Escaping backslash in AWK in command substituion
Solution 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"
Solution 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.
Solution 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
![Bill](https://lh4.googleusercontent.com/-aozJEg9VUWw/AAAAAAAAAAI/AAAAAAAAAAA/Fv-MMjSrXH4/photo.jpg?sz=256)
Bill
Updated on June 19, 2022Comments
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Bill about 2 years
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.
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Birei about 11 years@Lorkenpeist: From the man page of
bash
: When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by $, `, or \.. So, when you found two backslashes their meaning is the usual. Otherwise it will be interpreted literally, so either with 5 or 6 will work. In the former because each two of them, one is output, and the last one has no special meaning. And with the latter because all of them are interpreted and will output three of them, just what you want. -
Uwe about 11 yearsThe correct way in this particular case is
q=\\\"$a\"
. There is no need to use command substitution, echo, awk, or a pipeline. -
Chris Seymour about 11 yearsSo the correct way use shell variables and backslashes with awk is not to use awk? I don't think so... Don't take the question literary, it's just an example.
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Uwe about 11 yearsSorry, I missed that this was not the OP's real problem but a simplified version.