C# 拼接Json串的几种方法与解析

C#定义多行字符串的方式

在定义的前面加上@符号：

1             string aa = @"asdfsdfsd
2                         fsdsfsdfsdfsdfsdfsdfs
3                         safasfsadfsdfasfsfsdfsd ";

在C#中拼接字符串有几种方法

1. 利用  JsonConvert.SerializeObject方法 (Nuget获取Newtonsoft.Json Package)，需要Newtonsoft.Json 支持。

string uid = "22";

var abcObject = new
{
  AccessKey = 11,
  CustomerNo = uid,
  mc = "33",
  qd = "44",
  mr = "55",
  insertDate = DateTime.Now
};
  string serJson = JsonConvert.SerializeObject(abcObject);

2. 利用StringBuilder

StringBuilder str = new StringBuilder();
            str.Append("{");
            str.Append("AccessKey:\"" + 11 + "\",");
            str.Append("mc:\"" + 22 + "\",");
            str.Append("qd:\"" + 33 + "\"");
            str.Append("}");
            string serJson = str.ToString();

3. 直接拼接字符串

1、

string json = "{\"speed\":" + speed + "," + "\"direction\":" + direction + "}";

TODO：输出
{
    "speed": 591,
    "direction": 0
}

"{\"Bool_Type\":\"Bool\",\"Int_Type\":6666666,\"Float_Type\": 66.99,\"String_Type\":\"这是String类型\",\"Vector2_Type\":{\"x\":666.0,\"y\":666.0},\"Vector3_Type\":{\"x\":666.0,\"y\":666.0,\"z\":666.0}}";

4. 利用StringFormat

string mc = "22";
string id = "11";
string serJson = string.Format("[{{ AccessKey:\"{0}\",mc:\"{1}\"}},{{ AccessKey:\"{2}\",mc:\"{3}\"}}]", id, mc, "33", "44");

Jobject 数据结构的解析：

首先下载Newtonsoft.Json，增加引用using Newtonsoft.Json.Linq;

把jobject的内容提取出来，

//Jobject的内容格式如下:
{
 "code": 200,
 "msg": "SUCCESS",
 "data": {
  "id": "12345678",
  "name": "张三",
  "sex": "男",
  "result": {
   "access_token": "49d58eacd7811e463429a1ae10b42173",
   "user_info": [{
    "school": "社会大学",
    "major": "软件开发",
    "education": "本科",
    "score": 97
   }, {
    "school": "湖南大学",
    "major": "软件工程",
    "education": "研究生",
    "score": 100
   }]
  }
 }
}

可放到json官网在线JSON校验格式化工具里解析。

代码如下：

1，新建类：
        public class UserInfo
        {
            public string id { get; set; }
            public string name { get; set; }
            public string sex { get; set; }
            public string access_token { get; set; }
            public string school { get; set; }
            public string major { get; set; }
            public string education { get; set; }
            public string score { get; set; }
        }


2，获取值：
            JObject result = new JObject();//假设result为数据结构
            UserInfo userinfo = new UserInfo();
            userinfo.id = result["data"].Value<string>("id");//id
            userinfo.name = result["data"].Value<string>("name"); //name
            userinfo.sex = result["data"].Value<string>("sex"); //sex
            userinfo.access_token= result["data"]["result"]["access_token"].ToString();//access_token
            JArray res = result["data"]["result"].Value<JArray>("user_info");
            JObject obj = JObject.Parse(res[0].ToString());//只获取数据结构中第一个userinfo里的数据信息
            userinfo.school = obj.Value<string>("school"); //schoool
            userinfo.major = obj.Value<string>("major");//major
            userinfo.education = obj.Value<string>("education");//education
            userinfo.score= obj.Value<string>("score");//score