CodeIgniter - 如何从控制器返回 Json 响应

php javascript codeigniter

2022-08-30 12:31:27

如何将来自控制器的响应返回到Jquery Javascript？

Javascript

$('.signinform').submit(function() { 
   $(this).ajaxSubmit({ 
       type : "POST",
       url: 'index.php/user/signin', // target element(s) to be updated with server response 
       cache : false,
       success : onSuccessRegistered,
       error: onFailRegistered
   });        
   return false; 
});

数据返回空（空白）！

function onSuccessRegistered(data){
    alert(data);
};

控制器 -

public function signin() {
    $arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);    
    echo json_encode( $arr );
}

答案 1

return $this->output
            ->set_content_type('application/json')
            ->set_status_header(500)
            ->set_output(json_encode(array(
                    'text' => 'Error 500',
                    'type' => 'danger'
            )));

答案 2

//do the edit in your javascript

$('.signinform').submit(function() { 
   $(this).ajaxSubmit({ 
       type : "POST",
       //set the data type
       dataType:'json',
       url: 'index.php/user/signin', // target element(s) to be updated with server response 
       cache : false,
       //check this in Firefox browser
       success : function(response){ console.log(response); alert(response)},
       error: onFailRegistered
   });        
   return false; 
}); 


//controller function

public function signin() {
    $arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);    

   //add the header here
    header('Content-Type: application/json');
    echo json_encode( $arr );
}