neprasa Posted January 4, 2007 Report Posted January 4, 2007 $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
andrisp Posted January 4, 2007 Report Posted January 4, 2007 Pārbaudi, kas ir iekš $aaa. Uztaisi print_r($aaa) tieši pirms kverija.
bubu Posted January 4, 2007 Report Posted January 4, 2007 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.
neprasa Posted January 4, 2007 Author Report Posted January 4, 2007 ui, kluda pavisam citur ir nebija dazi $ pareiz uzrakstiti tapec ari $aaa nebij pareizs
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