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How to take the file name in run time using shell.?

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Old Unix and Linux 02-04-2013
praveen265 praveen265 is offline
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RedHat How to take the file name in run time using shell.?

I want to take the file name as an input to the program and copy that file into new location using shell. Below program is not working properly.

Code:
#!/bin/sh
if [ $file != '']; then
`/usr/bin/perl -pi -e's/(notifications_enabled\s*)(\d+)/$sub = "$1" . ("$2"== "0" ? "1":"0")/e; ' $file`
`cp /script/test/123/$file /etc/hosts/`
fi
case $1 in
'file') echo $file ;;
*) echo "Usage: $0 specify file name" ;;
esac


Last edited by Franklin52; 02-04-2013 at 05:11 AM.. Reason: Please use code tags for data and code samples
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Old Unix and Linux 02-04-2013
Corona688 Corona688 is offline Forum Staff  
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Why are you putting everything in backticks `` ? That is not correct, they are pointless here and probably spew errors too.

You probably want positional parameters here. If someone calls your script with myscript filename then filename will be $1.


Code:
if [ -z "$1" ]
then
        echo "Usage:  $0 filename" >&2
        exit 1
fi

cp "/script/test/123/$1" /etc/hosts/

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Old Unix and Linux 02-05-2013
praveen265 praveen265 is offline
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RedHat

Thanks for correcting me Linux
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