Trying to pass a bash variable to an awk script to insert a value in front of each line, currently the script runs but prints 0 in each line. Thank you :).
file
2022/.../.../...
2022/.../.../...
awk
id=1234
awk -v r="$id" '{print /"id"/"/"$0}'
current
0/2022/.../.../...
0/2022/.../.../...
desired
1234/2022/.../.../...
1234/2022/.../.../...
You are passing variable correctly but not using correctly inside the awk.
/"id"/" is a regex expression that will attempt to match "id" string in each line and since it doesn't exist it returns 0 and that's what you get in output.
You may just use:
awk -v r="$id" '{print r "/" $0}' file
1234/2022/.../.../...
1234/2022/.../.../...
btw this sed solution would also work:
sed "s~^~$id/~" file
1234/2022/.../.../...
1234/2022/.../.../...
this should make it clear why it appended "0/" :
0/
# gawk profile, created Tue Aug 30 14:49:38 2022
# Rule(s)
1 {
1 print (/"id"/) "/" $0
}
It's a regex non-match, concat with a string containing ASCII "/" (\057 | 0x2F)`, then concat with input row
awk -v r="$id" '{print r "/" $0}'