Package: zstd
Version: 1.3.8+dfsg-2
Severity: normal
Dear Maintainer(s),
the search command zstdgrep always returns an exit code of 1 when
called with file name(s).
The logic with EXIT_CODE and CUR_EXIT_CODE in that shell script seems
bogus to me. Shouldn't something simple like
EXIT_CODE=0
...
[ "$?" -ne 0 ] && EXIT_CODE=1
...
work?
I attach a patch which also corrects a minor nit regarding the *egrep and *fgrep
variants.
Regards,
Jörg.
--- zstdgrep.orig 2019-01-03 14:47:31.000000000 +0100
+++ zstdgrep 2019-01-05 19:16:01.697399037 +0100
@@ -35,8 +35,8 @@
# handle being called 'zegrep' or 'zfgrep'
case "${prg}" in
- *zegrep) grep_args="-E";;
- *zfgrep) grep_args="-F";;
+ *zstdegrep) grep_args="-E";;
+ *zstdfgrep) grep_args="-F";;
esac
# skip all options and pass them on to grep taking care of options
@@ -113,16 +113,11 @@
if [ "${silent}" -lt 1 ] && [ "$#" -gt 1 ]; then
grep_args="-H ${grep_args}"
fi
- CUR_EXIT_CODE=0
- EXIT_CODE=1
set -f
while [ "$#" -gt 0 ]; do
# shellcheck disable=SC2086
"${zcat}" -fq -- "$1" | "${grep}" --label="${1}" ${grep_args} -- "${pattern}" -
- CUR_EXIT_CODE=$?
- if [ "${CUR_EXIT_CODE}" -eq 0 ] && [ "${EXIT_CODE}" -ne 1 ]; then
- EXIT_CODE=0
- fi
+ [ "$?" -ne 0 ] && EXIT_CODE=1
shift
done
set +f
