PHP using IF to display error
i have a MySQL query and i want to display 1 thing only if the number of affected rows is >=1
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If the program exits normally the por
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Hi,
I have a script with some options.
I use regex to replace patterns in strings, but
Simple image grab script.
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Error when call dll from oracle
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I have dll write C language, I want call it from oracle procedur
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$variable = $variable
Hello
Sorry if this is a really simple question but I have run out of ideas. Is there a reall
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
what is wrong with this
Code: [Select]<?php
if ($_SERVER['HTTP_REFERER'])
{