Added: 9 months ago
From: phpacademy
Views: 1,300
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  • You can also do

    mysql_query("INSERT INTO `urls` (url, code) VALUES ('$url', '$code')");

  • keep getting an error on the return function for code_exists function in func.inc.php file. It says "Warning: mysql_result(): supplied argument is not a valid MySQL result resource" Help?

  • Comment removed

  • Comment removed

  • nice and awesome

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