How to define multi-line strings in C#
Add the @ symbol in front of the definition:
1 string aa = @"asdfsdfsd
2 fsdsfsdfsdfsdfsdfsdfs
3 safasfsadfsdfasfsfsdfsd ";
There are several ways to concatenate strings in C#
1. Use the JsonConvert.SerializeObject method (Nuget to obtain Newtonsoft.Json Package), which requires Newtonsoft.Json support.
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. Use 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. Directly concatenate strings
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. Use StringFormat
string mc = "22";
string id = "11";
string serJson = string.Format("[{
{ AccessKey:\"{0}\",mc:\"{1}\"}},{
{ AccessKey:\"{2}\",mc:\"{3}\"}}]", id, mc, "33", "44");
Analysis of Jobject data structure:
First download Newtonsoft.Json and add reference using Newtonsoft.Json.Linq;
Extract the contents of jobject,
//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
}]
}
}
}
It can be parsed in the online JSON verification and formatting tool on the json official website .
code show as below:
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