Bash syntax help: WHILE loop with multiple comparison operators

dz-015 · 04-21-2009, 10:27 AM

Hi,

I'm trying to code the following WHILE loop in Bash:

While [string1 is non zero] and [string2 is non zero] and [counter < max]
Do
...
Done

I can't get the syntax right. I've tried:

Code:

while [ -n "$STRING1"] -a [ -n "$STRING2"] -a [$COUNTER -lt $MAX ]

But I get 'Too may arguments'

Can anyone help?

Thanks,

DZ

colucix · 04-21-2009, 10:54 AM

Too many square brackets!

Code:

while [ -n "$STRING1" -a -n "$STRING2" -a $COUNTER -lt $MAX ]
do
  commands
done

kentyler · 04-21-2009, 10:57 AM

Try like:

while [ -n "$STRING1" && -a -n "$STRING2" && -a $COUNTER && -lt $MAX ]

colucix · 04-21-2009, 11:04 AM

Quote:

Originally Posted by kentyler

Try like:

while [ -n "$STRING1" && -a -n "$STRING2" && -a $COUNTER && -lt $MAX ]

Nope. && and -a would have the same meaning. It should be

Code:

while [ -n "$STRING1" ] && [ -n "$STRING2" ] && [ $COUNTER -lt $MAX ]

but I prefer the -a operator.

Code:

while [[ -n "$STRING1"  &&  -n "$STRING2"  &&  $COUNTER -lt $MAX ]]

also works.

dz-015 · 04-21-2009, 11:13 AM

Thanks guys.
Shell scripts are powerful, but the syntax takes some getting used to!!!!