Question

在CakePHP 2.4中，为了从我的控制器自定义JSON输出的内容和结构，通过我的数据foreach的正确语法是什么？

我目前有以下方法，它创建一个直接镜像我所有帖子的Cake内部数组的.json：

public function points() {
$this->autoRender = false; // We don't render a view in this example
$this->Post->recursive = -1; //Don't return stuff we don't need

return json_encode($this->Post->find('all'));
}

这会创建一个看起来像这样的JSON，每个帖子的数据都是它自己的Post对象的子句（用回车来解决，所以你可以读取它）：

[{"Post":{"id":"1",
          "user_id":"1",
          "organism_id":"0",
          "title":"Title Text",
          "lat":"44.54401744186992",
          "lng":"-68.26070404052734",
          "body":"Body Text",
          "created":"2014-01-19 07:13:29",
          "modified":"2014-01-19 07:13:29"}
 },
 {"Post":{"id":"2",
          "user_id":"1",
          "organism_id":"0",
          "title":"Title Text",
          "lat":"44.54401744186992",
          "lng":"-68.26070404052734",
          "body":"Body Text",
          "created":"2014-01-19 07:13:29",
          "modified":"2014-01-19 07:13:29"}
 }]

这是一个问题，因为（A）出于性能原因，我可能不希望将每个帖子的所有数据都转储到JSON中，并且（B）输出到Google Maps，我需要将每个帖子的数据输出为一个Posts对象的子对象，如下所示：

{"Posts":[
    {"id":"1",
    "user_id":"1",
    "organism_id":"0",
    "title":"Title Text",
    "lat":"44.54401744186992",
    "lng":"-68.26070404052734",
    "body":"Body Text",
    "created":"2014-01-19 07:13:29",
    "modified":"2014-01-19 07:13:29"
},
    {"id":"2",
    "user_id":"1",
    "organism_id":"0",
    "title":"Title Text",
    "lat":"44.54401744186992",
    "lng":"-68.26070404052734",
    "body":"Body Text",
    "created":"2014-01-19 07:13:29",
    "modified":"2014-01-19 07:13:29"}]}

我知道我需要foreach通过数据并构建一个数组。这是如何运作的？ foreach ($posts as $post):在控制器内部无效。

Answer 1

试试这个......

$data = array();
foreach ($posts as $post) {
      $data[] = $post['Posts'];
}

$postdata['Posts'] = $data;

 echo json_encode($postdata);

Answer 2

我找到了解决方案！

在控制器中：

public function points() {
    $this->request->onlyAllow('ajax'); //Don't respond to non-AJAX requests
    $this->Post->recursive = -1; //don't return info from other controllers
    $this->set('posts', $this->Post->find('all'));
}

在/view/posts/json/points.ctp中：

foreach ($posts as $post) {
$markers[] = array(
    'id' => $post['Post']['id'],
    'lat' => $post['Post']['lat'],
    'lng' => $post['Post']['lng']
    ); //make an array of the contents of each post I want to include

}

$data = array(
'Posts' => $markers); //Make everything the child of a 'Posts' object

 echo json_encode($data); //Turn it into JSON!

Answer 3

您还可以使用Hash实用程序：

    $posts = $this->Post->find('all');
    $posts = array('Posts' => Hash::extract($posts, '{n}.Post'));
    $this->set('posts', posts);

在CakePHP中自定义JSON输出

3 个答案: