php 如何从 CakePHP 2.2 控制器返回 JSON？
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How to return JSON from a CakePHP 2.2 controller?
提问by sergserg
I'm invoking a controller function:
我正在调用一个控制器函数：
$.get("http://localhost/universityapp/courses/listnames", function(data){
alert("Data Loaded: " + data);
});
And in my Controller:
在我的控制器中：
public function listnames() {
$data = Array(
"name" => "Sergio",
"age" => 23
);
$this->set('test', $data);
$this->render('/Elements/ajaxreturn'); // This View is declared at /Elements/ajaxreturn.ctp
}
And in that View:
在该视图中：
<?php echo json_encode($asdf); ?>
However, the Action is returning the entire page including the Layout content (header, footer, navigation).
但是，Action 将返回整个页面，包括布局内容（页眉、页脚、导航）。
What am I missing here? How can I return just the JSON data without the Layout content?
我在这里缺少什么？如何仅返回 JSON 数据而没有 Layout 内容？
回答by Guilherme Ferreira
Set autoRender=falseand return json_encode($code):-
设置autoRender=false并返回json_encode($code)：-
public function returningJsonData($estado_id){
$this->autoRender = false;
return json_encode($this->ModelBla->find('first',array(
'conditions'=>array('Bla.bla_child_id'=>$estado_id)
)));
}
回答by Moyed Ansari
You need to disable layout like this
您需要像这样禁用布局
$this->layout = null ;
Now your action will become
现在你的行动将变成
public function listnames() {
$this->layout = null ;
$data = Array(
"name" => "Sergio",
"age" => 23
);
$this->set('test', $data);
$this->render('/Elements/ajaxreturn'); // This View is declared at /Elements/ajaxreturn.ctp
}
回答by Anil kumar
You can try any of the following to return json response (I've taken failure case here to return json response) :
您可以尝试以下任何一种方法来返回 json 响应（我在这里采用了失败案例来返回 json 响应）：
public function action() {
$this->response->body(json_encode(array(
'success' => 0,
'message' => 'Invalid request.'
)));
$this->response->send();
$this->_stop();
}
OR
或者
public function action() {
$this->layout = false;
$this->autoRender = false;
return json_encode(array(
'success' => 0,
'message' => 'Invalid request.'
));
}
