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#1
04-13-2008
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testing if files exist
I am trying to test arguments to see if they are files in any directory.
I have : [ $# -f 0 ] but it's not working 
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#2
04-13-2008
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Is your script supposed to be invoked with the files you want to process, and terminate if they don't exist? $# indicates how many arguments there were (presumably you want to abort if there were none) but to check whether the named file(s) exist, you give them as arguments to "test -f".
Code:
case $# in 0) echo no files, try again >&2 ;; esac
for f in "$@"; do
if [ -f "$f" ] ; then
... handle file
else
echo "$0: file $f not found -- skipping" >&2
fi
done
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#3
04-13-2008
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when I run my program and enter 2 arguments at the command line it is not checking to see if the 2 arguments are files or not.
I have tried if [ -f filename ] elif [ -f $1 -o $2 ] I guess I don't know how to make it so it uses the 2 arguments as filenames and then searches for them.
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#4
04-13-2008
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That's the loop. If you always want exactly two arguments and they both need to exist, you can do that too. The -o doesn't "remember" what you did before so you need to tell it again to look for a file.
Code:
if [ -f "$1" -o -f "$2" ]
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#5
04-30-2008
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How to check if a file exists
I need to check following three things:
a) a file exists but is zero byte b) a file exists with non-zero bytes c) a file doesn't exist at all What switches do we use? Is it -f, -s, -z. Please explain the difference in these! Many Thanks Shalua
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#6
04-30-2008
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You know enough to be better served by finding the documentation and reading that. It's more precise and more authoritative than any of us here, on a typical day. Read man sh and maybe man test if you have that.
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