Question

我有以下格式的JSON对象：

temp:[
        {
           test:'test 1',
           testData:  [ 
                       {testName: 'do',testId:''}
                         ],
           testRcd:'value'                             
        },
        {
            test:'test 2',
           testData:  [
                            {testName: 'do1',testId:''}
                         ],
           testRcd:'value'                           
        }
      ],

如何在jquery中为上述格式创建JSON对象。我想创建一个动态JSON对象。

Answer 1

只需将您的数据放入这样的对象：

var myObject = new Object();
myObject.name = "John";
myObject.age = 12;
myObject.pets = ["cat", "dog"];

然后通过：

进行字符串化

var myString = JSON.stringify(myObject);

你不需要jQuery。这是纯粹的JS。

Answer 2

“JSON对象”没有意义：JSON是基于Javascript对象声明结构的交换格式。

如果要将javascript对象转换为json字符串，请使用JSON.stringify(yourObject);

如果你想创建一个javascript对象，只需这样做：

var yourObject = {
          test:'test 1',
          testData: [ 
                {testName: 'do',testId:''}
          ],
          testRcd:'value'   
};

Answer 3

我相信他要求将新的json写入目录。你需要一些Javascript 和 PHP。所以，撇开其他答案：

的script.js

var yourObject = {
  test:'test 1',
  testData: [ 
    {testName: 'do',testId:''}
   ],
   testRcd:'value'   
};
var myString = 'newData='+JSON.stringify(yourObject);  //converts json to string and prepends the POST variable name
$.ajax({
   type: "POST",
   url: "buildJson.php", //the name and location of your php file
   data: myString,      //add the converted json string to a document.
   success: function() {alert('sucess');} //just to make sure it got to this point.
});
return false;  //prevents the page from reloading. this helps if you want to bind this whole process to a click event.

buildJson.php

<?php
    $file = "data.json";  //name and location of json file. if the file doesn't exist, it   will be created with this name

    $fh = fopen($file, 'a');  //'a' will append the data to the end of the file. there are other arguemnts for fopen that might help you a little more. google 'fopen php'.

    $new_data = $_POST["newData"]; //put POST data from ajax request in a variable

    fwrite($fh, $new_data);  //write the data with fwrite

    fclose($fh);  //close the dile
?>

Answer 4

如何获取像json这样的追加输入字段值

temp:[
        {
           test:'test 1',
           testData:  [ 
                       {testName: 'do',testId:''}
                         ],
           testRcd:'value'                             
        },
        {
            test:'test 2',
           testData:  [
                            {testName: 'do1',testId:''}
                         ],
           testRcd:'value'                           
        }
      ],

Answer 5

嵌套JSON对象

var data = {
        view:{
            type: 'success', note:'Updated successfully',
        },
    };

您可以解析此data.view.type和data.view.note

JSON对象和内部数组

var data = {
          view: [ 
                {type: 'success', note:'updated successfully'}
          ],  
     };

您可以解析此data.view[0].type和data.view[0].note

Answer 6

var model = {"Id": "xx", "Name":"Ravi"};
$.ajax({    url: 'test/set',
                        type: "POST",
                        data: model,
                        success: function (res) {
                            if (res != null) {
                                alert("done.");
                            }
                        },
                        error: function (res) {

                        }
                    });

如何使用jQuery创建JSON对象

6 个答案: