問題描述

我想將查詢的結果提取到一個類中(到一個類的實例數組中).但我收到以下錯誤消息:致命錯誤:在...中找不到類類別"這是我的數據庫管理器類中涉及的兩個函數的代碼:

I want to fetch the result of a query into a class (into an array of instances of a class). But I get the following error message: Fatal error: Class 'Category' not found in ... This is the code of the two functions in my database manager class that are involved:

public function prepareStatement($_Statement)
{
    $this->preparedStmt = $this->pdo->prepare($_Statement);

    if($this->preparedStmt === FALSE)
        throw new PreparedStmtException ("Fehler: Statement konnte nicht prepared werden.");
    else
        return TRUE;
}


public function execute($_Params = array(), $_FetchMode = NULL, $_Class = NULL)
{
    # Cancel execution if no statement prepared
    if($this->preparedStmt === null) 
        throw new PreparedStmtException ("Fehler: Statement wurde vor execute nicht prepared.");

    try
    {
        # Execute PDO call with params
        $this->preparedStmt->execute($_Params);

        # If no data is returned throw NoDataException
        if($this->preparedStmt->columnCount() == 0)
            throw new NoDataException;

        // else
        // Determine which fetch mode should be called, if NULL or something != 1 || != 0 just call
        // fetchAll without params
        if ($_FetchMode == 1)
             $result = $this->preparedStmt->fetchAll(PDO::FETCH_ASSOC); 

        else if ($_FetchMode == 2)
             $result = $this->preparedStmt->fetchAll(PDO::FETCH_CLASS, $_Class); 

        else 
            $result = $this->preparedStmt->fetchAll();
    }
    catch (PDOException $e)
    {
        # Errormanagement --> Message im live Betrieb rausnehmen
        echo '<div style="color: red;">'.$e->getMessage().'</div>';
        $result = FALSE;
    }

    // If result is null throw Instance Exception, if result emtpy throw NoDataException
    if ($result == null)
        throw new InstanceException;
    else if (empty($result))
        throw new NoDataException;

    return $result;
}

這是一個類中的測試函數來調用它們:

This is a test function in a class to call them:

public function test () 
{
    $stmt = "SELECT * FROM tx_exhibition_category WHERE uid = 1 OR uid = 2";
    $this->mysql->prepareStatement($stmt);
    return $this->mysql->execute (array(), 2, "Category");
}

這就是我調用測試函數的方式:

This is how i call test function:

$comment = CommentQuery::getInstance();
$result = $comment->test();
var_dump($result); // should be an array with instances of Category

這是它應該被提取到的類:

And this is the class where it should be fetched into:

class Category {

private $id;
private $name;
private $projectId;

// getter and setter...
}

一些附加信息:

• 我使用自動加載器來包含我的類.
• 我使用命名空間
• 是的，可以在所有三個函數中創建類的實例，所以
  將包含類并使用命名空間
• $_Mode == 1 工作正常

有什么想法嗎?

推薦答案

如果您的 Category 類在命名空間中，您需要將完全限定的類名傳入 fetchAll.

If your Category class is in a namespace, you'll need to pass in a fully qualified class name into fetchAll.

現在，PDO 正在嘗試獲取根命名空間中的 Category 類.它不存在.你需要告訴 PDO 關于命名空間:

Right now, PDO is trying to fetch into the class Category in the root namespace. It doesn't exist. You need to tell PDO about the namespace:

$stm->fetchAll(PDO::FETCH_CLASS, 'Vendor\Package\Category');

或者使用 __NAMESPACE__ 常量，如果這樣更容易(并且正確):

Or use a __NAMESPACE__ constant if that makes it easier (and is correct):

$stm->fetchAll(PDO::FETCH_CLASS, __NAMESPACE__ . '\Category');

或者，更好的是，使用 PHP 5.5+ 的 ::class 常量來獲取完全限定的類名.

Or, even better, use PHP 5.5+'s ::class constant to ge the fully qualified class name.

use AcmePackageCategory;
$stm->fetchAll(PDO::FETCH_CLASS, Category::class);

這篇關于PHP - 將準備好的 stmt 提取到類中:致命錯誤“找不到類"的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！