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
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.

you can use continue to skip the loop

another way to do:
if [ $size -ge 630 ]; then
echo $list
fi

"thomasriise" <thomasriise@gmail.com>
??????:1156245931.171886.305230@m79g2000cwm.google groups.com...
> 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
> 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.
>

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.

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
> 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.
>

Instead of negative logic:

if [ $size -lt 630 ]
then
# break out of loop
fi
echo $list

just use positive:

if [ $size -ge 630 ]
then
echo $list
fi

By the way, your script will fail for files whose names contain white
space. See question 14,
http://home.comcast.net/~j.p.h/cus-faq-2.html#14, in the FAQ.

Regards,

Ed.

fengrui wrote:
> you can use continue to skip the loop

This worked. I just changed "done" to "continue" and then it only
processed the files that had a proper size. Thanks fengrui.

> By the way, your script will fail for files whose names contain white
> space. See question 14,
> http://home.comcast.net/~j.p.h/cus-faq-2.html#14, in the FAQ.
>
> Regards,
>
> Ed.

That's right, but there's no filenames with whitespaces in this
fileflow. But thanks for the useful link anyway!

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
> 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?

Yes, you should do it without loop.
See the -S option of the ls command, if available, or use the -s and
-l option and postprocess the output using awk like this

ls -sl | awk '$5>=630'

I have no Unix system at hand, so you need to replace the $5 by the
column number of ls where the size is displayed.

Janis

> Thanks.

On 22 Aug 2006 04:25:31 -0700, 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"


Ouch. That's ugly. If you're using ksh, you could as well use
its extensions. Like arrays.

set -A files -- fil*
printf "there are %d files to process\n" "${#files}"

> doit(){
> for i in $files

for i in "${files[@]}"

> do
> list=`ls -ut $i`

What's that for?

> size=`ls -lut $i 2>/dev/null | awk '{print $5}'`
size=`ls -lut -- "$i" 2>/dev/null | awk '{print $5; exit}'`

What is the -t for?

What about -u as you're only interested in the size.

> if [ $size -lt 630 ]

if [ "$size" -lt 630 ]

> then
> done

continue instead of done

> fi
> echo $list

printf '%s\n' "$i"


> done
> }

This script would more reasonably be written:

find . \( -name . -o -prune \) -name 'fil*' \! -size -630c -print

Note zsh's:

print -rl fil*(^L-630)

--
Stephane