Cakephp3:如何返回json数据?
发布时间:2020-12-13 18:24:23 所属栏目:PHP教程 来源:网络整理
导读:我正在调用cakePhp控制器的ajax调用: $.ajax({ type: "POST",url: 'locations/add',data: { abbreviation: $(jqInputs[0]).val(),description: $(jqInputs[1]).val() },success: function (response) { if(response.status === "success") { // do somethin
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我正在调用cakePhp控制器的ajax调用:
$.ajax({
type: "POST",url: 'locations/add',data: {
abbreviation: $(jqInputs[0]).val(),description: $(jqInputs[1]).val()
},success: function (response) {
if(response.status === "success") {
// do something with response.message or whatever other data on success
console.log('success');
} else if(response.status === "error") {
// do something with response.message or whatever other data on error
console.log('error');
}
}
});
当我尝试这个时,我收到以下错误消息:
在AppController中我有这个 $this->loadComponent('RequestHandler');
启用. Controller函数如下所示: public function add()
{
$this->autoRender = false; // avoid to render view
$location = $this->Locations->newEntity();
if ($this->request->is('post')) {
$location = $this->Locations->patchEntity($location,$this->request->data);
if ($this->Locations->save($location)) {
//$this->Flash->success(__('The location has been saved.'));
//return $this->redirect(['action' => 'index']);
return json_encode(array('result' => 'success'));
} else {
//$this->Flash->error(__('The location could not be saved. Please,try again.'));
return json_encode(array('result' => 'error'));
}
}
$this->set(compact('location'));
$this->set('_serialize',['location']);
}
我在这里想念什么?是否需要其他设置?
我在这里看到的大多数答案要么过时,要么重载不必要的信息,要么依赖于withBody(),感觉解决方法而不是CakePHP方式.
以下是对我有用的东西: $my_results = ['foo'=>'bar'];
$this->set([
'my_response' => $my_results,'_serialize' => 'my_response',]);
$this->RequestHandler->renderAs($this,'json');
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