JSON序列化 求大神

Question

[2,{"0":{"1":{"9.934":[0],"9.937":[1,4,5,7],"9.938":[2,3,8,9],"9.936":[6]},"6":{"1.983":-1},"11":{"1.983":-1},"2":{"9.934":[0],"9.937":[1,4,5,7],"9.938":[2,3,8,9],"9.936":[6]},"7":{"1.983":-1},"12":{"1.983":-1},"3":{"9.934":[0],"9.937":[1,4,5,7],"9.938":[2,3,8,9],"9.936":[6]},"8":{"1.983":-1},"13":{"1.983":-1},"4":{"9.934":[0],"9.937":[1,4,5,7],"9.938":[2,3,8,9],"9.936":[6]},"9":{"1.983":-1},"14":{"1.983":-1},"5":{"9.934":[0],"9.937":[1,4,5,7],"9.938":[2,3,8,9],"9.936":[6]},"10":{"1.983":-1},"15":{"1.983":-1}},"2":0}]这样一串JSON如何能序列化后在C#中取值

Answer 1

这好像不是一个正常的json。[2,{

Answer 2

            var jsonStr = @"[2, {
""0"": {
""1"": {
""9.934"": [0],
""9.937"": [1, 4, 5, 7],
""9.938"": [2, 3, 8, 9],
""9.936"": [6]
},
""6"": {
""1.983"": -1
},
""11"": {
""1.983"": -1
},
""2"": {
""9.934"": [0],
""9.937"": [1, 4, 5, 7],
""9.938"": [2, 3, 8, 9],
""9.936"": [6]
},
""7"": {
""1.983"": -1
},
""12"": {
""1.983"": -1
},
""3"": {
""9.934"": [0],
""9.937"": [1, 4, 5, 7],
""9.938"": [2, 3, 8, 9],
""9.936"": [6]
},
""8"": {
""1.983"": -1
},
""13"": {
""1.983"": -1
},
""4"": {
""9.934"": [0],
""9.937"": [1, 4, 5, 7],
""9.938"": [2, 3, 8, 9],
""9.936"": [6]
},
""9"": {
""1.983"": -1
},
""14"": {
""1.983"": -1
},
""5"": {
""9.934"": [0],
""9.937"": [1, 4, 5, 7],
""9.938"": [2, 3, 8, 9],
""9.936"": [6]
},
""10"": {
""1.983"": -1
},
""15"": {
""1.983"": -1
}
},
""2"": 0
}]";
            var json = JsonConvert.DeserializeObject(jsonStr) as JArray;

Answer 3

转换成json对象很简单，一句话就行
            Newtonsoft.Json.Linq.JArray o= (Newtonsoft.Json.Linq.JArray) JsonConvert.DeserializeObject(strJ);但是你想读取其中数据的话会麻烦一点，这是由这个json对象的复杂度决定的：多层嵌套的json数据，只能一步步去获取，这点比较麻烦，但是基本语法都一样，你应该能实现

Answer 4

不知道你要取这里面哪个值，最近才用到C#获取json嵌套的数组，做了个学习记录，共勉。https://blog.csdn.net/m0_37722484/article/details/90258132