[SOLVED] pass file-name as command line argument

vikas027 · 10-14-2008, 10:17 PM

Hi all,

I have made a script which has a necessity that the file name should be "OVO_Report".

How can I modify the script so that it could take any file name provided as the first argument in the command line (WITHOUT making much changes to my present script).

I need something like capture filename given in the command line in a variable and use that variable in the script.

For e.g.

Code:

./my_script "any_filename"

Pls help, I haven't worked on scripts which take command line arguments.

Regards,
VIKAS

born4linux · 10-14-2008, 10:43 PM

you can use $1 - $1 is the first argument:

so, in your script, you can set:

nfileSource=$1

then use nfileSource throughout your script.

if you can only need 1 argument (the filename), you can also:

if [[ $# != 1 ]]
then
echo "\nUSAGE: $0 filename\n"
exit 0
fi

nfileSource=$1

echo "target file is ${nfileSource}."

vikas027 · 10-14-2008, 11:52 PM

Thanks born4linux,
works perfect, will make a note for future.

Regards,
vikas