Question

以下是用户列表webservice的json响应的代码。

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class Webservice extends CI_Controller 
{
        function  list_user()
        {   
                $result_login = $this->db->get('user_registration')->result();
                $response = array();
                $response ["success"] = 1;          
                $response ["message"] = "User List.";
                foreach($result_login as $row)
                {
                        $data = array();
                        $data['User Id'] = $row->user_id;
                        $data['Name'] = $row->name;
                        $data['Email'] = $row->email;
                        $data['mobile_number'] = $row->mobile_number;
                        $data['Password'] = $row->password;
                        $output2 = json_encode(array('responsedata' => $data));
                         echo $output2;
                }

        }
}
?>

在我的代码中，如果我在json_encode中用$ response替换$ data，那么我就无法获得$ data的值。我以这种格式得到了json的回应。 JSON响应。

    {
        "responsedata": {
                  "User Id": "7",
                  "Name": "bhavin",
                  "Email": "bhavin123@gmail.com",
                  "mobile_number": "123456789",
                  "Password": "abc"
  }
}

但我想以这种格式回复json。

{
       "responsedata":
        {
        "success": 1,
        "data": [
          {
                      "User Id": "7",
                      "Name": "test",
                      "Email": "test1@gmail.com",
                      "mobile_number": "123456789",
                      "Password": "abc"
          },
          {
                      "User Id": "8",
                      "Name": "test2",
                      "Email": "test2@gmail.com",
                      "mobile_number": "123456789",
                      "Password": "abc"
          }
        ]
      }
    }

Answer 1

您需要像这样排列您的数组

我更新以下代码

$array_of_event = array()
foreach($result_login->result_array() as $row)
{
$array_of_event[] = $row;
}
    $data['success'] = "1";
    $data['data'] = $array_of_event; //need to assign event here
    $response['responsedata'] = $data;

    echo json_encode($response);

Json在Codeigniter中的回应

1 个答案: