是否有方法将JSON内容反序列化为c#动态类型?为了使用DataContractJsonSerializer,最好跳过创建一堆类。


当前回答

使用Cinchoo ETL -一个开源库,可将JSON解析为动态对象:

string json = @"{
    ""key1"": [
        {
            ""action"": ""open"",
            ""timestamp"": ""2018-09-05 20:46:00"",
            ""url"": null,
            ""ip"": ""66.102.6.98""
        }
    ]
}";
using (var p = ChoJSONReader.LoadText(json)
    .WithJSONPath("$..key1")
    )
{
    foreach (var rec in p)
    {
        Console.WriteLine("Action: " + rec.action);
        Console.WriteLine("Timestamp: " + rec.timestamp);
        Console.WriteLine("URL: " + rec.url);
        Console.WriteLine("IP address: " + rec.ip);
    }
}

输出:

Action: open
Timestamp: 2018-09-05 20:46:00
URL: http://www.google.com
IP address: 66.102.6.98

样本提琴:https://dotnetfiddle.net/S0ehSV

有关更多信息,请访问codeproject文章

声明:我是这个库的作者。

其他回答

如何解析简单的JSON内容与动态& JavaScriptSerializer

请添加System.Web.Extensions的引用,并使用System.Web.Script.Serialization添加此命名空间;在前:

public static void EasyJson()
{
    var jsonText = @"{
        ""some_number"": 108.541,
        ""date_time"": ""2011-04-13T15:34:09Z"",
        ""serial_number"": ""SN1234""
    }";

    var jss = new JavaScriptSerializer();
    var dict = jss.Deserialize<dynamic>(jsonText);

    Console.WriteLine(dict["some_number"]);
    Console.ReadLine();
}

如何解析嵌套和复杂的json与动态和JavaScriptSerializer

请添加System.Web.Extensions的引用,并使用System.Web.Script.Serialization添加此命名空间;在前:

public static void ComplexJson()
{
    var jsonText = @"{
        ""some_number"": 108.541,
        ""date_time"": ""2011-04-13T15:34:09Z"",
        ""serial_number"": ""SN1234"",
        ""more_data"": {
            ""field1"": 1.0,
            ""field2"": ""hello""
        }
    }";

    var jss = new JavaScriptSerializer();
    var dict = jss.Deserialize<dynamic>(jsonText);

    Console.WriteLine(dict["some_number"]);
    Console.WriteLine(dict["more_data"]["field2"]);
    Console.ReadLine();
}

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);

我在我的代码中使用这样的代码,它工作得很好

using System.Web.Script.Serialization;
JavaScriptSerializer oJS = new JavaScriptSerializer();
RootObject oRootObject = new RootObject();
oRootObject = oJS.Deserialize<RootObject>(Your JSon String);

使用Json非常简单。NET:

dynamic stuff = JsonConvert.DeserializeObject("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");

string name = stuff.Name;
string address = stuff.Address.City;

同样使用Newtonsoft.Json.Linq:

dynamic stuff = JObject.Parse("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");

string name = stuff.Name;
string address = stuff.Address.City;

文档:使用动态查询JSON

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" }
...