Re: Testing for a value not sent to a shell script

Quoting Scott Klement <midrange-l@xxxxxxxxxxxxxxxx>:

Pete,

That code doesn't do what you think it does!!

#!/bin/bash
if [ $1 ] ; then
echo 'Yes!'
else
echo 'No!'
fi

What this is doing is trying to execute $1 as an expression for the
"test" utility to evaluate. If test evaluates to true, it'll print yes,
if not, it'll print no. So this isn't really testing to see if a
parameter is passed! It's evaluating the parameter as an expression!

For example, try this:

$ ./myscript '1 -eq 0'
No!

See what I mean? I *did* pass it a parameter, but it says no. Why?
Because it's running my parameter as an expression. Since 1 is not
equal to 0, it says "No!"

If you just want to test if a parameter is empty, do this instead:

if [ -z "$1" ] ; then
echo 'No!'
else
echo 'Yes!'
fi

I would've given that answer at first if I had known that was what John
was asking. But there are so many things he could've meant...

Scott,

Thanks for the explanation of the issue with Pete's example.

The moment I saw the -z in your final example, it hit me like a brick. I used
that so much twenty years ago I thought I couldn't forget it. It is like the
obsure stuff I can remember about the IBM 1130 and other stuff I can't
remember. Weird how memory works.

Thanks for the example, and your encouragement.

John McKee