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Loops Create New Processes?

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ImNtReal
 PostPosted: Mon Nov 14, 2011 1:14 pm   

Joined: Mon Nov 14, 2011 1:08 pm
Posts: 2
I'm confused. I have created a script called aslkgsdklh to ensure that it's name won't match any other running processes. If I execute this script only once, why would this code ever return anything other than ones?

Code:
#!/bin/bash
while [ 0 -lt 1 ]; do
        echo $(ps -A | grep -v defunct | grep -c $(basename $0))
done


Thanks,
=-Jameson
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choroba
 PostPosted: Mon Nov 14, 2011 5:06 pm   
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Joined: Tue Apr 27, 2010 2:28 pm
Posts: 172
Location: Czech Republic
See the difference between
Code:
ps -A | grep bash

and
Code:
echo $(ps -A | grep bash)

or, even
Code:
echo $(ps -A | grep bash) | cat
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ImNtReal
 PostPosted: Mon Nov 14, 2011 7:09 pm   

Joined: Mon Nov 14, 2011 1:08 pm
Posts: 2
What I I'm actually trying to accomplish is:
Code:
if [[ $(ps -A | grep -c $(basecode $0) -gt 1) ]]; then
     exit 1
fi


I know I've used something similar before, but that explains why it's not working. Thanks.
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jsz
 PostPosted: Tue Nov 15, 2011 11:36 am   
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Joined: Wed Jun 08, 2011 8:27 am
Posts: 159
Location: outer Shpongolia
What you're trying to do is called mutual exclusion. It rejoins this topic.

So, you can use a lock file and flock(1).

Code:
#!/bin/bash

exec 3> /tmp/my_super_script_name.lock

if ! flock -n 3; then
    echo 'error: another instance is running.' >&2
    exit 1
fi

# script's code goes here and beyond...
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