LinuxQuestions.org - [SOLVED] pgrep "program" and if statement

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pgrep "program" and if statement

Hi, I don't know why this always returns the false boolean:

Code:

if [ ! -z $(pgrep "mediatomb") ] ; then

                        mtstatus="Mediatomb is running!"

                else

                        mtstatus="Mediatomb is not running!"

                fi

I tried:

Code:

echo $(pgrep "mediatomb")

1222 1223 1224 1225 1230 1232 1234 1236 1237 1238 1975 1979 1980 1981

Thanks,
Ted

The command substitution is performed before the test take places, so that it actually is something like:

Code:

if [ ! -z 1222 1223 1224 1225 1230 1232 1234 1236 1237 1238 1975 1979 1980 1981 ] ; then

This should throw an error for too many arguments or binary operator expected. Anyway, what you really need is

Code:

if pgrep bash; then

that checks the exit status of the pgrep command: when processes are found the exit status is 0 (success) and the test is true, when processes don't exist, the exit status is 1 (failure) and the test is false. Anyway, I cannot explain why your test doesn't throw an error message and returns always false.

Moreover note that ! -z is the same as -n (the length of string is nonzero).

Quote:

Originally Posted by colucix (Post 4524689)

The command substitution is performed before the test take places, so that it actually is something like:

Code:

if [ ! -z 1222 1223 1224 1225 1230 1232 1234 1236 1237 1238 1975 1979 1980 1981 ] ; then

This should throw an error for too many arguments or binary operator expected. Anyway, what you really need is

Code:

if pgrep bash; then

Thanks for reminding, I realized that double quote should be made:

Code:

if [ ! -z "$(pgrep "mediatomb")" ] ; then

                        mtstatus="Mediatomb is running!"

                else

                        mtstatus="Mediatomb is not running!"

                fi