UNIX for Dummies Questions & Answers

Hi
I have problem with my script.
I dont now why but i don't change anything and script stop working.

this is my code:

Code:

#!/bin/sh
for i in `ps -A | grep pocron.sh | grep -v grep | awk '{print $2}'`
do
        COUNT=$((COUNT+1))
done

ostatnie_wykonanie=`cat porader.log`
aktualna_data=`date +%s`
wynik=$(($aktualna_data - $ostatnie_wykonanie))

if [ $wynik -gt 63 ]
then

if [ $COUNT -z ]
then
./pocron.sh
fi
elif [ $COUNT -gt 1 ]
then
killall pocron.sh
./pocron.sh
fi
exit

when i run script error is:
[: -gt: unary operator expected

on line 18

line 18 is:

Code:

elif [ $COUNT -gt 1 ]

can you help me fix problem?

I suppose the problem comes from "$COUNT" not being initialised. If the expression driving the for-loop in the first line of your script produces no output the loops body will never be executed and this leaves COUNT uninitialised.

The error-message comes from "test" ("[" is just another name for this program) and says that "-gt" is a clause which expects two operands: one before and one behind. One of these required operands is missing.

Insert a debug line before the line

Code:

echo \"$COUNT\"
if [ $COUNT -z ]

to make sure COUNT does indeed have a value - i suppose it doesn't.

Furthermore, there are severe problems with your code:

Code:

if [ $COUNT -z ]

is a syntax error: "-z" expects an operand after, not before. It tests for a variable being uninitialised or empty and because no variable is given this is always true. Still, this is probably not what you want. (Syntactically) Correct is

Code:

if [ -z $COUNT ]

Still, using variables which are sometimes initialised, sometimes not is always a hazard. Add to this that you don't even need the for-loop for counting (use "grep -c" instead) and you shouldn't use backticks at all (really! - they are B-A-D).

Further, i suggest you reindent your code properly. The following syntax error would have stood out immediately:

Code:

if ...
     ....
fi
elif ....
     ....
fi

The "elif" should be an "if", because the first "if" is already closed by a "fi". The "elif" is not the "else"-part of the "if". As there are no other clauses you test the "elif" is superlfuous and you should simply use "else".

Write your script this way:

Code:

#! /bin/sh

typeset -i COUNT=0               # you make sure the variable is declared

COUNT=$(ps -A | grep pocron.sh | grep -vc grep)

...

if [ $COUNT -eq 0 ] ; then
     ....
else
    .....

I hope this helps.

bakunin