How do I take only the first occurrence of a hyphen in sed?

Question

I have a string, for example home/JOHNSMITH-4991-common-task-list, and I want to take out the uppercase part and the numbers with the hyphen between them. I echo the string and pipe it to sed like so, but I keep getting all the hyphens I don't want, e.g.:

echo home/JOHNSMITH-4991-common-task-list | sed 's/[^A-Z0-9-]//g'

gives me:

JOHNSMITH-4991---

I need:

JOHNSMITH-4991

How do I ignore all but the first hyphen?

Answer 1

You can use

sed 's,.*/\([^-]*-[^-]*\).*,\1,' 

POSIX BRE regex details:

• .* - any zero or more chars
• / - a / char
• \([^-]*-[^-]*\) - Group 1: any zero or more chars other than -, a hyphen, and then again zero or more chars other than -
• .* - any zero or more chars

The replacement is the Group 1 placeholder, \1, to restore just the text captured.

See the online demo:

#!/bin/bash
s="home/JOHNSMITH-4991-common-task-list"
sed 's,.*/\([^-]*-[^-]*\).*,\1,' <<< "$s"
# => JOHNSMITH-4991

Answer 2

1st solution: With awk it will be much easier and we could keep it simple, please try following, written and tested with your shown samples.

echo "echo home/JOHNSMITH-4991-common-task-list" | awk -F'/|-' '{print $2"-"$3}'

Explanation: Simple explanation would be, setting field separator as / OR - and printing 2nd field - and 3rd field of current line.

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2nd solution: Using match function of awk program here.

echo "echo home/JOHNSMITH-4991-common-task-list" | 
awk '
match($0,/\/[^-]*-[^-]*/){
  print substr($0,RSTART+1,RLENGTH-1)
}'

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3rd solution: Using GNU grep solution here. Using -oP option of grep here, to print matched values with o option and to enable ERE(extended regular expression) with P option. Then in main program of grep using .*/ followed by \K to ignore previous matched part and then mentioning [^-]*-[^-]* to make sure to get values just before 2nd occurrence of - in matched line.

echo "echo home/JOHNSMITH-4991-common-task-list" | grep -oP '.*/\K[^-]*-[^-]*'