(18 ноя 2012, 01:16) (
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if(mysql_result(mysql_query(\"SELECT COUNT(*) FROM \'kolhoz_cellground\' WHERE \'id_user \' = \' $ku[id]\'\"
,0)==0)
(18 ноя 2012, 01:10) (
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Miledi, А как исправить?
(18 ноя 2012, 01:07) (
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Warning: mysql_result(): supplied argument is not a valid MySql result resourse in:home/u764179945/public_html/inc/start.php on line 24