escape dollar sign in bashscript (which uses awk)

Question

I want to use awk in my bashscript, and this line clearly doesn't work:

line="foo bar"
echo $line | awk '{print $1}'

How do I escape $1, so it doesn't get replaced with the first argument of the script?

Answer 1

Your script (with single quotes around the awk script) will work as expected:

$ cat script-single
#!/bin/bash
line="foo bar"
echo $line | awk '{print $1}'

$ ./script-single test
foo

The following, however, will break (the script will output an empty line):

$ cat script-double
#!/bin/bash
line="foo bar"
echo $line | awk "{print $1}"

$ ./script-double test
​

Notice the double quotes around the awk program.

Because the double quotes expand the $1 variable, the awk command will get the script {print test}, which prints the contents of the awk variable test (which is empty). Here's a script that shows that:

$ cat script-var
#!/bin/bash
line="foo bar"
echo $line | awk -v test=baz "{print $1}"

$ ./script-var test
baz

Related reading: Bash Reference Manual - Quoting and Shell Expansions