UNIX for Advanced & Expert Users

With the syntax you used, set overwrites the positional parameters just as the example shows. Sometimes you want to do that. (admittedly, not often.)

Thanks, the set observation fixed the problem almost at all !!!

The script now works !!!

Here I go again...

So, the script is working pretty well now, but I'm testing a case that...
In my opinion should not work, but the program "understands" it anyway and fortunately acts correctly !

Here it is:
When I search for a pattern like "proc.sh" in lines like
"more proc.sh"
"grep proc.sh"
"/bin/ksh ./proc.sh" ---> using grep, it returns all this lines, and that's ok, 'cause in fact they all have this "proc.sh" pattern.

But searching for a pattern like "./proc.sh"
in "/bin/ksh proc.sh", for example, the grep still returns this line as result ! Why this happens ? I'm not passing any parameter, so
wasn't grep supposed to find for an occurance of the ENTIRE pattern ./proc.sh ? I thaught it was ...
Well, here's the current code for the script:

Code:

#!/bin/ksh

CUR_SHELL="/bin/ksh"
V1="VAR1"
set -x

ps -e -o args | grep "$1" | tr -s " " | cut -d" " -f1 |
while read -r V1
do
    #V1=`echo $V1 | cut -d'/' -f2`
    if [ "$V1" = "$CUR_SHELL" ]; then
    echo "Error - Program is already running."
    exit 1
    fi 
done

If anyone could help...
Thanks in advance,

435 Gavea.

added code tags for readability --oombera

I don't follow you. If you think that the grep command is broken, try this as a way to test it:
echo "/bin/ksh proc.sh" | grep "./proc.sh"
this does not return the echo line for me. But if I remove the ./ from the grep, I do get the line.

I wasn't clear enough in last question, forget it...
Now I'm stuck in the following problem:

Code:

#!/bin/ksh

CUR_SHELL = "/bin/ksh"
V1="VAR1"

...

ps -e -o args | grep "$1" | tr -s " " | cut -d" " -f1 |
while read -r V1 V2
do
    if [ "$V1" = "$CUR_SHELL" ]; then
        echo "Error - Program is already running."
        exit 1
    fi 
done

Every word I use as the $1 parameter makes the code enter into the if - although the ps -e -o args | grep "$1" , executed alone, doesn't return any lines beggining with "/bin/ksh" !

There's something you think could be bugging the script ?
This is so strange... because the script was working properly last friday ! Can exist any process with the $1 in the args list that isn't shown with the "ps -e ..." ? I don't think so...
Maybe my user in the Unix server here doesn't have access to view all the processes running ?

Thanks again for any help.

=| 435 Gavea |=

added code tags for readability --oombera

I've found the mistake...
The fact was that when that code was run as a script, that script turned into a process... as usual, so far...

And running the grep at this point would find a line correspondent to this script, as it contained the $1 , because it was passed as argument...
And this line actually starts with "/bin/ksh", because it is a running script... so it entered the if...
And it will allways do so, that way

I'm working on to fix this now

I'll post the final script - one that works properly - when I'm done !

Thanks for all.

=| 435 Gavea |=

How can I count equal lines from an input ?