AWK and Bash Scripting

Question

How to use a value of a variable in awk? Something like this:

    filename = "test.txt"
    ls -l | awk '{ if ($9 == filename) print("File exists")}'

I can't use $ in awk to access the value of that variable.

@ABC; note that your code is not robust; if the filename contains spaces then `$9` will only compare against the first part of the filename. But this can be easily fixed by using plain `ls` (and comparing against `$1` in awk) instead of using `ls -l`. — Janis, Mar 22 '15 at 05:41
Yeah. You are right, but I need besides name and size. That is just a part of code. — ABC, Mar 22 '15 at 05:46
Note, there's also the `stat` command available (instead of `ls`); e.g. `stat -c "%s %n" files...`, and you can use your own formatting (incl. delimiters, quoting) so that any subsequent (awk-) processing becomes more robust. — Janis, Mar 22 '15 at 06:06

score 2 · Answer 1 · answered Mar 22 '15 at 05:37

2

Here's the syntax to pass variables (and a few awk-style issues fixed):

awk -v filename="${filename}" '$9 == filename { print "File exists" }'

answered Mar 22 '15 at 05:37

Janis

2

Note that you cannot use -v for arbitrary content as awk expands backslash escape sequences in them. best is to use `ENVIRON` instead – Stéphane Chazelas Mar 22 '15 at 08:38

1 Answers1