需要語法parseJSON

Question

我試圖解析如下：

[{"Student":{"student_id":"19","student_name":"Test"},"Teacher":{"teacher_id":"1"}}] 

但我不知道的語法來獲取值19，測試和1，請幫助。

Answer 1

嘗試，

var test = '[{"Student":{"student_id":"19","student_name":"Test"},"Teacher":{"teacher_id":"1"}}]'; 

var obj = jQuery.parseJSON(test); 

console.log(obj[0].Teacher.teacher_id) 

http://jsfiddle.net/ZsUrN/

Answer 2

在PHP：

$json = '[{"Student":{"student_id":"19","student_name":"Test"},"Teacher":{"teacher_id":"1"}}] 
'; 

$data = json_decode($json); 

// You can access values like this: 
$student = $data[0]->Student; 
$student_id = $student->student_id; 
$student_name = $student->student_name; 
$teacher_id = $data[0]->Teacher->teacher_id; 

在Javascript中：

var json = '[{"Student":{"student_id":"19","student_name":"Test"},"Teacher":{"teacher_id":"1"}}] 
'; 

var data; 
if (window.JSON) { 
    data = JSON.parse(json); 
} else { 
    data = eval(json); 
} 

// You can access values like this: 

var student = data[0].Student; 
var student_id = student.student_id; 
var student_name = sudent.student_name; 
var teatcher_id = data[0].Teacher.teacher_id; 

Answer 3

var persons = JSON.parse('[{"Student":{"student_id":"19","student_name":"Test"},"Teacher":{"teacher_id":"1"}}]'); 
var myStudentID = persons[0].Student.student_id; 
var myStudentTest = persons[0].Student.Test; 
var myTeacherID = persons[0].Teacher.teacher_id; 

_{我得到一個奇怪的感覺，這是功課}