如何操作嵌套的Json数据以在java中去嵌套

I have been looking solution for this problem but could not find one so asking this question.
I have some data which looks like this

我一直在寻找解决这个问题的办法，但是没有找到，所以就问这个问题。我有一些像这样的数据

{
       "data": [
         {
            "id": "5ab892c71810e201e81b9d39",
            "isSignedUpUsingFb": false,
            "personalInformation": {
                "firstName": "jio",
                "lastName": "g",
                "mobileNumber": "1234567890",
             },
            "accountBalance": 0,
           }
         ]
},

I want to write a java code to change the data structure to this

我想编写一个java代码将数据结构更改为这个

{
           "data": [
              {
                "id": "5ab892c71810e201e81b9d39",
                "isSignedUpUsingFb": false,
                "personalInformation_firstName":"jio",
                 "personalInformation_lastNAme":"g",
                 "personalInformation_mobileNumber":"1234567890",

                "accountBalance": 0,
               }
             ]
},

I am getting data from db as:

我从db得到的数据是:

       @Override
        public List<User> getAllUsers() {
            logger.debug("entering all users method");
            List<User> allUsers=mongoOperations.findAll(User.class);
            for (User user : allUsers) {
                PersonalInformation info=user.getPersonalInformation());
                //manipulation code here
                user.setPersonalInformation(info);

            }
            return allUsers;
        }

So I want to write a logic so that i can convert the data in desired format and send it a return type. I know how to do same thing using J query but I want to do it in backend so any code for the above or any link will help.

所以我想写一个逻辑，这样我就可以把数据转换成想要的格式并发送一个返回类型。我知道如何使用J查询做同样的事情，但是我想在后端做它，所以上面的任何代码或任何链接都会有所帮助。

2 个解决方案

#1

I have fond one solution which is very simple.So, basically when we create object for nested data we create it like this in JAVA.

我有一个很简单的解。基本上，当我们为嵌套数据创建对象时我们在JAVA中创建它。

public MyClass{

public String name;
public String contact;
public PersonalInformation personalinformation;
//setters and getter here
}

this will give me data as

这会给我数据a

"MyClass":{
"name": "abc",
"contact": "12345",
"personalInformation":{
"address": "asdasdasdad",
"city":"asdadad",
"pin": "asdfg",
}
}

so to remove this nested data we need to use @JsonUnwrapped which removes all the nested object and add it to our main object.

因此，要删除这个嵌套数据，我们需要使用@ jsonunwrap，它将删除所有嵌套对象并将其添加到我们的主对象中。

public MyClass{

    public String name;
    public String contact;
    @JsonUnwrapped
    public PersonalInformation personalinformation;
    //setters and getter here
    }

which will change the data structure as:

将改变数据结构为:

"MyClass":{
    "name": "abc",
    "contact": "12345",
    "address": "asdasdasdad",
    "city":"asdadad",
    "pin": "asdfg",
    }

for more reference you can check this link http://fasterxml.github.io/jackson-annotations/javadoc/2.0.0/com/fasterxml/jackson/annotation/JsonUnwrapped.html
Hope this helps.

要获得更多的引用，可以查看这个链接http://fasterxml.github.io/jackson- annotations/javadoc/2.0.0/com/fasterxml/jackson/annotation/jsonunwrap。html希望这有助于。

#2

There are multiple possible solutions. As Prabhav has mentioned the most intuitive one would be to create a new class and from there a object which can be transformed with a library to a JSON.

有多种可能的解决方案。正如Prabhav所提到的，最直观的方法是创建一个新类，然后从这个类创建一个对象，该对象可以用库转换为JSON。

Variant one:

变体:

The new class would look like your data structure you want and access would be:

新的类看起来像您想要的数据结构，访问的对象是:

PersonalInformationJson pf = new PersonalInformationJson();
pf.setFirstName = info.getPersonalInformation_firstName
//... setting the rest of the object
//using jackson
ObjectMapper mapper = new ObjectMapper();
try {
    // convert user object to json string and return it 
   String jsonString = mapper.writeValueAsString(u);
}

The other easier version to create a string, either per hand or use a lib:

另一个更简单的版本是创建一个字符串，或者是用手或者是使用lib:

// using org.json.JSONObject
  String jsonString = new JSONObject().put("personalInformation_firstName", info.value())
            .put("personalInformation_lastNAme", info.value());

#1