Csh to check for existence of file - Page 2 - Shell Programming and Scripting

drl · #8 12-05-2007

Hi.

Looking at the man page:

Code:

#!/usr/bin/env csh

# @(#) s1       Demonstrate file existence test in csh.

echo
echo "(Versions displayed with local utility version)"
sh -c "version >/dev/null 2>&1" && version tcsh
echo

rm -f t1 t2
touch t1 t2

if ( -e t1 ) then
  echo " File t1 exists."
else
  echo " File t1 does not exist."
endif

rm t2
if ( -e t2 ) then
  echo " File t2 exists."
else
  echo " File t2 does not exist."
endif

exit 0

Producing:

Code:

% ./s1

(Versions displayed with local utility version)
tcsh 6.13.00

 File t1 exists.
 File t2 does not exist.

Adjust as necessary for your situation, see man page for details (this is done with csh as a link to tcsh) ... cheers, drl

_____
Obligatory scripting advice: "use Bourne family, not csh family, for scripting".

reborg · #9 12-05-2007

Quote:

Originally Posted by Raynon

Code:

% ls *myfile
xxx1_myfile  xxx2_myfile
% ls *myfile|echo $status
0
% Reset tty pgrp from 7873 to 11434

I'm not really sure why you piped ls into echo $status, but as indicated previous status is the same as $? in bourne based shells.

Raynon · #10 12-05-2007

Quote:

Originally Posted by reborg

I'm not really sure why you piped ls into echo $status, but as indicated previous status is the same as $? in borne based shells.

Hi Reborg,

But what's your purpose of this statement " ls *_myfile >& /dev/null " before the statement " if ( $status == 0 ) then ". My code " ls *myfile|echo $status " was emulating this.

I tried to perform "echo $status" in a directory with files and in a directory without files, the output is always " 0 ". So what does $status really mean ?

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