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mysqli bind_param() fatal error

开发者 https://www.devze.com 2023-03-17 15:22 出处:网络
I Have an Error at my Code could someone help me? <?php $db = new mysqli(\"localhost\",\"root\",\"\",\"karmintalender\");

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", $开发者_如何学Pythonowner_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"


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;
     }
 }


it should be $stmt->bind_result($name, $kalender_ID);

drop the s

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