drdl18946 2011-07-05 12:18
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mysqli bind_param()致命错误

I Have an Error at my Code could someone help me?

<?php
  $db = new mysqli("localhost","root","","karmintalender");

  $owner_ID = 1;

  $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
  $stmt = $db->prepare($sql);
  $stmt->bind_param("i", $owner_ID);
  $stmt->execute();
  $stmt->bind_results($name, $kalender_ID);

  while ($stmt->fetch()) {
    echo $name . " " . $kalender_ID;
  }
?>

When I open it this error appears "Fatal error: Call to a member function bind_param() on a non-object in G:\xampp\htdocs\Karmintalender\test.php on line 8"

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2条回答 默认 最新

  • drtohng5613 2011-07-05 12:23
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    One of your fields on this line doesn't exist,check them.

    $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
    

    Also, you should be checking for $stmt.

    $db = new mysqli("localhost","root","","karmintalender");
    
     $owner_ID = 1;
    
     $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
     $stmt = $db->prepare($sql);
     if($stmt){
         $stmt->bind_param("i", $owner_ID);
         $stmt->execute();
         $stmt->bind_results($name, $kalender_ID);
    
         while ($stmt->fetch()) {
           echo $name . " " . $kalender_ID;
         }
     }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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