I want to use AWK in my Bash script, 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?
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