How to remove string between two characters and before the first occurrence using sed

user_13628404 0
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I would like to remove the string between ":" and the first "|" using sed.

input: |abc:1.2.3|def|

output from sed: |abc|def|

I managed to come up with sed 's|\(:\)[^|]*|\1|', but this sed command does not remove the first character (":"). How can I modify this command to also remove the colon?

sed bash
2 Answers

You don't need to group : in your pattern and use it in substitution.

You should keep it simple:

s='|abc:1.2.3|def|'
sed 's/:[^|]*//' <<< "$s"

|abc|def|

: matches a colon and [^|]* matches 0 or more non-pipe characters

user_5482250

1st solution: With awk you could try following awk program.

awk 'match($0,/:[^|]*/){print substr($0,1,RSTART-1) substr($0,RSTART+RLENGTH)}' Input_file

Explanation: Using match function of awk, where matching from : to till first occurrence of | here. So what match function does is, whenever a regex is matched in it, it will SET values for its OOTB variables named RSTART and RLENGTH, so based on that we are printing sub-string to neglect matched part and print everything else as per required output in question.

2nd solution: Using FPAT option in GNU awk, try following, written and tested with your shown samples only.

awk -v FPAT=':[^|]*' '{print $1,$2}'  Input_file
user_58665800
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