How to replace regex capture group content using SED?
I’m trying to edit my nginx.conf file programmatically, which contains a line like this: I believe the regex ^\\s*access_log ( [^;]*); will capture the /var/log/nginx/access.log part in a capture group, but I’m not sure how to correctly replace the capture group with sed?
How to replace multiple lines using the sed command?
Create a sed file named to replace.sed with the following content to replace the multiple lines based on the search pattern. Here, the word ‘ CSE ‘ will be searched in the text file, and if the match exists, then it will again search the number 35 and 15. If the second match exists in the file, then it will be replaced by the number 45.
Why do I need to escape regex characters in SED?
The -r switch enables extended regular expressions which is why you don’t get the error. See Why do I need to escape regex characters in sed to be interpreted as regex characters?. Second, you are putting the capture group in the wrong place. If I understand you correctly, you can do this:
How to replace a regex group in Bash?
If you would like to preserve a portion of the left regex, you must explicitly copy it to the right or make it a capturing group as well: will correctly replace all of the match, id=’42’, with 421 since \\1 is 42.
How to output only captured groups with SED?
You can use ripgrep, which also seems to be a sed replacement for simple substitutions, like this where ripgrep uses -o or –only matching and -r or –replace to output only the first capture group with $1 (quoted to be avoid intepretation as a variable by the shell) two times due to two matches.
Is the capture group in the wrong place?
Second, you are putting the capture group in the wrong place. If I understand you correctly, you can do this: Note the use of ! as regex delimiters to avoid having to escape the forward slashes in /dev/stdout.