Shell Programming and Scripting

Hi,

I am using Linux and sh shell

Code:

count=7
find * -prune -type d | sort -r -n | ( while read d; do
if [ "$count" == "7" ]; then
echo "FOUND COUNTER1 is: $count"
break 2;
fi
done
echo "FOUND COUNTER2 is: $count" )
if [ "$count" != "7" ]; then
echo "Problem:  Multiple or NO records...Please CHECK !!"
fi

Output:
FOUND COUNTER1 is: 7
FOUND COUNTER2 is: 7
Problem: Multiple or NO records...Please CHECK !!

Why is "Problem: Multiple or NO records...Please CHECK !!" being printed and how can i avoid it ? Can some explain ?

What are you trying to achieve?

EDIT:
Changed doublepost ??: Unable to set Global variable outside while loop

All im trying is to check the value of variable count is equal to 7 or not.

I was expecting it to be equal to 7 and the if statement should fail. But that is not what is happening. The if statement is showhow passing and is being printed when it should not have.

What could i be doing wrong ?

---------- Post updated at 05:01 AM ---------- Previous update was at 04:43 AM ----------

I don't know if they are two different issues or the same.

Running your code snippet, I can't reproduce your problem. sh has a problem with the == test, though...