在JSON中反序列化。NET可以使用包含在该库中的JObject类来实现动态。我的JSON字符串表示这些类:

public class Foo {
   public int Age {get;set;}
   public Bar Bar {get;set;}
}

public class Bar {
   public DateTime BDay {get;set;}
}

现在我们在不引用上述类的情况下反序列化字符串:

var dyn = JsonConvert.DeserializeObject<JObject>(jsonAsFooString);

JProperty propAge = dyn.Properties().FirstOrDefault(i=>i.Name == "Age");
if(propAge != null) {
    int age = int.Parse(propAge.Value.ToString());
    Console.WriteLine("age=" + age);
}

//or as a one-liner:
int myage = int.Parse(dyn.Properties().First(i=>i.Name == "Age").Value.ToString());

或者如果你想深入一点:

var propBar = dyn.Properties().FirstOrDefault(i=>i.Name == "Bar");
if(propBar != null) {
    JObject o = (JObject)propBar.First();
    var propBDay = o.Properties().FirstOrDefault (i => i.Name=="BDay");
    if(propBDay != null) {
        DateTime bday = DateTime.Parse(propBDay.Value.ToString());
        Console.WriteLine("birthday=" + bday.ToString("MM/dd/yyyy"));
    }
}

//or as a one-liner:
DateTime mybday = DateTime.Parse(((JObject)dyn.Properties().First(i=>i.Name == "Bar").First()).Properties().First(i=>i.Name == "BDay").Value.ToString());

完整的示例请参见文章。

你可以在Newtonsoft.Json的帮助下实现这一点。从NuGet安装它，然后:

using Newtonsoft.Json;

dynamic results = JsonConvert.DeserializeObject<dynamic>(YOUR_JSON);

使用DataSet(c#)和JavaScript。一个创建带有DataSet输入的JSON流的简单函数。创建JSON内容，如(多表数据集):

[[{a:1,b:2,c:3},{a:3,b:5,c:6}],[{a:23,b:45,c:35},{a:58,b:59,c:45}]]

只是客户端，使用eval。例如,

var d = eval('[[{a:1,b:2,c:3},{a:3,b:5,c:6}],[{a:23,b:45,c:35},{a:58,b:59,c:45}]]')

然后使用:

d[0][0].a // out 1 from table 0 row 0

d[1][1].b // out 59 from table 1 row 1

// Created by Behnam Mohammadi And Saeed Ahmadian
public string jsonMini(DataSet ds)
{
    int t = 0, r = 0, c = 0;
    string stream = "[";

    for (t = 0; t < ds.Tables.Count; t++)
    {
        stream += "[";
        for (r = 0; r < ds.Tables[t].Rows.Count; r++)
        {
            stream += "{";
            for (c = 0; c < ds.Tables[t].Columns.Count; c++)
            {
                stream += ds.Tables[t].Columns[c].ToString() + ":'" +
                          ds.Tables[t].Rows[r][c].ToString() + "',";
            }
            if (c>0)
                stream = stream.Substring(0, stream.Length - 1);
            stream += "},";
        }
        if (r>0)
            stream = stream.Substring(0, stream.Length - 1);
        stream += "],";
    }
    if (t>0)
        stream = stream.Substring(0, stream.Length - 1);
    stream += "];";
    return stream;
}

你想要的DynamicJSONObject对象包含在ASP. web . helpers .dll中。NET Web Pages包，它是WebMatrix的一部分。

我需要的是返回一个带有不同字段的json模型。 我的模型是这样的，但它可以改变。

{
    "employees":
    [
        { "name": "Darth", "surname": "Vader", "age": "27", "department": "finance"},
        { "name": "Luke", "surname": "Skywalker", "age": "25", "department": "IT"},
        { "name": "Han", "surname": "Solo", "age": "26", "department": "credit"}
    ]
}

获取数据值的列表

    JObject array = JObject.Parse(model.JsonData);
    var tableData = new List<JsonDynamicModel>();

    foreach (var objx in array.Descendants().OfType<JProperty>().Where(p => p.Value.Type != JTokenType.Array && p.Value.Type != JTokenType.Object))
            {
                var name = ((JValue)objx.Name).Value;
                var value = ((JValue)objx.Value).Value;
                if (tableData.FirstOrDefault(x => x.ColumnName == name.ToString()) == null)
                {
                    tableData.Add(new JsonDynamicModel
                    {
                        ColumnName = name.ToString(),
                        Values = new List<string> { value.ToString() },
                    });
                }
                else
                {
                    tableData.FirstOrDefault(x=>x.ColumnName == name.ToString()).Values.Add(value.ToString());
                }
            }

输出如下所示。然后我把结果模型转换成一个html表，我用这个方法创建了一个html表

// output
tableData[0].ColumnName -> "name";
tableData[0].Values -> {"Darth", "Luke", "Han" }
tableData[1].ColumnName -> "surname";
tableData[1].Values -> {"Vader", "Skywalker", "Solo" }
...

最简单的方法是:

只需包含这个DLL文件。

像这样使用代码:

dynamic json = new JDynamic("{a:'abc'}");
// json.a is a string "abc"

dynamic json = new JDynamic("{a:3.1416}");
// json.a is 3.1416m

dynamic json = new JDynamic("{a:1}");
// json.a is

dynamic json = new JDynamic("[1,2,3]");
/json.Length/json.Count is 3
// And you can use json[0]/ json[2] to get the elements

dynamic json = new JDynamic("{a:[1,2,3]}");
//json.a.Length /json.a.Count is 3.
// And you can use  json.a[0]/ json.a[2] to get the elements

dynamic json = new JDynamic("[{b:1},{c:1}]");
// json.Length/json.Count is 2.
// And you can use the  json[0].b/json[1].c to get the num.