Why I have a integer expression expected error with this:
at=`echo $1 | grep -q "@"`
if [ $at -ne 0 ]; then
echo "blabla"
else
echo "bloblo"
fi
$at is set, and the test working fine outside the script
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2 Answers
When testing the result of grep -q, you want to test $? not the output of grep, which will be empty
at=$(echo "$1" | grep -q "@")
if [ $? -ne 0 ]; then ...
or simply
if echo "$1" | grep -q "@"; then ...
or, more bash-ly
if grep -q "@" <<< "$1"; then ...
or, without calling grep:
if [[ "$1" == *@* ]]; then ...
or
case "$1" in
*@*) echo "match" ;;
*) echo "no match" ;;
esac
1 Comment
user767990
When testing the result of grep -q, you want to test $? : EXACTLY Thanks for all the way you purpose, I think I learn something today :) particulary like the "no grep" syntax. And thanks by the way
-ne is for comparing integers. Use != to compare strings.
1 Comment
user767990
Thanks a lot man ! it's working fine, but it needed [ "$at" != 0 ]; Thanks for the very quick reply :) –
