Jeffrey Ross wrote:
> The other more general method is:
>
> ls | # to get all the files in the current
directory
> grep "\.class$" | # to filter out all except ones ending ".class"
> xargs ls -l # to append those files to "ls -l" in batches
keeping
> under the maximum argument size
>
> Regards,
> Jeffrey.


does this work under csh as well? i'm going crazy trying to find a
workaround for this error. i work for a vfx film house in los angeles
and im constantly needed to grab info from a range of frames
(represented as individual files), like so:

092_comp_01_shp.0935.cin
092_comp_01_shp.0936.cin
092_comp_01_shp.0937.cin
092_comp_01_shp.0938.cin
092_comp_01_shp.0939.cin
092_comp_01_shp.0940.cin


so for instance, to find the 25th file in from the end of the range i'd
do normally do this:

set last = `ls -1 *.*.cin | tail -25 | head -1`

which usually works fine - until i get a shot with a LOT of frames
(like the above example), in which case i get the "arg list too long"
error.

so i attempted to implement jeffrey's solution thusly:

set last = `ls -1 *.*.cin | xargs ls | tail -25 | head -1`

and i get this:

father 16% set last = `ls -1 *.*.cin | xargs ls | tail -25 | head -1`
/sbin/ls: Arg list too long.
father 17% echo $last
092_comp_01_shp.0916.cin

so it works, but i still get the error msg. what am i doing wrong? i
freely admit i have no idea how xargs works and the man page (like most
of them), is depressingly cryptic and dense. any help is appreciated
(even if i am in the wrong group - is there a csh group?)

cheers
christopher

[email]deepstructure@gmail.com[/email] wrote in
news:1111470754.951758.38930@f14g2000cwb.googlegro ups.com:

> Jeffrey Ross wrote:
>> The other more general method is:
>>
>> ls | # to get all the files in the current
> directory
>> grep "\.class$" | # to filter out all except ones ending ".class"
>> xargs ls -l # to append those files to "ls -l" in batches
> keeping
>> under the maximum argument size
>>
>> Regards,
>> Jeffrey.
>
>
> does this work under csh as well? i'm going crazy trying to find a
> workaround for this error. i work for a vfx film house in los angeles
> and im constantly needed to grab info from a range of frames
> (represented as individual files), like so:
>
> 092_comp_01_shp.0935.cin
> 092_comp_01_shp.0936.cin
> 092_comp_01_shp.0937.cin
> 092_comp_01_shp.0938.cin
> 092_comp_01_shp.0939.cin
> 092_comp_01_shp.0940.cin
>
>
> so for instance, to find the 25th file in from the end of the range i'd
> do normally do this:
>
> set last = `ls -1 *.*.cin | tail -25 | head -1`
>
> which usually works fine - until i get a shot with a LOT of frames
> (like the above example), in which case i get the "arg list too long"
> error.
>
> so i attempted to implement jeffrey's solution thusly:
>
> set last = `ls -1 *.*.cin | xargs ls | tail -25 | head -1`
>
> and i get this:
>
> father 16% set last = `ls -1 *.*.cin | xargs ls | tail -25 | head -1`
> /sbin/ls: Arg list too long.
> father 17% echo $last
> 092_comp_01_shp.0916.cin
>
> so it works, but i still get the error msg. what am i doing wrong? i
> freely admit i have no idea how xargs works and the man page (like most
> of them), is depressingly cryptic and dense. any help is appreciated
> (even if i am in the wrong group - is there a csh group?)

you get the same error because "ls -1 *.*.cin" will give you the error.
Keep in mind that what you type on the command line IS NOT nessessary what
the shell will execute.

With a ksh shell just set ( set -x ) on the command line and you will see
what a "ls /tmp/* " is really.

But to solve our problem:

ls -l | grep "\.cin$" | tail -25 | head -1

should do the job

ls -l :will show everything but you do not get the " arg list to long "
error
grep "\.cin$" : will match everything that ends with ".cin"

HTH
Hajo

hey thanks. i just have one question - what does the $ stand for in
that line? im using csh, and $ is the variable call. what is it doing
here?

thanks for your help! much appreciated.

[email]deepstructure@gmail.com[/email] wrote in
news:1111561124.878648.44410@z14g2000cwz.googlegro ups.com:

> hey thanks. i just have one question - what does the $ stand for in
> that line? im using csh, and $ is the variable call. what is it doing
> here?
>
> thanks for your help! much appreciated.
>

The "$" means End of File
The "^" means Begin of File

Example:
You have two files in your directory
../start.end
../start.end.bkp
../next.start

Now you would like to match only files that ENDS with ".end"


ls | grep "\.end" would match both files ( lines )
ls | grep "\.end$" would match ONLY files (lines )that ENDS with ".end"

ls | grep "^start\." would match ONLY files (lines) begining with "start."

So the $ means for your shell a variable ( If a letter follows it ) but it
does has a different meaning for grep,sed a.s.o


Happy Programming
Hajo