问题描述
我想将查询结果提取到一个类中(放入一个类实例的数组中).但是我收到以下错误消息:致命错误:在...中找不到类'Category'这是我的数据库管理器类中涉及的两个函数的代码:
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;
}
这是一个在类中调用它们的测试函数:
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");
}
这就是我调用测试函数的方式:
This is how i call test function:
$comment = CommentQuery::getInstance();
$result = $comment->test();
var_dump($result); // should be an array with instances of Category
这是应将其提取到的类:
And this is the class where it should be fetched into:
class Category {
private $id;
private $name;
private $projectId;
// getter and setter...
}
一些其他信息:
- 我使用自动加载器来包含我的课程.
- 我使用名称空间
- 是的,可以在所有三个函数中创建该类的实例,所以
类将被包括在内并使用命名空间 - $ _ Mode == 1可以正常工作
- I use an autoloader to include my classes.
- I use namespaces
- yes, it is possible to create an instance of the class in all of the three functions, so
class would be included and namespace used - $_Mode == 1 works fine
有什么想法吗?
推荐答案
如果您的Category
类位于名称空间中,则需要将完全限定的类名称传递给fetchAll
.
If your Category
class is in a namespace, you'll need to pass in a fully qualified class name into fetchAll
.
现在,PDO正在尝试获取根名称空间中的类Category
.它不存在.您需要告诉PDO有关名称空间的信息:
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(如果这样更容易(并且正确)):
Or use a __NAMESPACE__
constant if that makes it easier (and is correct):
$stm->fetchAll(\PDO::FETCH_CLASS, __NAMESPACE__ . '\\Category');
或者甚至更好的是,使用PHP 5.5+的::class
常量来生成完全限定的类名.
Or, even better, use PHP 5.5+'s ::class
constant to ge the fully qualified class name.
use Acme\Package\Category;
$stm->fetchAll(\PDO::FETCH_CLASS, Category::class);
这篇关于PHP-将准备好的stmt提取到类中:致命错误“未找到类".的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!