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Old 07-16-2011, 12:54 PM   #1
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Registered: May 2011
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How to pass file content through URL using shell script

Hi All,

I am calling a URL from shell script and passing few argumants,
Here i have to pass file content as one argument.

How can i pass file content through URL.

content=`cat /Users/test1.txt`
open http://localhost:8080?filecontent=$content

it is not working.

Please give me solution for this.

Old 07-16-2011, 02:50 PM   #2
Registered: Oct 2004
Location: USA
Distribution: Ubuntu-Server, Mac OS X, Arch Linux
Posts: 356

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I'm guessing you're passing it to a Php script? I'm not sure how it would work on the shell script side, but here is what you would do on the Php side:

content=`cat /Users/test1.txt`

  $content = $_GET['filecontent'];

  echo $content;

We can at least start at this point if you haven't gotten this far.


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