Question

I spent a whole day trying to process some files with backslashes and spaces inside their names. No matter what I do awk (gawk) refuses to print backslashes:

echo "this/pathname/contains/spa ces/and/back\\slashes" | xargs -d'\n' -n1 -I{} bash -c 'echo "{}"; echo whatever | gawk "{printf {}}"'
this/pathname/contains/spa ces/and/back\slashes
gawk: {printf this/pathname/contains/spa ces/and/back\slashes}
gawk:                                           ^ syntax error
gawk: {printf this/pathname/contains/spa ces/and/back\slashes}
gawk:                                                ^ backslash not last character on line

This didn’t work since the backspace gets directly into awk code.

echo "this/pathname/contains/spa ces/and/back\\slashes" | xargs -d'\n' -n1 -I{} bash -c 'echo "{}"; echo whatever | gawk "{printf \"{}\"}"'
this/pathname/contains/spa ces/and/back\slashes
gawk: warning: escape sequence `\s' treated as plain `s'
this/pathname/contains/spa ces/and/backslashes

This worked, but awk eats the backslash. As you can see above, echo prints it but awk doesn’t.

echo "this/pathname/contains/spa ces/and/back\\slashes" | ./escape.sh | xargs -d'\n' -n1 -I{} bash -c 'echo "{}"; echo whatever | gawk "{printf \"{}\"}"'
this/pathname/contains/spa\ ces/and/back\slashes
gawk: warning: escape sequence `\ ' treated as plain ` '
gawk: warning: escape sequence `\s' treated as plain `s'

Next I tried escaping the filenames using escape.sh

#!/bin/bash
xargs -d'\n' -n1 -I{} bash -c 'echo $(printf "%q" "{}")'

Now there’s a double backslash in there but awk still complains.

echo "this/pathname/contains/spa ces/and/back\\slashes" | ./escape.sh | xargs -d'\n' -n1 -I{} bash -c 'echo "{}"; echo whatever | gawk -v VAR=$(printf "%q" "{}") "{printf VAR}"'
this/pathname/contains/spa\ ces/and/back\slashes
gawk: ces/and/back\\slashes
gawk:        ^ syntax error
gawk: ces/and/back\\slashes
gawk:         ^ unterminated regexp

Now awk said some nonsense about some unterminated regexp.

Any ideas? Thanks!

Answer 1

The fix is just to double every backslash that is fed into mawk, either in the input or via variables.
Like this:

# awk needs escaped backslashes
VAR=$(echo "$1" | sed -r 's:\\:\\\\:g')

mawk -v VAR="$VAR" -f "script.awk"

Therefore, if a filename containing backslashes is passed inside $1, this is how you obtain the expected result.