如何从CakePHP 2.2控制器返回JSON?
我正在调用一个控制器函数:如何从CakePHP 2.2控制器返回JSON?,php,json,cakephp,controller,cakephp-2.2,Php,Json,Cakephp,Controller,Cakephp 2.2,我正在调用一个控制器函数: $.get("http://localhost/universityapp/courses/listnames", function(data){ alert("Data Loaded: " + data); }); 在我的控制器中: public function listnames() { $data = Array( "name" => "Sergio", "age" => 23 );
$.get("http://localhost/universityapp/courses/listnames", function(data){
alert("Data Loaded: " + data);
});
在我的控制器中:
public function listnames() {
$data = Array(
"name" => "Sergio",
"age" => 23
);
$this->set('test', $data);
$this->render('/Elements/ajaxreturn'); // This View is declared at /Elements/ajaxreturn.ctp
}
因此,我认为:
<?php echo json_encode($asdf); ?>
但是,该操作将返回整个页面,包括布局内容(页眉、页脚、导航)
我错过了什么?如何在没有布局内容的情况下仅返回JSON数据?您需要像这样禁用布局
$this->layout = null ;
现在,你的行动将变得
public function listnames() {
$this->layout = null ;
$data = Array(
"name" => "Sergio",
"age" => 23
);
$this->set('test', $data);
$this->render('/Elements/ajaxreturn'); // This View is declared at /Elements/ajaxreturn.ctp
}
阅读手册上的相关内容。设置
autoRender=false
并返回json\u encode($code)
:-
您可以尝试以下任何一种方法来返回json响应(我在这里使用了failure case来返回json响应): 或
public function action() {
$this->layout = false;
$this->autoRender = false;
return json_encode(array(
'success' => 0,
'message' => 'Invalid request.'
));
}
设置$this->layout=null;在里面listnames@MoyedAnsari:如果你把它写下来作为你的答案,我会接受它作为解决方案。我希望这对u+1有效,这是这个简单情况下最简单的解决方案。
public function action() {
$this->response->body(json_encode(array(
'success' => 0,
'message' => 'Invalid request.'
)));
$this->response->send();
$this->_stop();
}
public function action() {
$this->layout = false;
$this->autoRender = false;
return json_encode(array(
'success' => 0,
'message' => 'Invalid request.'
));
}