Вот возникла такая проблема: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/all4ucoz/public_html/test_site/index.php on line 4
Код:
Code
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<?php
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include ("blocks/db.php"); /*Соединяюсь с БД*/
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$result = mysql_query("SELECT title,meta_d,meta_t,text FROM settings WHERE page='index'",$db);