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Bug #22994 if ( $something == "NAME") { print $something } prints "0"
Submitted: 2003-03-31 16:45 UTC Modified: 2003-03-31 17:02 UTC
From: hermandepauw at hotmail dot com Assigned:
Status: Not a bug Package: Unknown/Other Function
PHP Version: 4.3.0 OS: Red Hat Linux
Private report: No CVE-ID: None
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 [2003-03-31 16:45 UTC] hermandepauw at hotmail dot com 
if ( $something == "NAME") { print $something } prints "0"

the main problem is in the summary ... example code and source on http://www.deparadox.be/php/ i don't know the explain it in words so get a view ... code is only ~10 lines

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 [2003-03-31 17:02 UTC] rasmus@php.net 
Use === if you want to check both type and value.  The numerical value of "NAME" is indeed 0, so if $something is 0 the condition will be true.
 
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