Question

我的代码有误，有人可以帮助我吗？

<?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;
  }
?>

当我打开它时，出现此错误“致命错误：在第8行的G：\ xampp \ htdocs \ Karmintalender \ test.php中的非对象上调用成员函数bind_param（）”

Answer 1

此行上的某个字段不存在，请检查它们。

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

此外，您应该检查$ 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;
     }
 }

Answer 2

应该是$ stmt-＆gt; bind_result（$ name，$ kalender_ID）;

放弃s

mysqli bind_param（）致命错误

2 个答案: