Confused about escaping in backticks

[djeyewater]

I have a line in a shell script that looks like this:

running=`ps -p $pid -o args= | sed -n "s?^$php_fpm_BIN\(\ .*\)*\$?&?p"`

But this gives an error about an unterminated `s' command. If I double escape the closing $ like

running=`ps -p $pid -o args= | sed -n "s?^$php_fpm_BIN\(\ .*\)*\\$?&?p"`

then it works.

Why does the $ need double escaping?

[cfajohnson]

What does $php_fpm_BIN contain?

[djeyewater]

What does $php_fpm_BIN contain?

The path and filename of a binary executable e.g.
/path/to/php/bin/php_fpm

[Manko10]

You have to double-escape this because you are passing the dollar sign to an environment, which interprets it. If you write \$, it is interpreted as a raw $ and then passed to the expression in backticks. Therefore, the backtick environment tries to interpret it again.
To avoid this behavior either use $(…) (which is the newer and recommended way) or use single-quotes for your sed command.

You can easily check the different behaviors with these commands:

export FOO=bar

echo `echo $FOO`
echo `echo \$FOO`
echo `echo "$FOO"`
echo `echo "\$FOO"`
echo `echo '$FOO'`
echo `echo '\$FOO'`
echo $(echo $FOO)
echo $(echo \$FOO)