I have the following command, like:
command1 | grep "xxxx"
command1 will output some line text.
I hope that execute exit 1 when the line includes xxxx, or output the line to console, like:
command1 | if [ grep "xxxx" ] then; exit 1; else echo;
How should I write the shell? thanks!
Since you want to test the exit status of grep, and you want to see the line if it does not contain xxxx, you can adapt the following code to suit our requirements (where I'm using echo "$line" as a surrogate for your command1, and I'm echoing exit 1 rather than actually exiting when your code would execute exit 1):
for line in 'hello xxxx here' 'hello yyyy here'
do
if ! echo "$line" | grep -v 'xxxx'
then echo "exit 1"
fi
done
The output is:
exit 1
hello yyyy here
See How can I negate the return value of a process? for a discussion of the ! operator in POSIX shells (such as Bash).
Since you don't want to see the output when it contains xxxx but do otherwise, the -v option to grep inverts the logic, only printing lines that do not match xxxx. The status of grep -v 'xxxx' is 0 when it only finds lines that don't match xxxx, and 1 if it finds a line that matches. The ! operator inverts this status, and the if tests whether the inverted status is 0, executing the then code if the inverted status is 0.
Note that [ is just a command that returns a suitable exit status which the if tests. On many systems, you can find /bin/[ as an executable, maybe as a (symbolic?) link to /bin/test — but modern shells invariably treat [ and test as built-in commands. Note that [ checks that its last argument is ] and objects if it isn't; test does not.
