Extract first match in line with sed

Question

I have a series of lines form.

Agenda HR-1 Presented by XYZ
HR-2 Debate-1 - All
HR-3 Debate-2 - All
(Cov-4) Conclusion 

Each line has an ID of the (sed) pattern [A-Za-z]\+-[0-9]\+, i.e. one or more alphabets follows by a dash (-) follows by one or more numbers. They occur anywhere in the line.

I need to extract the IDs. My thought was to stick a .* at the beginning and end and print \1, but I can't get it to work.

This reply says sed replaces the first match only and that is correct:

$ cat /tmp/scratch/x | sed -n   's/\<\([A-Za-z]\+-[0-9]\+\)/ID:\1/p'
Agenda ID:HR-1 Presented by XYZ
ID:HR-2 Debate-1 - All
ID:HR-3 Debate-2 - All
(ID:Cov-4) Conclusion

But of course a .* at start would greedily go to the last match:

$ cat /tmp/scratch/x | sed -n   's/.*\<\([A-Za-z]\+-[0-9]\+\).*/ID:\1/p'
ID:HR-1
ID:Debate-1
ID:Debate-2
ID:Cov-4

The only way I can think of doing this in sed is by adding markers around the ID in one command and extract using another, as follows.

Is there a better way to do this in sed?

$ cat x | sed -n   's/\<\([A-Za-z]\+-[0-9]\+\)/<id>\1<~id>/;s/.*<id>\(.*\)<~id>.*/\1/;p'
HR-1
HR-2
HR-3
Cov-4

Answer 1

With GNU awk, try:

gawk -v FPAT='[A-Za-z]+-[0-9]+' '$1{print $1}' FILE

or:

gawk -v FPAT='[A-Za-z]+-[0-9]+' '$0=$1' FILE

Answer 2

With sed, try:

<FILE sed -En '/([[:alpha:]]+-[0-9]+).*/{ s//:\1/; s/.*://p; }'

• Remove everything after the desired match
• Now the match is at the end of every line – this can be handled in various ways
  • The above is easily readable – place a char which isn't part of the the match before each match, then remove everything up to the last char with the second s command
  • Another option for the second s command is to remove everything up to a character which is not part of the match:

<FILE sed -En '/([[:alpha:]]+-[0-9]+).*/{ s//\1/; s/.*[^[:alnum:]-]//; p; }'

Answer 3

With Perl we can do it as follows:

$ perl -lne 'print /([a-z]+-\d+)/i' file

output:

HR-1
HR-2
HR-3
Cov-4