I think several lines are not quite right:
- regular expression in gawk should be inside the {}, but
you will have to pass the argument to gawk.
- next line after if should be then, it is usually used like
if [ -z "$processid" ] ; then
the script will only echo first time it finds the string, if
you have several instances of the same string, then you need
a for loop like:
=======================
#!/bin/bash
processid=`lsof -i | grep $1 | gawk '{print $1}'`
for proc in $processid ; do
[ -z "$proc" ] && "echo not running"
done
========================
hope it helps
raymundo
Ryan Babchishin wrote:
[EMAIL PROTECTED] wrote:
I'm trying to get a pidof a php script by capturing the path:
like /home/somebody/my_script which will be the first arg. to the bash
script
Could anybody tell me what's wrong with this script ?
I'm new to scripting in general so any help would be apreciated !
#!/bin/bash
1st_arg=$1
prosesses=`lsof -i`
prosessid=`echo $prosesses | gawk /1st_arg/'{print $2}'`
echo $prosessid
if [ -z "$prosessid" ]
echo "not running"
fi
exit 0
Regards
Lars Sorensen
It would write it like this (although yours almost works):
--- CUT ---
#!/bin/bash
pid=`lsof -i | grep "$1" | awk '{print $2}'`
if [ -z "$pid" ]; then
echo "not running"
exit 1
fi
echo "$pid is running"
--- CUT ---
This would handle characters that would normaly have to be escaped a
little better and provides a non-zero error level if your process isn't
running. You could then use this script within other scripts if you
wanted, like this:
--- CUT ---
if ! /somewhere/checkpid ; then
dosomething
fi
--- CUT ---
Although, I think the only thing wrong with your script is that it
doesn't have a "; then" after the "if [ xxx ]" statement.
Good luck!
- Ryan
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