如何在Angularjs中更改json数据 [英] How to change json data in Angularjs

问题描述

我在一个变量中将以下数据作为JSON，我在一个控制器中构建。我可以使用工厂/服务在其他控制器中访问此json数据。现在我想修改这个json数据就像输出json数据一样。

I have following data as a JSON in one variable which I am building in one controller. I can access this json data in other controller using factory/service. Now I want to modify this json data as like output json data.

输入Json

[  
   {  
      "text":"Identity",
      "checked":true,
      "timestamp":1435862483093
   },
   {  
      "text":"Calendar",
      "checked":true,
      "timestamp":1435862483443
   },
]

输出：

{  
   "myname":{  
      "Facebook":{  
         "trackdata":[  
            {  
               "text":"Identity",
               "checked":true,
               "timestamp":1435862483093
            },
            {  
               "text":"Calendar",
               "checked":true,
               "timestamp":1435862483443
            }
         ],
         "selecteddata":[  
            {  
               "text":"Identity",
               "checked":true,
               "timestamp":1435862483093
            },
            {  
               "text":"Calendar",
               "checked":true,
               "timestamp":1435862483443
            }
         ]
      }
   }
}

我在尝试什么：

        var trackdata = JSON.stringify(DataService.getTrackedData());
        var selecteddata = JSON.stringify(DataService.getSelectedData());

        var userJson = {};
        userJson["trackdata"] = trackdata;
        userJson["selecteddata"] = selecteddata;
        userJson["Facebook"] = ???
        userJson["myname"] = ???

我可以在最后一行写什么。我之所以这么说是因为未来的myname和Facebook将按用户输入。

What Can I write in last lines. The reason I put like is this in future "myname" and "Facebook" will be as per user input.

更新：2

 pmApp.controller('FooterController', function ($scope, $state, DataService) {

    $scope.infunc = function () {
        console.log("Username : " + DataService.username);
        console.log("Application Name : " + DataService.applicationName);

        var username = DataService.username;
        var applicationName = DataService.username;

        $scope.outputJson = {
            username: {
                applicationName: {
                    "trackdata": DataService.getTrackedData(),
                    "selecteddata": DataService.getSelectedData()
                }
            }
        }

        /* $scope.outputJson.myname.Facebook.trackdata = ;
         $scope.outputJson.myname.Facebook.selecteddata = DataService.getSelectedData();*/

        console.log(JSON.stringify($scope.outputJson));

    };

});

它给出了这样的输出：

 "username":{  
      "applicationName":{  
         "trackdata":[  

它应该打印那些变量的实际值，而不是username和applicationName。你能否告诉我这里做错了什么。

Instead of username and applicationName it should print actual value of those variable. Can you please tell me what I am doing wrong here.

提前致谢。

任何帮助都会是赞赏。

推荐答案

好的，我想出了一些简单的事情：

OK, here's something simple that I came up with:

// parse input in order to modify it
var inputObj = JSON.parse(input);

// create a new object based on your data from your input
var outputObj = {
    'myname': {
        'Facebook': {
            'trackdata': inputObj,
            'selecteddata': inputObj
        }
    }
};

// create new JSON output
var output = JSON.stringify(outputObj);

var input = '[     {        "text":"Identity",      "checked":true,      "timestamp":1435862483093   },   {        "text":"Calendar",      "checked":true,      "timestamp":1435862483443   }]';

var inputObj = JSON.parse(input);

var outputObj = {
  'myname': {
    'Facebook': {
      'trackdata': inputObj,
      'selecteddata': inputObj
    }
  }
};
  
var output = JSON.stringify(outputObj);
  
$('#input').html(JSON.stringify(inputObj,null,2));
$('#output').html(JSON.stringify(outputObj,null,2));

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
Input: <pre id="input"></pre><br>
Output: <pre id="output"></pre>

编辑：

我认为对于什么是 JSON.parse（）和 JSON.stringify有些困惑（）。 JSON.parse（）将json字符串作为输入并输出javascript对象。 JSON.stringify（）将javascript对象作为输入并输出字符串。在您尝试的内容中，当您可能希望将对象分配给字段时，您将字符串分配给javascript对象字段。这有帮助吗？

I think the is some confusion as to what JSON.parse() and JSON.stringify(). JSON.parse() takes a json string as input and outputs a javascript object. JSON.stringify() takes as input a javascript object and outputs a string. In what you tried, you are assigning a string to a javascript object field, when you probably want to assign an object to the field instead. Does that help?

编辑：

要完成问题中列出的示例，请执行以下操作。

To finish your example listed in the question, do the following.

var trackdata = JSON.stringify(DataService.getTrackedData());
var selecteddata = JSON.stringify(DataService.getSelectedData());

var fieldName1 = "Facebook";
var fieldName2 = "myname";

var userJson = {};
userJson["trackdata"] = trackdata;
userJson["selecteddata"] = selecteddata;
var userJson2 = {};
userJson2[fieldName1] = userJson;
var userJson3 = {};
userJson3[fieldName2] = userJson2;

注意：我不建议这样做，因为它使用更多变量，令人困惑，不容易维护。从root到child构建对象要比反之亦然，这是你的模板代码试图做的事情。

Note: I would not recommend doing it this way, as it uses more variables, is confusing, and isn't easy to maintain. Building the object from root to children is much easier than vice versa, which is what your template code was attempting to do.

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