I have a file that contains directory names:
my_list.txt
:
/tmp
/var/tmp
I'd like to check in Bash before I'll add a directory name if that name already exists in the file.
grep -Fxq "$FILENAME" my_list.txt
The exit status is 0 (true) if the name was found, 1 (false) if not, so:
if grep -Fxq "$FILENAME" my_list.txt
then
# code if found
else
# code if not found
fi
Here are the relevant sections of the man page for grep
:
grep [options] PATTERN [FILE...]
-F
,--fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched.
-x
,--line-regexp
Select only those matches that exactly match the whole line.
-q
,--quiet
,--silent
Quiet; do not write anything to standard output. Exit immediately with zero status if any match is found, even if an error was detected. Also see the
-s
or--no-messages
option.
As rightfully pointed out in the comments, the above approach silently treats error cases as if the string was found. If you want to handle errors in a different way, you'll have to omit the -q
option, and detect errors based on the exit status:
Normally, the exit status is 0 if selected lines are found and 1 otherwise. But the exit status is 2 if an error occurred, unless the
-q
or--quiet
or--silent
option is used and a selected line is found. Note, however, that POSIX only mandates, for programs such asgrep
,cmp
, anddiff
, that the exit status in case of error be greater than 1; it is therefore advisable, for the sake of portability, to use logic that tests for this general condition instead of strict equality with 2.
To suppress the normal output from grep
, you can redirect it to /dev/null
. Note that standard error remains undirected, so any error messages that grep
might print will end up on the console as you'd probably want.
To handle the three cases, we can use a case
statement:
case `grep -Fx "$FILENAME" "$LIST" >/dev/null; echo $?` in
0)
# code if found
;;
1)
# code if not found
;;
*)
# code if an error occurred
;;
esac