I am trying to match with the following regex.
\d{11}(.*)
Which is any 11 digits followed by a string. I want to extract the tailing string whatever it is.
I used RE2::FullMatch but it gives the first half (the 11 digits). How to get the sub-string matched with (.*) ?
string subStr
RE2::FullMatch("<sip:+19073381121#216.67.108.201:5060;user=phone>;npi=ISDN",(<sip:\+(\d{11}))(.*), &subStr);
I am trying to extract everything starting from # in above string. Basically I want what matches to (.*) but the above function returns <sip:+19073381121.
I am not very familiar with regex but I looked at different APIs to extract substrings and found this one usefull
Remove the extra capturing groups from your regular expression.
<sip:\+\d{11}(.*)
To get the sub string matched with (.*) use $1. That is the first capturing group that you specified with the brackets.
Related
I am using Regular Expressions to find very simple patterns.
However, I want to insert a hyphen character between the matches.
I'm very familiar with writing RegEx Match patterns, but struggling with how to use RegEx replace to insert characters.
My RegEx is:
(\d{1,2})([A-Z]{1,3})(_)?(\d{3,4})
which matches:
03EM0109
03EM0112
03EM0151
3V204
02SDV_0900
I would like the output, using RegEx Replace, to input hyphens between the matches to give me:
03-EM-0109
03-EM-0112
03-EM-0151
3-V-204
02-SDV-0900
I tried changing the RegEx and entering numbered capture groups for null patterns between, but when using a replace function this returns only hyphens. Presumably because the null capture group is not actually capturing anything?
Using:
(\d{1,2})()([A-Z]{1,3})()(_)?()(\d{3,4})
And replacing with $2-$4-$5-
Returns 3 hyphens - - -
Could someone please help....
If you use the RegExp (\d{1,2})([A-Z]{1,3})_?(\d{3,4}), and replace with $1-$2-$3 then it seems to produce the desired results. I removed the capture group around the underscore
I am trying to use Regex to grab a substring of a large string.
The overall string has certain text, 'cow/', then any number of characters or spaces that are not digits. The first digit hit is the start of the desired substring I want.
This desired substring consists of only digits and periods, the first character or space seen that is not a digit or period indicates the end of the desired substring.
For example:
'cow/ a12.34 -123'
The desired substring is '12.34'.
So far I have this regex that partially works (I think the '| .' is not entirely correct):
(?<=([A-z]|[0-9])/\s*).?(?=\s[^0-9 |.])
Thanks in advance.
This should be easy to achieve by relying on capturing groups:
cow/[^0-9]*([0-9.]+)
The group will contain the text that you want to extract, in Java group(index), in C# with Groups[index]. Other languages provide similar features.
Don't try to solve everything inside the regular expression, but leverage the power of your runtime :)
Edit after comment on the OP:
Azure Kusto has the extract(regex, captureGroup, text [, typeLiteral]) function to extract groups from regular expression matches:
extract("cow/[^0-9]*([0-9.]+)", 1, "cow/ a12.34 -123") == "12.34";
The argument 1 tells Kusto to extract the first capturing group (the expression inside the parentheses).
I'm trying to match multiple strings in a single line using regex in Sublime Text 3.
I want to match all values and replace them with null.
Part of the string that I'm matching against:
"userName":"MyName","hiScore":50,"stuntPoints":192,"coins":200,"specialUser":false
List of strings that it should match:
"MyName"
50
192
200
false
Result after replacing:
"userName":null,"hiScore":null,"stuntPoints":null,"coins":null,"specialUser":null
Is there a way to do this without using sed or any other substitution method, but just by matching the wanted pattern in regex?
You can use this find pattern:
:(.*?)(,|$)
And this replace pattern:
:null\2
The first group will match any symbol (dot) zero or more times (asterisk) with this last quantifier lazy (question mark), this last part means that it will match as little as possible. The second group will match either a comma or the end of the string. In the replace pattern, I substitute the first group with null (as desired) and I leave the symbol matched by the second group unchanged.
Here is an alternative on amaurs answer where it doesn't put the comma in after the last substitution:
:\K(.*?)(?=,|$)
And this replacement pattern:
null
This works like amaurs but starts matching after the colon is found (using the \K to reset the match starting point) and matches until a comma of new line (using a positive look ahead).
I have tested and this works in Sublime Text 2 (so should work in Sublime Text 3)
Another slightly better alternative to this is:
(?<=:).+?(?=,|$)
which uses a positive lookbehind instead of resetting the regex starting point
Another good alternative (so far the most efficient here):
:\K[^,]*
This may help.
Find: (?<=:)[^,]*
Replace: null
I am having issues matching regex to pull the follow out of the text below.
23d63443-47d5-4b19-9fce-5a0b151526a0
Output will always look like below but what I'm looking to match above varies slightly.
"C:\Program Files\ScreenConnect Client (bd5ecacad274bdc6)\Elsinore.ScreenConnect.ClientService.exe" "?e=Access&y=Guest&h=screenconnect.com&p=8041&s=23d63443-47d5-4b19-9fce-5a0b151526a0&k=BgIAAACkAABSU0ExAAgAAAEAAQDvDCdQGcu%2fuKP5cPvdclGMBYhhdI0zIC3oNwkJnNmUCbrd%2bAgugzNThBGHoR8mu30zR6nYVJbqYrtjMgxvhC7b2MJptUanf5mLh%2fMpmdQE1rGMtTqCWDH%2fpXQa4DN5QUbz66UcJ%2bdpCQ5TUax8oSw%2fX1I2x1llgax4jCk%2fWc6%2fpcj3JQIODej0z85X%2f1LJhELki2eNcD1QMMN0t%2fR7GZICw7HlL%2ftqOnZyF%2fnr9d62LQQ37n4L5Ra9S5VDk1B9V8umOx9aTkeXuhcRE88e6uGXkuNSQfXjqaAlwSV1xNkPJA8aJvS%2bkkMSNCWfi5chKhGyU4CXaldWPDcsPpA05XKw&t=&c=&c=&c=&c=&c=&c=&c=&c="
How to achieve this?
You can use the following regex to capture what you want:
&s=([^&]+)&
&s= matches the literal characters &s=, [^&]+ (a character class) matches any character other than & one or more times. It is enclosed in a pair of parentheses (a capturing group), meaning the matched text is captured to group 1 (as it's the first pair of parentheses in the regex).
Visualization:
Group 1 will contain the string you're looking for.
RegEx Demo
I need to do a non greedy match and hope someone can help me. I have the following, and I am using JavaScript and ASP
match(/\href=".*?\/pdf\/.*?\.pdf/)
The above match, matches the first start of an href tag. I need it to only match the last href that is part of the /pdf/ folder.
any ideas ?
You need to use capturing parenthesis for sub-expression matches:
match(/\href=".*?(\/pdf\/.*?\.pdf)/)[1];
Match will return an array with the entire match at index 0, all sub expression captures will be added to the array in the order they matched. In this case, index 1 contains the section matching \/pdf\/.*?\.pdf.
Try and make your regex more specific than just .*? if it's matching too broadly. For instance:
match(/\href="([^"]+?\/pdf\/[^\.]+?\.pdf)"/)[1];
[^"]+? will lazily match a string of characters that doesn't contain the double quote character. This will limit the match to staying within the quotes, so the match won't be too broad in the following string, for instance:
TestSome PDF