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22/08/2006, 13h31

thomasriise wrote:
> I e.g. have this script:
>
> ---------------------------------
> #!/bin/ksh
>
> numfiles=`ls fil* | wc -w | sed 's/ *//'`
> files=`ls fil*`
> echo "there are $numfiles files to process"
>
> doit(){
> for i in $files
> do
> list=`ls -ut $i`
> size=`ls -lut $i 2>/dev/null | awk '{print $5}'`
> if [ $size -lt 630 ]
> then
> done

If you want to skip the file then use

continue

if you want to leave the loop then use

break

instead of done.

Janis

> fi
> echo $list
> done
> }
>
> doit
> ---------------------------------
>
> My shell reports that the done in line 15 is unexpected. Is there
> another way to skip the files smaller than the specified size? Or
> should I just do that before the loop should begin?
>
> Thanks.

22/08/2006, 13h31	#3
Janis Messages: n/a Hébergeur:	Re: How to skip small files in a for loop thomasriise wrote: > I e.g. have this script: > > --------------------------------- > #!/bin/ksh > > numfiles=`ls fil* \| wc -w \| sed 's/ //'` > files=`ls fil` > echo "there are $numfiles files to process" > > doit(){ > for i in $files > do > list=`ls -ut $i` > size=`ls -lut $i 2>/dev/null \| awk '{print $5}'` > if [ $size -lt 630 ] > then > done If you want to skip the file then use continue if you want to leave the loop then use break instead of done. Janis > fi > echo $list > done > } > > doit > --------------------------------- > > My shell reports that the done in line 15 is unexpected. Is there > another way to skip the files smaller than the specified size? Or > should I just do that before the loop should begin? > > Thanks.