in a group as Sebastian Proske said in the comments:
Defect:(.*?)(?:Action|$)
I'm trying to complete a regular expression that will pull out matches based on their opening and closing characters, the closest I've gotten is
^(\[\[)[a-zA-Z.-_]+(\]\])
Which will match a string such as "[[word1]]" and bring me back all the matches if there is more than one, The problem is I want it to pick up matchs where there may be a space in so for example "[[word1 word2]]", now this will work if I add a space into my pattern above however this pops up a problem that it will only get one match for my entire string so for example if I have a string
"Hi [[Title]] [[Name]] [[surname]], How are you"
then the match will be [[Title]] [[Name]] [[surname]] rather than 3 matches [[Title]], [[Name]], [[surname]]. I'm sure I'm just a char or two away in the Regex but I'm stuck, How can I make it return the 3 matches.
Thanks
You just need to make you regex non-greedy by using a ? like:
^(\[\[)[a-zA-Z.-_ ]+?(\]\])
Also there is a bug in your regex. You've included - in the char class thinking of it as a literal hyphen. But - in a char class is a meta char. So it effectively will match all char between . (period) and _ (underscore). So you need to escape it as:
^(\[\[)[a-zA-Z.\-_ ]+?(\]\])
or you can put is in some other place in the regex so that it will not have things on both sides of it as:
^(\[\[)[a-zA-Z._ -]+?(\]\])
or
^(\[\[)[-a-zA-Z._ ]+?(\]\])
You need to turn off greedy matching. See these examples for different languages:
asp.net
java
javascript
You should use +? instead of +.
The one without the question mark will try to match as much as possible, while the one with the question mark as little as possible.
Another approach would be to use [^\]] as your characters instead of [a-zA-Z.-_]. That way, a match will never extend over your closing brackets.