Quick and easy if then script to determine if an application is running?

Question

I'm getting confused on syntax with an if else statement. I just want see if an app is running and exit 0 if so. If not, then exit 1. I tried a test using TextEdit and I always get a syntax error in the console when running my shell script. Here's what I've got:
ps -axo pid,ucomm | awk '{ if ( $2 == "TextEdit" ) exit 0 [ else exit 1 ] }'
I've been looking at this too long. What am I missing?
Thanks.

Answer 1

Hi Mills,
   You must have stared at it too long. The syntax error is just the square brackets; erase them. If there's only one command "else" works without any grouping symbol. If there is more than one command, the required grouping symbols are braces, "{}" and they would follow the "else" command. By the way Michael, "exit" is an awk command. I swear that I didn't know that either but it is. However oddly, it doesn't do what it says but more about that in a sec.
   The real problem with the awk script is the logic. It would only detect TextEdit if that's the first process listed. This awk script is only going to test the first line because the script is run once for each line and since there's an exit command in each branch of the if statement, it exits the first time it's run. You want something like the following:
ps -axo pid,ucomm | awk '/TextEdit/{ exit 0 } END { exit 1 }'
However, this doesn't work either because the END block is executed despite the "exit" command. Therefore you'd have to do something like:
ps -axo pid,ucomm | awk 'BEGIN{ STAT = 1 } /TextEdit/{ STAT = 0; exit } END { exit STAT }'
That kindof makes Michael's suggestion of "grep" seem more appealing, eh? The exit status of "grep" depends on whether there is a match so it's tailor made to do what you want.
Gary
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