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Posted

$result = mysql_query("SELECT * FROM $aaa ");

while ($row = mysql_fetch_assoc($result))

 

man izmet erroru - Supplied argument is not a valid MySQL result uz to otro rindu

Posted

Nu cik reizes jāsaka...

$result = mysql_query("SELECT * FROM $aaa");
if (!$result)
{
 echo mysql_error();
 die(); // vai kā savādāk nobeidzējies..
}

Izvadi arī pašu kveriju: echo "SELECT * FROM $aaa"; un paskaties, kas tur ir nepareizs.

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