shell programming question: help with expr command

[Devin Teske]

On Apr 9, 2011, at 6:31 AM, Dino Vliet <dino_vliet at yahoo.com> wrote:

> Hi folks,
> I'm having trouble with a little shell script. Can somebody explain me why I get 3 times "expr: syntax  error" in my console after I run this little script?
> 
> #! /usr/local/bin/bash
> # testscript 
> 
> var1="trees.J48" #other value will be rules.Jrip, rules.DecisionTable
> len=${#var1}
> ind=`expr index $var1 s`
> pos=`expr $len - $ind`
> out=`expr substr $var1 $ind $pos` 
> 
> I would expect (and want the following to happen):
> 
> $ind should contain 6
> $pos should contain 3
> $out should contain J48 (other values will have to be Jrip,DecisionTable)
> 
> Can anyone help me with this?

This would be a /bin/sh compatible (read: portable) way to accomplish the above:

#!/bin/sh
# testscript 
var1="trees.J48" #other value will be rules.Jrip, rules.DecisionTable
len=${#var1}
ind=`echo "$var1" | awk '{print index($0,"s")+1}'`
pos=$(( $len - $ind ))
out=`echo "$var1" | awk -vind="$ind" -vpos="$pos" '{print substr($0,ind+1,pos)}'`

Though, there are certainly easier ways to get at what it is that I assume your after:

#!/bin/sh
# testscript
var1="trees.J48" #other value will be rules.Jrip, rules.DecisionTable
out="${var1##*.}"

-- 
Devin

> 
> Thanks
> 
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