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Full Discussion: [Solved] take name of directory and files as variables
Top Forums UNIX for Dummies Questions & Answers [Solved] take name of directory and files as variables Post 302575329 by me. on Monday 21st of November 2011 10:28:44 AM
Old 11-21-2011
Quote:
Originally Posted by vbe
You will have to explain to me what you are trying to do in your script (I don't get you here... especially $1 $2 ...).

What are you trying to achieve? An almost exact copy of a directory content to a new directory? If so why not use appropriate unix commands ( but I can understand you also want to perfect your understanding and improve in shell scripting ! SmilieSmilie
hi vbe, exactly!!!
I want to improve here, I would say need to improve a lot Smilie

what I am trying to write a script that show where you are ( $pwd ) name of directory and all files within that directory then each file copy to other directory ( so my understanding was that $1 $2 ... would be those files ) but dont want to use
PHP Code:
cp * /new/direcotry 
, instead I want to have each file to be copied one by one with echo message showing which file was copied .
PHP Code:
echo " file one (name of file ) has been copied "
echo " file two ( name of file )has been copied " etc 
so I think, variables needs to be used for each file but need a help on that
#hope it sounds more clearly now
 

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