Question

我有这个PHP代码返回一些json

public function my_account(){
             header("Access-Control-Allow-Origin: *");
           header("Access-Control-Allow-Methods: PUT, GET, POST");
           header("Access-Control-Allow-Headers: Origin, X-Requested-With, Content-Type, Accept");
           $email_posted = $this->input->post('email');
           $email = preg_replace('/\s+/', '', $email_posted);

            $where = "email='$email'";

            $this->db->where($where);


            $query = $this->db->get('users');

            if ($query->num_rows() > 0)
            {
               foreach ($query->result() as $row)
               {
                  $json = array("id"=>$row->id, "email"=>$row->email,"names"=>$row->names,"country"=>$row->country,"password"=>$row->password,"telephone"=>$row->telephone);
                  echo json_encode($json);
               }
            }

         }

返回的json是这种格式

{"id":"15","email":"corn64@gmail.com","names":"Cern 64","country":"","password":"cern!768","telephone":"00000"}

获取jquery中的数据

$.each(data, function(i, obj) {
        alert(obj.email);
});

并且有这个小提琴https://jsfiddle.net/c5b4o6yh/2/

我收到错误

Uncaught TypeError: Cannot use 'in' operator to search for 'length' in

{ “ID”： “15”， “电子邮件”： “corn64@gmail.com”， “名称”：“欧洲核子研究中心 64" ， “国家”： “”， “密码”： “CERN 768！”， “电话”： “00000”}

使用已知密钥获取值的正确方法是什么？。

Answer 1

这肯定是一个问题： -

1.您在$json

中覆盖了foreach()变量

2.您只在foreach()内发送单一回复。

更改以下代码（更改已注释）： -

public function my_account(){
    header("Access-Control-Allow-Origin: *");
    header("Access-Control-Allow-Methods: PUT, GET, POST");
    header("Access-Control-Allow-Headers: Origin, X-Requested-With, Content-Type, Accept");
    $email_posted = $this->input->post('email');
    $email = preg_replace('/\s+/', '', $email_posted);

    $where = "email='$email'";

    $this->db->where($where);


    $query = $this->db->get('users');
    $json = array(); //define variable as an array variable
    if ($query->num_rows() > 0)
    {
       foreach ($query->result() as $row)
       {
          $json []= array("id"=>$row->id, "email"=>$row->email,"names"=>$row->names,"country"=>$row->country,"password"=>$row->password,"telephone"=>$row->telephone);//assign values to array

       }
    }
    echo json_encode($json); // send response after loop completion
 }

现在您的原始脚本代码将起作用。

Answer 2

首先，将值保存在数组中。（（使用array_push（）函数可以将一个或多个元素插入到数组的末尾））

dlmwrite('/Users/amar/Desktop/geenrated/rate.txt', [newTime intHeartRate intRespRate]);

Answer 3

在这种情况下，您的响应应该是数组，而不是简单的对象。因此，尝试重新编写代码以返回

[{"id":"15","email":"corn64@gmail.com","names":"Cern 64","country":"","password":"cern!768","telephone":"00000"}]

这jsfiddle是正确的。

Answer 4

使用JSON.parse(data)将JSON字符串从您的端点转换为JS对象，然后获取各个键值：

var obj = JSON.parse(data);
var id = obj.id;

使用已知密钥获取返回的json的值

4 个答案: