如何在C#中将两个不同的JSON连接成单个JSON [英] how to join two different JSON into single JSON in C#
本文介绍了如何在C#中将两个不同的JSON连接成单个JSON的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我的第一个JSON如下:
[{
用户ID:4,
姓:rupesh,
名字:ABC,
电子邮件:abc@gmail.com,
性别:男
}]
我的第二JSON如下:
[{
ACCOUNTID:2,
帐户名:rupeshinfo,
AccountDomain:空,
角色ID:1,
用户ID:4
},{
ACCOUNTID:3,
帐户名:Rameshinfo,
AccountDomain :空,
角色ID:2,
用户ID:4
}]
其结果必然是
{
的UserDetails:[{
用户ID:4,
姓:rupesh,
名字:ABC,
电子邮件:abc@gmail.com,
性别:男
}],
AccountDetails:[{
ACCOUNTID:2,
帐户名:rupeshinfo,
AccountDomain:空,
角色ID:1,
用户ID:4
},{
ACCOUNTID:3,
帐户名: Rameshinfo,
AccountDomain:空,
角色ID:2,
用户ID:4
}]
}
解决方案
如果你不想用绳子惹插入即可使用动态对象一起去(我推荐的话):
VAR的JavaScriptSerializer =新的JavaScriptSerializer();
变种为userDetails = javaScriptSerializer.DeserializeObject(json1);
VAR accountDetails = javaScriptSerializer.DeserializeObject(json2);
VAR resultJson = javaScriptSerializer.Serialize(新{=的UserDetails为userDetails,AccountDetails = accountDetails});
My first JSON is as follows
[{
"UserId": 4,
"FirstName": "rupesh",
"LastName": "Abc",
"Email": "abc@gmail.com",
"Gender": "Male"
}]
My Second JSON is as follows
[{
"AccountId": 2,
"AccountName": "rupeshinfo",
"AccountDomain": null,
"RoleId": 1,
"UserId": 4
}, {
"AccountId": 3,
"AccountName": "Rameshinfo",
"AccountDomain": null,
"RoleId": 2,
"UserId": 4
}]
the result must be
{
"UserDetails": [{
"UserId": 4,
"FirstName": "rupesh",
"LastName": "Abc",
"Email": "abc@gmail.com",
"Gender": "Male"
}],
"AccountDetails": [{
"AccountId": 2,
"AccountName": "rupeshinfo",
"AccountDomain": null,
"RoleId": 1,
"UserId": 4
}, {
"AccountId": 3,
"AccountName": "Rameshinfo",
"AccountDomain": null,
"RoleId": 2,
"UserId": 4
}]
}
解决方案
If you don't want to mess with string inserts you can go with (and I recommend so) using dynamic objects:
var javaScriptSerializer = new JavaScriptSerializer();
var userDetails = javaScriptSerializer.DeserializeObject(json1);
var accountDetails = javaScriptSerializer.DeserializeObject(json2);
var resultJson = javaScriptSerializer.Serialize(new {UserDetails = userDetails, AccountDetails = accountDetails});
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