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Top Forums Shell Programming and Scripting Pass command line argument to variable Post 302456437 by Franklin52 on Friday 24th of September 2010 08:22:42 AM
Old 09-24-2010
Quote:
Originally Posted by Poonamol
Thanks for reply.
But the error still exists, after changing
input_file = " $1 "

echo $input_file

Please help me out.
Please reread the 2nd line of the post of danmero.
 

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base64(1)						    BSD General Commands Manual 						 base64(1)

NAME
base64 -- Encode and decode using Base64 representation SYNOPSIS
base64 [-d | -h | -v | -D] [-b count] [-i input_file] [-o output_file] DESCRIPTION
base64 encodes and decodes Base64 data, as specified in RFC 4648. With no options, base64 reads raw data from stdin and writes encoded data as a continuous block to stdout. OPTIONS
The following options are available: -b count --break=count Insert line breaks every count characters. Default is 0, which generates an unbroken stream. -d --debug Print verbose log messages during processing. -D --decode Decode incoming Base64 stream into binary data. -h --help Print usage summary and exit. -i input_file --input=input_file Read input from input_file. Default is stdin; passing - also represents stdin. -o output_file --output=output_file Write output to output_file. Default is stdout; passing - also represents stdout. -v --version Print build version and exit. SEE ALSO
openssl(1), wikipedia page <http://en.wikipedia.org/wiki/Base64>, RFC 4648 <http://tools.ietf.org/html/rfc4648> Mac OS X 10.7 February 8, 2011 Mac OS X 10.7
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