mysql_fetch_assoc（）：6不是有效的MySQL結果資源

-5

嘗試將某些信息查詢到數據庫中，但似乎無法找到該錯誤。我一直在PHP的MySQL函數上遇到一些麻煩，但我會說大約90％的修復都是簡單的語法錯誤。mysql_fetch_assoc（）：6不是有效的MySQL結果資源

 $TableName = "opportunities"; 
     $Opportunities = array(); 
     $SQLString = "SELECT opportunity_ID, company, city, " . 
     "start_date, end_date, position, description" . " FROM $TableName;";  
     $QueryResult = mysql_query($SQLString); 

     if (mysql_num_rows($QueryResult) > 0) 
     { 
      while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) 
      { 
       $Opportunities[] = $Row; 
       mysql_free_result($QueryResult); 
      } 
     }

來源

2014-04-08 guyg

http://stackoverflow.com/questions/1858304/mysql-fetch-assoc-supplied-argument-is-not-a-valid-mysql-result-resource-in-p?rq=1 – Marty

http：///stackoverflow.com/questions/3268691/mysql-num-fields-supplied-argument-is-not-a-valid-mysql-result-resource?rq=1 – Marty

http://stackoverflow.com/questions/3568478/mysql -mysql-fetch-assoc-supplied-argument-is-not-a-valid-mysql-result-resour？rq = 1 – Marty

確保讓您連接時添加檢查以及查詢

$server = '127.0.0.1'; 
    $username = 'root'; 
    $password = ''; 
    $database = 'test'; 


mysql_connect($server, $username, $password) or die(mysql_error()); 
mysql_select_db($database) or die(mysql_error()); 

$TableName = "opportunities"; 
$Opportunities = array(); 
$SQLString = "SELECT opportunity_ID, company, city, " . 
"start_date, end_date, position, description" . " FROM $TableName;";  
$QueryResult = mysql_query($SQLString); 

if(mysql_error()) { 
    die(mysql_error(); 
} 

if($QueryResult) { 

     while($Row = mysql_fetch_assoc($QueryResult)) 
     { 
      $Opportunities[] = $Row; 
     } 
      mysql_free_result($QueryResult); 
} 

mysql_close();

而且，你釋放循環內的結果$ QueryResult中後，使下一次迭代就沒有資源從中獲取數據。

來源

2014-04-08 03:48:56

這很好。現在錯誤消失了。謝謝 – guyg

樂於協助。 –

mysql_fetch_assoc（）：6不是有效的MySQL結果資源

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