如何在JavaScript中将Json字符串附加到另一个Json字符串

Question

我有几个使用JSON.stringify的json字符串，现在必须将所有jsonString合并为一个json。

以下是我拥有的3个json字符串

json 1= '{
   "authSeqNo" : 14,
  "flagNewEdit": "EDIT",
  "groupId": "AD0009",
  "groupName": "IT-Updated",
  "roleId": "Admin-Updated",
  "recordStatus": "A"
}'
json 2=
 {"userList": [
    {
      "userId": "x",
      "email": "x@gamail.com",
      "isDelete" : "TRUE"
    },
    {
      "userId": "y",
      "email": "y@gmail.com",
      "isDelete" : "FALSE"
    }
  ]}
json 3=
{"authMenuList": [
    {
    "menuId" : "ATHMGT",
    "viewFlag": "1",
    "createFlag": "1",
    "editFlag": "0",
    "deleteFlag": "1",
    "creditnoteFlag": "0",
    "cancelFlag": "1"
  }]}

现在必须将所有这3个加入1，我尝试了以下方式

var completeDetails = json1.concat(json2);
completeDetails=completeDetails.concat(json3);

但它没有提供所需的输出。

我的预期结果应该如下

 {
   "authSeqNo" : 14,
  "flagNewEdit": "EDIT",
  "groupId": "AD0009",
  "groupName": "IT-Updated",
  "roleId": "Admin-Updated",
  "recordStatus": "A",

  "userList": [
    {
      "userId": "x",
      "email": "x@gmail.com",
      "isDelete" : "TRUE"
    },
    {
      "userId": "y",
      "email": "y@gmail.com",
      "isDelete" : "FALSE"
    }
  ],
  "authMenuList": [
    {
    "menuId" : "ATHMGT",
    "viewFlag": "1",
    "createFlag": "1",
    "editFlag": "0",
    "deleteFlag": "1",
    "creditnoteFlag": "0",
    "cancelFlag": "1"
  }]
}

但我得到的输出是

{
    "authSeqNo": "0",
    "flagNewEdit": "NEW",
    "groupId": "TEST",
    "groupName": "GroupN",
    "roleId": "Administrator",
    "recordStatus": ""
} {
    "userList": "[{"
    userId ":"
    x ","
    email ":"
    x @v.com ","
    delete ":"
    "},  {
        "userId": "asdkl",
        "email": "x@sd.com",
        "delete": ""
    }]
"}  {
    "authMenuList[{"
    menuId ":"
    ATHMGT ","
    viewFlag ":"
    1 ","
    createFlag ":"
    1 ","
    editFlag ":"
    0 ","
    deleteFlag ":"
    1 ","
    creditnoteFlag ":"
    0 ","
    cancelFlag ":"
    1 "}]}

我是JavaScript的新手，正在学习它。 请帮我解决这个问题。

Answer 1

您可以将它们转换为JS对象并将其组合

所以：

const obj1 = JSON.parse(json1);
const obj2 = JSON.parse(json2);
const obj3 = JSON.parse(json3);

const mergedObj = Object.assign(obj1, obj2, obj3);

const jsonStr = JSON.stringify(mergedObj);

jsonStr应该结合三个JSON

---

我刚刚看到您使用JSON.stringify（）来获取字符串，因此，如果您有三个对象，则只需执行Object.assign（）部分就可以了。

Answer 2

如果要使用jQuery，则可以使用$.extend函数将所有已解析的json字符串合并为一个javascript对象，并对结果进行字符串化。 下面的代码示例。

    const object1 = JSON.parse(json_1);
    const object2 = JSON.parse(json_2);
    const object3 = JSON.parse(json_3);

    const mergedObject = $.extend({}, object1, object2, object3);

    const mergedJSON = JSON.stringify(mergedObject);

    console.log(mergedJSON);

通过传递一个空对象作为target（first）参数，可以保留两个对象。 但是，如果要合并第二个和第三个对象，则可以像$.extend(object2, object3).

在此处查看文档-jQuery.extend API

Answer 3

 const json1= `{ "authSeqNo" : 14, "flagNewEdit": "EDIT", "groupId": "AD0009", "groupName": "IT-Updated", "roleId": "Admin-Updated", "recordStatus": "A" }` const json2=` {"userList": [ { "userId": "x", "email": "x@gamail.com", "isDelete" : "TRUE" }, { "userId": "y", "email": "y@gmail.com", "isDelete" : "FALSE" } ]}` const json3=` {"authMenuList": [ { "menuId" : "ATHMGT", "viewFlag": "1", "createFlag": "1", "editFlag": "0", "deleteFlag": "1", "creditnoteFlag": "0", "cancelFlag": "1" }]}` const extendJSON = (...objs) => objs.reduce((result, current) => ({...result, ...JSON.parse(current)}), {}); console.log(extendJSON(json1, json2, json3));