[PHP-DEV] PHP 4.0 Bug #9444 Updated: Allow show_source output to be set as a variable From: derick <email protected>
Date: 02/25/01

ID: 9444
Updated by: derick
Reported By: brad <email protected>
Old-Status: Open
Status: Closed
Bug Type: Feature/Change Request
Assigned To:
Comments:

You can easily do this with the outputbuffering functions:
ob_start();
show_source("filename");
$contents = ob_get_contents();
ob_end_flush();

Previous Comments:
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[2001-02-25 13:21:06] brad <email protected>
Can you make it so that when you call show_source that you can set its output as a variable:

ie: $var = show_source( $file );

Thanks!

Brad Taylor
Senior Web Content Developer
BradCom Industries LTD

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