pass shell parameters to awk does not work 1

[kristo5747]

Why does this work

    for myfile in `find . -name "R*VER" -mtime +1`
    do
       SHELLVAR=`grep ^err $myfile || echo "No error"`
       ECHO $SHELLVAR
    done

and outputs

    No error
    err ->BIST Login Fail 3922 err 
    No error
    err ->IR Remote Key 1 3310 err

But this does not

    for myfile in `find . -name "R*VER" -mtime +1`
    do
       SHELLVAR=`grep ^err $myfile || echo "No error"`
       awk -v awkvar=${SHELLVAR} '{print awkvar}'
    done

and outputs

    awk: cmd. line:1: fatal: cannot open file `{print awkvar}' for reading (No such file or directory)

What am I missing?

 

[Annihilannic]

This should fix it.

awk -v awkvar=[red]"[/red]"${SHELLVAR}[red]"[/red] ' ... '

${SHELLVAR} expands to multiple words, the second of which awk assumes is your script. Quoting it forces it to be treated as one parameter to the variable assignment.

Annihilannic.

 

[Annihilannic]

I managed an extra double quote in there somehow... please ignore it.

Annihilannic.

 

[kristo5747]

I changed to a while loop and fed $myfile to the awk command. It worked. Thank you for your time.