Question

我收到错误声明：

在第16行的/home/content/26/11794426/html/practice/home.php中调用非对象的成员函数fetch（）

基于此，我相信mysql查询不是一个对象。但是，我直接从一本书中复制我的代码所以我很困惑为什么我收到错误。它是fetch（）部分本身还是查询？

另外，数据库我将sql查询复制并粘贴到phpmyadmin中，所以我不认为那里有错误，但我不排除它。

有什么想法吗？

<?php
require_once('database.php');

// Get category ID
if(!isset($category_id)) {
    $category_id = $_GET['category_id'];
    if (!isset($category_id)) {
        $category_id = 1;
    }
}

// Get name for current category
$query = "SELECT * FROM categories
          WHERE categoryID = $category_id";
$category = $db->query($query);
$category = $category->fetch();
$category_name = $category['categoryName'];

// Get all categories
$query = 'SELECT * FROM categories
          ORDER BY categoryID';
$categories = $db->query($query);

// Get products for selected category
$query = "SELECT * FROM products
          WHERE categoryID = $category_id
          ORDER BY productID";
$products = $db->query($query);

?>

这就是我的database.php文件：

<?php

 $dsn = 'mysql:host=guitarshop27.db.11794426.hostedresource.com;dbname = guitarshop27';
 $username = 'changed';
 $password = 'changed';

 try {
     $db = new PDO($dsn, $username, $password);
 }catch (PDOException $e) {
     $error_message = $e->getMessage();
     include('database_error.php');
     exit();
     }
  ?>

Answer 1

$category = $db->query($query);
$categorys = $category->fetch();
$category_name = $categorys['categoryName'];

Answer 2

您可以使用fetch_assoc（）;

进行尝试

$query = "SELECT * FROM categories
          WHERE categoryID = $category_id";
$category = $db->query($query);
$category = $category->fetch_assoc();
$category_name = $category['categoryName'];

但问题是数据库连接。在哪里定义？

Answer 3

尝试调试您的代码，并添加此添加此代码

// Get name for current category
$query = "SELECT * FROM categories
          WHERE categoryID = $category_id";
$category = $db->query($query);
echo mysql_nun_rows($category);
$categories = $category->fetch();
$category_name = $categories['categoryName'];

如果发现仍有错误，请检查数据库连接

fetch（）方法出错

3 个答案: