Awk variable replacement

Shell Programming and Scripting
1

I have a function

awkvarrep() {
awk -F'|' '$1~/$1/{printf "%-10s %-30s %-15s %-30s %-15s\n", $2,$3,$4,$5,$6}' testfile
}

I'm calling it by this
VARREP=XYZ
awkvarrep $VARREP

since i'm passing $VARREP to the awkvarrep() function I want to use this with $1 but it dosen't seem to be working inside the awk in the Shell Script. Do I need to use an escape sequence for it to work pls advice or else I need to do this like 10 times!

2

Try :

awkvarrep() {
awk -F'|' '$1~/'"$1"'/{printf "%-10s %-30s %-15s %-30s %-15s\n", $2,$3,$4,$5,$6}' testfile
}
3

Thanks !!! Way to go it works :slight_smile:

4

Can you explain why whe need to put it in double quotes pls.

5

Actually, the single quotes are making the difference, not the double ones:

awk -F'|' '$1~/'"$1"'/{ code... }'

All the awk code is enclosed in single quotes, when you want to "break" the awk code because you need the value held by an external shell environment variable, you simply close your code with a quote and reopen it (with another quote) when you've finished the external variable evaluation.

Putting double quotes around a shell variable is only a good scripting practice (so you avoid possible problem with null values, etc...).

6

Thanks its very clear explanation!