php - 如何从 MySQL 中获取 PHP 对象

Question

我正在尝试生成在数据库中获取的对象数组。

我尝试这样做的代码是这样的:

public function findAllObjects() {
    $conn = $this->openConnection();
    $result = $conn->query('SELECT * FROM content');
    while ($content = $result->fetch_object($class = 'Content')) {
        var_dump($content);
        $contents[] = $content;
    }
    $conn = $this->closeConection();
    return $contents;
}

如果我使用 foreach 循环内打印“Content”的值var_dump ，结果如下:

object(Content)[4]
protected 'id_content' => string '1' (length=1)
public 'title' => null
public 'description' => null
public 'category' => null
public 'date' => string '2015-01-01' (length=10)
public 'townReceiver' => null
public 'author' => null

显然，表 Content 具有完整的信息，但此代码仅获取 id 和 date。有什么建议可以解决这个问题吗？谢谢!

编辑
$class = 'Content' 是参数class_name，来自官方文档:

The name of the class to instantiate, set the properties of and return. If not specified, a stdClass object is returned.

您可以在以下位置找到更多信息:

http://php.net/manual/en/mysqli-result.fetch-object.php
这是 class Content 的属性:

class Content extends Model {

protected $id_content;
public $title;
public $description;
public $category;
public $date;
public $townReceiver;
public $author;

function __construct($title = null, $description = null, $category = null, $townReceiver = null, $author = null) {
    $this->title = $title;
    $this->description = $description;
    $this->category = $category;
    $this->townReceiver = $townReceiver;
    $this->author = $author;
}

//Methods of the class...

编辑2
问题在于 __construct 方法。如果我删除此方法，fetch_object 工作正常。 原因是什么？我不能将 __construct 与 fetch_object 一起使用吗？

编辑3
根据 Ryan Vincent 的评论，我找到的解决方案是:

function __construct($title = null, $description = null, $category = null, $townReceiver = null, $author = null) {
    if (!isset($this->id_content)) {
        $this->title = $title;
        $this->description = $description;
        $this->category = $category;
        $this->townReceiver = $townReceiver;
        $this->author = $author;
    }
}

Answer 1

最后，问题在于该对象是在调用 __construct 之前设置的。方法，因此 __construct 将对象重写为 null，如定义的那样。

为了避免这种情况，我检查对象的 id 是否已设置； 2种可能性:

1) 如果我想生成一个新内容并将其存储到数据库中，则 id 将是 null因为id在DB中被分配了自动增量选项。

2) 如果我想获取对象，将设置 id，并且数据不会被覆盖。

所以解决方案是这段代码:

function __construct($title = null, $description = null, $category = null, $townReceiver = null, $author = null) {
    if (!isset($this->id_content)) {
        $this->title = $title;
        $this->description = $description;
        $this->category = $category;
        $this->townReceiver = $townReceiver;
        $this->author = $author;
    }
}