I'm an older programmer who is trying to figure out shell scripts on Linux. I'm trying to do something very simple but having a very difficult time figure out what I'm doing wrong. I've looked for answers here ane elsewhere but nothing has really shown me the folly of my ways.

I'm running Fedora 17 (I know I should upgrade) on a Compaq laptop and have created the following shell script in vi:

#!/bin/bash

a=0
while [ $a -lt 3 ]
do
echo "$a"
a=a+1
done


When I run it I get the following error:

./whileexample: line 4: [: a+1: integer expression expected

I've tried it with $a+1 and both change the error message to 0+1 from a+1

If someone could please explain what I'm doing wrong and why it's wrong I'd really appreciate it.

Thank you,

dinoprogrammer

Hi...

Welcome to the forum

There's nothing funny about finding the difficulties of others amusing. Please don't think that I or others on this forum feel that way.

Regards...

This has some examples of how to increment in bash. You need to use arithmetic expansion.
http://askubuntu.com/questions/38552...riable-in-bash

No one is laughing, we all know the feeling!

You can't do simple arithmetic in a shell script as you might expect.

For details see the bash man page, man bash, and find the syntax and usage for ((expression)) and "let" and the section on ARITHMETIC EVALUATION...

What this means for your example is try this instead...

...where I have commented out the original line.

In

Welcome aboard.

Nobody's laughing, but we may just chuckle... when we think of the fact that we've done it too.

Just wanted to second this.