Shell script search for a literal dollar sign next to a variable name

Question

When searching through code, I'm trying to write a regular expression that needs to search for a literal dollar sign immediately preceding a variable.

$ cat file1
a
b
c
$c1
$c2
d
$c-falaffel
$f

I want a grep expression that, when passed "c", returns:

$c1
$c2
$c-falaffel

What I have so far:

mode="c"
grep "\\$${mode}" file1

But the shell keeps interpreting $$ as asking for the pid. How in hell do you search an expression that is defined in a variable with a literal dollar sign in front??

Oh, and it has to be good in POSIX-compliant shells (it needs to run on the busybox shell), so bashisms are not acceptable.

Answer 1

you've got one too many backslashes

$ mode="c"
$ grep "\$$mode" file1
$c1
$c2
$c-falaffel
$ 

tested in bash 4.2.20, ksh 93u+, and zsh 5.0.5

Answer 2

This might be close to what you need:

awk -v RS= -v var="$mode" 'sub(".*if [$]" var "_mode","")' "$0"

or maybe this:

awk -v RS= -v var="$mode" 'match($0,"if [$]" var "_mode"){print substr($0,RSTART)}' "$0"

Once you post some input and output we'll have a better idea.

Answer 3

Very similar to Ed Morton's answer: you have to build the string.

search_string='$'
search_string="${search_string}$mode"
grep "$search_string" file1