Inline if shell script

Question

Is it possible to execute shell script in command line like this :

counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ] then; echo "true"; > 

Above example is not working I get only > character not the result I'm trying to get, that is "true"

When I execute ps -ef | grep -c "myApplication I get 1 output. Is it possible to create result from single line in a script ? thank you

Answer 1

It doesn't work because you missed out fi to end your if statement.

counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ]; then echo "true"; fi 

You can shorten it further using:

if [ $(ps -ef | grep -c "myApplication") -eq 1 ]; then echo "true"; fi 

Also, do take note the issue of ps -ef | grep ... matching itself as mentioned in @DigitalRoss' answer.

update

In fact, you can do one better by using pgrep:

if [ $(pgrep -c "myApplication") -eq 1 ]; then echo "true"; fi 

Answer 2

Other responses have addressed your syntax error, but I would strongly suggest you change the line to:

 test $(ps -ef | grep -c myApplication) -eq 1 && echo true 

If you are not trying to limit the number of occurrences to exactly 1 (eg, if you are merely trying to check for the output line myApplication and you expect it never to appear more than once) then just do:

 ps -ef | grep myApplication > /dev/null && echo true 

(If you need the variable counter set for later processing, neither of these solutions will be appropriate.)

Using short circuited && and || operators is often much clearer than embedding if/then constructs, especially in one-liners.