neprasa Posted January 4, 2007 Report Share 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 Link to comment Share on other sites More sharing options...
andrisp Posted January 4, 2007 Report Share Posted January 4, 2007 Pārbaudi, kas ir iekš $aaa. Uztaisi print_r($aaa) tieši pirms kverija. Link to comment Share on other sites More sharing options...
bubu Posted January 4, 2007 Report Share 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. Link to comment Share on other sites More sharing options...
neprasa Posted January 4, 2007 Author Report Share Posted January 4, 2007 ui, kluda pavisam citur ir nebija dazi $ pareiz uzrakstiti tapec ari $aaa nebij pareizs Link to comment Share on other sites More sharing options...
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