Escaping backslash in AWK in command substituion

Question

I am trying to escape backslash in AWK. This is a sample of what I am trying to do.

Say, I have a variable

$echo $a
hi

The following works

$echo $a | awk '{printf("\\\"%s\"",$1)'}
\"hi"

But, when I am trying to save the output of the same command to a variable using command substitution, I get the following error:

$ q=`echo $a | awk '{printf("\\\"%s\"",$1)'}`
awk: {printf("\\"%s\"",$1)}
awk:               ^ backslash not last character on line

I am unable to understand why command substitution is breaking the AWK. Thanks a lot for your help.

Answer 1

Try this:

q=$(echo $a | awk '{printf("\\\"%s\"",$1)}')

Test:

$ a=hi
$ echo $a
hi
$ q=$(echo $a | awk '{printf("\\\"%s\"",$1)}')
$ echo $q
\"hi"

Update:

It will, it just gets a littler messier.

q=`echo $a | awk '{printf("\\\\\"%s\"",$1)}'`

Test:

$ b=hello
$ echo $b
hello
$ t=`echo $b | awk '{printf("\\\\\"%s\"",$1)}'`
$ echo $t
\"hello"

Reference

Answer 2

Quoting inside backquoted commands is somewhat complicated, mainy because the same token is used to start and to end a backquoted command. As a consequence, to nest backquoted commands, the backquotes of the inner one have to be escaped using backslashes. Furthermore, backslashes can be used to quote other backslashes and dollar signs (the latter are in fact redundant). If the backquoted command is contained within double quotes, a backslash can also be used to quote a double quote. All these backslashes are removed when the shell reads the backquoted command. All other backslashes are left intact.

The new $(...) avoids these troubles.

Answer 3

Don't get into bad habits with backticks, quoting and parsing shell variables to awk The correct way to do this is:

$ shell_var="hi"

$ awk -v awk_var="$shell_var" -v c='\' 'BEGIN{printf "%s%s\n",c,awk_var}'
\hi

$ res=$(awk -v awk_var="$shell_var" -v c='\' 'BEGIN{printf "%s%s\n",c,awk_var}')

$ echo "$res"
\hi