This question already has an answer here:
How to insert the whole matched text in the replacement in Perl?
(1 answer)
Closed 3 years ago.
I am trying to put every last element of each column of a csv into quotation marks by using regex in Visual Studio Code.
I am matching the string using [^,;]+$ and trying to replace it by using "$1".
After replacing, the strings are not in order anymore and some vanish.
Can anybody help me out here?
My csv is shaped like this:
SOME_ID,SOME_ID2,SOME_ID3,NUM,CODE
1234,100,1723,1,403
1235,101,1723,2,486
1236,101,1723,3,5822
To refer to capture you require to put the expression in a capture group (...)
Your regex is correct and simply needs to be put in the capturing group, ([^,]+)$ , should fix your search and replace.
As Wiktor suggested in the comments $& will refer to the whole match and that can also be used.
Related
This question already has answers here:
Regex Match all characters between two strings
(16 answers)
RegEx match open tags except XHTML self-contained tags
(35 answers)
Closed 2 years ago.
I would like to mark all strings between 2 strings with the regular expression.
Example:
https://regex101.com/r/Etfpol/1
I want regular expression to mark follow text:
Solution changed from
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Updated By changed from
enter image description here
Thanks
You can use positive lookbehind and positive lookahead to check the tags.
(?<=<Name>Description<\/Name><Value>).*?(?=<\/Value>)
Match results
Solution changed from
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If you prefer not to use them, this will work as well, but the full match will include the strings before and after your desired string.
(?:<Name>Description<\/Name><Value>)(.*?)(?:<\/Value>)
This question already has answers here:
Regex to get string between curly braces
(16 answers)
Closed 2 years ago.
How do I use regex to get what is inside of a ${} enclosed value such as:
Dont care about this. ${This is the thing I want regex to return} Not this either.
Should return:
This is the thing I want regex to return
I've tried \${ }$
the best I got messing around on regex101.com
I'll be honest I have no Idea what I'm doing as far as regex goes
using on c++ but would also (if possible) like to use in geany text editor
I suggest \${[^}]*}. Note that $ have special meaning in regular expressions and need to be escaped with a \ to be read literary.
I use [^}]* instead of .* between the braces to avoid making a long match including the entire value of:
${Another} match, more then one ${on the same line}
[^}] means anything but }
What you want is matching the starting ${ and the ending } with any amount of characters in between: \$\{.*\}. The special part here is the .*, . means any character and * means the thing in front of it can be matched 0 or more times.
Since you want thre matched results, you might also want to wrap it in (): (\$\{.*\}). The parenthesis makes regex remember the stuff inside for later use.
See this stackoverflow on how to get the results back:
How to match multiple results using std::regex
This question already has answers here:
Regex to replace string not in quotes in IntelliJ
(2 answers)
Closed 3 years ago.
Given the following array:
R991,U847,L239,U883,L224,D359,L907,D944,L79,U265,L107
I'd like to add quotation around each individual instance of a letter followed by a number of characters using the 'search and replace' function.
Now in the search function I can target all the instances using the Regex option, and using [A-Z]\d*.
Now for the replace value, how can I use a certain wildcard character to keep the same Regex value intact?
Example:
The end result should be:
"R991","U847","L239","U883","L224","D359","L907","D944","L79","U265","L107"
Note: As a quick fix I searched for each , and replaced that with "," which almost fixed it entirely. But I'm wondering for more complex cases you could use a wildcard.
Search for: ([A-Z]\d+)
Replace by: "$1"
https://www.jetbrains.com/help/webstorm/tutorial-finding-and-replacing-text-using-regular-expressions.html#capture_groups_and_backreference
This question already has answers here:
Is it possible to check if two groups are equal?
(2 answers)
Closed 5 years ago.
I'm trying to match iff two capture groups are the same. I could manually check after the match, but I'm wondering if there is a way I can do this in the expression itself.
My expression is (\d+)\/(\d+), so I only want to accept strings where the two numbers are equal. Is there a nice way to check this in the regular expression, or do I have to manually check groups after?
EDIT: This was marked a duplicate but the supposed duplicate question is not related and does not in any way answer my question...
You can use this one in python : \b(\d+)\/+\1\b
Demo
This is the same usecase as checking for doubled words
When editing text, doubled words such as "the the" easily creep in. Using the regex \b(\w+)\s+\1\b in your text editor, you can easily find them. To delete the second word, simply type in \1 as the replacement text and click the Replace button.
Source
I assume you don't have any other capture groups, based on that:
\b(\d+)\/(\1)\b
Regex Demo
This question already has an answer here:
Notepad++ regex backreference doesn't seem to work
(1 answer)
Closed 3 years ago.
I have a text file that contains many numbers with quotation marks around them, and I want to remove all the quotation marks.
I tried this in Visual Studio's editor, EmEditor and Notepad++, and I just couldn't have it done.
In the the find box, I put in
"[0-9]+"
and in the replace box, I put in
\1
or
$1
All three editors could correctly find all those numbers with quotation marks, but failed to replace them with none quotation marks version. They just delete the original numbers, or replace them with "$1" literally.
Wrap the number matching in parentheses to create a capture group and you should be able to replace it with $1:
"([0-9]+)"
Okay, I found the answer.
find (")([0-9]+)(")
and replace with \2