Question

我在过去的六个小时里一直试图找出我的代码有什么问题。我的页面不断输出＆＃39; Catchable致命错误：PDOStatement类的对象无法转换为字符串。＆＃39;我不知道为什么！我看了以前的作业，尝试使用谷歌搜索错误的类型，没有任何改变输出。我想按类别ID显示我的菜单项。这是我的代码。

class Menu {
    public $conn;
    public function __construct() {
        $db = new Database();
        $this->conn = $db->conn;
    }

    public function __destruct() {
        $this-> conn = null;
    }

    /* Get function */
    public function __get($name) {
        return $this->$name;
    } // End get function

    public function __set($name, $value){
        $this->$name=$value;
    }

    public function menu_items($category_id = 0){
        try {
            $sql = "SELECT item_id, category, category_id, display_order, item_name, item_image, item_description, item_cost FROM menu_items WHERE category_id = $category_id";
          $result = $this->conn->query($sql);

            return $result;
        }
        catch (PDOException $e){
            echo 'Error: ' . $e->getMessage();
            exit();
        }
    }
}

更新：这是一切输出的地方。这个代码可能有问题吗？

include(ABSOLUTE_PATH . 'classes/menu.class.php');

// Create Menu object
$menu = new Menu();
$menu_categories = $menu->menu_categories();

include(ABSOLUTE_PATH . '_includes/header.inc.php');
?>

<hr />

<h2><?=$page?></h2>

<?
while($item = $menu_categories->fetch(PDO::FETCH_OBJ)) {
// Retrieve menu items
$menu_items = $menu->menu_items($menu_categories);
$item_count = $menu_items->rowCount();

if($item_count > 0) {
    echo '<h3>' . $item->category . '</h3>';
?>

    <table id="menu_items" class="listing">
    <?
        // Loop through menu records
        while($item = $menu_items->fetch(PDO::FETCH_OBJ)) {
            echo "\t<tr>\n";
            echo "\t\t" . '<td class="item"><h4>' . $item->item_name .     '</h4><p>' . $item->item_description . '</p></td>' . "\n";
            echo "\t\t" . '<td class="price">$' . $item->item_cost . '</td>'   . "\n";
            echo "\t\t" . '<td class="image"><img src="' . URL_ROOT . '_assets/images/menu/' . $item->item_image . '" alt="' . $item->item_name . '" /></td>' . "\n";
            echo "\t</tr>\n";
        }
    ?>
    </table>

<?
}}
?>

<hr />

<?
include(ABSOLUTE_PATH . '_includes/footer.inc.php');
?>

Answer 1

阿哈

这是你的问题：

while($item = $menu_categories->fetch(PDO::FETCH_OBJ)) {
// Retrieve menu items
$menu_items = $menu->menu_items($menu_categories); //<--Error
$item_count = $menu_items->rowCount();

你应该致电：

$menu_items = $menu->menu_items($item->id);

您正在将$menu_categories（PDOStatement）传递到menu_items()，然后尝试将其直接用作查询中的字符串。这就是错误实际发生的地方。

使此错误更明显的一个好习惯是键入check query parameters。所以在menu_items()：

public function menu_items($category_id = 0){

 if(!is_numeric($category_id)) {
    //Do something better here, but just for example
    echo "Error";
    return false;
  }  
  try {
     $sql = "SELECT item_id, category, category_id, display_order, item_name, item_image, item_description, item_cost FROM menu_items WHERE category_id = $category_id";
     $result = $this->conn->query($sql);

     return $result;
   ...

PDOStatement类的对象无法转换为字符串（帮助）

1 个答案: