Shell Programming and Scripting

I am getting a variable from a web page ($inst) which is looked at by the loop:

if ($inst == 2)
{
do stuff
}

the problem is that no matter what the value of $inst the code enters the loop. I know it is some thing small and stupid, but any help would be appreciated.
john

Please post the rest of your code and what shell and operating system you are using.

however, a couple of things you might like to try with the process of elimination

print (echo, printf, whateva) depending on what your coding in. You would firstly make sure that your condition is accurate with the variable. If it doesn't equal 2, you have a problem with your condition, probably the syntax. However, if it always equals 2 maybe you should look at what that variable is doing. Maybe something wrong elsewhere and if that is the case you will have to post more info.

if ($inst == 2)
{
do stuff ; print $inst
}

AND/ OR

print $inst
if ($inst == 2)
{
do stuff ;
}

I did not read the shell you are using (and that makes quite a difference)... on the otherhand, you might just try:

if ($inst == 2) then
{
do stuff
}