有没有办法将JSON内容反序列化为C＃4动态类型？为了使用DataContractJsonSerializer，跳过创建一堆类会很不错。
答案 0 :(得分：611)
如果您乐意依赖System.Web.Helpers程序集，那么您可以使用Json类：
dynamic data = Json.Decode(json);
它作为.NET 4框架的additional download包含在MVC框架中。如果有帮助的话，一定要给Vlad一个upvote！但是，如果您不能假设客户端环境包含此DLL，请继续阅读。
建议采用另一种反序列化方法here。我稍微修改了代码以修复错误并适合我的编码风格。您只需要这个代码以及对项目中System.Web.Extensions的引用：
using System;
using System.Collections;
using System.Collections.Generic;
using System.Collections.ObjectModel;
using System.Dynamic;
using System.Linq;
using System.Text;
using System.Web.Script.Serialization;
public sealed class DynamicJsonConverter : JavaScriptConverter
{
public override object Deserialize(IDictionary<string, object> dictionary, Type type, JavaScriptSerializer serializer)
{
if (dictionary == null)
throw new ArgumentNullException("dictionary");
return type == typeof(object) ? new DynamicJsonObject(dictionary) : null;
}
public override IDictionary<string, object> Serialize(object obj, JavaScriptSerializer serializer)
{
throw new NotImplementedException();
}
public override IEnumerable<Type> SupportedTypes
{
get { return new ReadOnlyCollection<Type>(new List<Type>(new[] { typeof(object) })); }
}
#region Nested type: DynamicJsonObject
private sealed class DynamicJsonObject : DynamicObject
{
private readonly IDictionary<string, object> _dictionary;
public DynamicJsonObject(IDictionary<string, object> dictionary)
{
if (dictionary == null)
throw new ArgumentNullException("dictionary");
_dictionary = dictionary;
}
public override string ToString()
{
var sb = new StringBuilder("{");
ToString(sb);
return sb.ToString();
}
private void ToString(StringBuilder sb)
{
var firstInDictionary = true;
foreach (var pair in _dictionary)
{
if (!firstInDictionary)
sb.Append(",");
firstInDictionary = false;
var value = pair.Value;
var name = pair.Key;
if (value is string)
{
sb.AppendFormat("{0}:\"{1}\"", name, value);
}
else if (value is IDictionary<string, object>)
{
new DynamicJsonObject((IDictionary<string, object>)value).ToString(sb);
}
else if (value is ArrayList)
{
sb.Append(name + ":[");
var firstInArray = true;
foreach (var arrayValue in (ArrayList)value)
{
if (!firstInArray)
sb.Append(",");
firstInArray = false;
if (arrayValue is IDictionary<string, object>)
new DynamicJsonObject((IDictionary<string, object>)arrayValue).ToString(sb);
else if (arrayValue is string)
sb.AppendFormat("\"{0}\"", arrayValue);
else
sb.AppendFormat("{0}", arrayValue);
}
sb.Append("]");
}
else
{
sb.AppendFormat("{0}:{1}", name, value);
}
}
sb.Append("}");
}
public override bool TryGetMember(GetMemberBinder binder, out object result)
{
if (!_dictionary.TryGetValue(binder.Name, out result))
{
// return null to avoid exception. caller can check for null this way...
result = null;
return true;
}
result = WrapResultObject(result);
return true;
}
public override bool TryGetIndex(GetIndexBinder binder, object[] indexes, out object result)
{
if (indexes.Length == 1 && indexes[0] != null)
{
if (!_dictionary.TryGetValue(indexes[0].ToString(), out result))
{
// return null to avoid exception. caller can check for null this way...
result = null;
return true;
}
result = WrapResultObject(result);
return true;
}
return base.TryGetIndex(binder, indexes, out result);
}
private static object WrapResultObject(object result)
{
var dictionary = result as IDictionary<string, object>;
if (dictionary != null)
return new DynamicJsonObject(dictionary);
var arrayList = result as ArrayList;
if (arrayList != null && arrayList.Count > 0)
{
return arrayList[0] is IDictionary<string, object>
? new List<object>(arrayList.Cast<IDictionary<string, object>>().Select(x => new DynamicJsonObject(x)))
: new List<object>(arrayList.Cast<object>());
}
return result;
}
}
#endregion
}
你可以像这样使用它：
string json = ...;
var serializer = new JavaScriptSerializer();
serializer.RegisterConverters(new[] { new DynamicJsonConverter() });
dynamic obj = serializer.Deserialize(json, typeof(object));
所以，给定一个JSON字符串：
{
"Items":[
{ "Name":"Apple", "Price":12.3 },
{ "Name":"Grape", "Price":3.21 }
],
"Date":"21/11/2010"
}
以下代码将在运行时运行：
dynamic data = serializer.Deserialize(json, typeof(object));
data.Date; // "21/11/2010"
data.Items.Count; // 2
data.Items[0].Name; // "Apple"
data.Items[0].Price; // 12.3 (as a decimal)
data.Items[1].Name; // "Grape"
data.Items[1].Price; // 3.21 (as a decimal)
答案 1 :(得分：559)
使用Json.NET非常简单：
dynamic stuff = JsonConvert.DeserializeObject("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");
string name = stuff.Name;
string address = stuff.Address.City;
另外using 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;
答案 2 :(得分：289)
您可以使用System.Web.Helpers.Json执行此操作 - 其Decode方法返回一个动态对象，您可以根据需要进行遍历。
它包含在System.Web.Helpers程序集（.NET 4.0）中。
var dynamicObject = Json.Decode(jsonString);
答案 3 :(得分：78)
.NET 4.0有一个内置库来执行此操作：
using System.Web.Script.Serialization;
JavaScriptSerializer jss = new JavaScriptSerializer();
var d = jss.Deserialize<dynamic>(str);
这是最简单的方法。
答案 4 :(得分：74)
简单的“字符串JSON数据”到没有任何第三方DLL文件的对象：
WebClient client = new WebClient();
string getString = client.DownloadString("https://graph.facebook.com/zuck");
JavaScriptSerializer serializer = new JavaScriptSerializer();
dynamic item = serializer.Deserialize<object>(getString);
string name = item["name"];
//note: JavaScriptSerializer in this namespaces
//System.Web.Script.Serialization.JavaScriptSerializer
注意：您也可以使用自定义对象。
Personel item = serializer.Deserialize<Personel>(getString);
答案 5 :(得分：28)
JsonFx可以将JSON内容反序列化为动态对象。
序列化到/从动态类型（.NET 4.0的默认值）：
var reader = new JsonReader(); var writer = new JsonWriter();
string input = @"{ ""foo"": true, ""array"": [ 42, false, ""Hello!"", null ] }";
dynamic output = reader.Read(input);
Console.WriteLine(output.array[0]); // 42
string json = writer.Write(output);
Console.WriteLine(json); // {"foo":true,"array":[42,false,"Hello!",null]}
答案 6 :(得分：18)
我制作了一个使用Expando Objects的DynamicJsonConverter的新版本。我使用了expando对象，因为我想使用Json.NET将动态序列化为JSON。
using System;
using System.Collections;
using System.Collections.Generic;
using System.Collections.ObjectModel;
using System.Dynamic;
using System.Web.Script.Serialization;
public static class DynamicJson
{
public static dynamic Parse(string json)
{
JavaScriptSerializer jss = new JavaScriptSerializer();
jss.RegisterConverters(new JavaScriptConverter[] { new DynamicJsonConverter() });
dynamic glossaryEntry = jss.Deserialize(json, typeof(object)) as dynamic;
return glossaryEntry;
}
class DynamicJsonConverter : JavaScriptConverter
{
public override object Deserialize(IDictionary<string, object> dictionary, Type type, JavaScriptSerializer serializer)
{
if (dictionary == null)
throw new ArgumentNullException("dictionary");
var result = ToExpando(dictionary);
return type == typeof(object) ? result : null;
}
private static ExpandoObject ToExpando(IDictionary<string, object> dictionary)
{
var result = new ExpandoObject();
var dic = result as IDictionary<String, object>;
foreach (var item in dictionary)
{
var valueAsDic = item.Value as IDictionary<string, object>;
if (valueAsDic != null)
{
dic.Add(item.Key, ToExpando(valueAsDic));
continue;
}
var arrayList = item.Value as ArrayList;
if (arrayList != null && arrayList.Count > 0)
{
dic.Add(item.Key, ToExpando(arrayList));
continue;
}
dic.Add(item.Key, item.Value);
}
return result;
}
private static ArrayList ToExpando(ArrayList obj)
{
ArrayList result = new ArrayList();
foreach (var item in obj)
{
var valueAsDic = item as IDictionary<string, object>;
if (valueAsDic != null)
{
result.Add(ToExpando(valueAsDic));
continue;
}
var arrayList = item as ArrayList;
if (arrayList != null && arrayList.Count > 0)
{
result.Add(ToExpando(arrayList));
continue;
}
result.Add(item);
}
return result;
}
public override IDictionary<string, object> Serialize(object obj, JavaScriptSerializer serializer)
{
throw new NotImplementedException();
}
public override IEnumerable<Type> SupportedTypes
{
get { return new ReadOnlyCollection<Type>(new List<Type>(new[] { typeof(object) })); }
}
}
}
答案 7 :(得分：15)
使用Newtonsoft.Json的另一种方式：
dynamic stuff = Newtonsoft.Json.JsonConvert.DeserializeObject("{ color: 'red', value: 5 }");
string color = stuff.color;
int value = stuff.value;
答案 8 :(得分：10)
您可以在Newtonsoft.Json的帮助下实现这一目标。从Nuget和：
安装Newtonsoft.Jsonusing Newtonsoft.Json;
dynamic results = JsonConvert.DeserializeObject<dynamic>(YOUR_JSON);
答案 9 :(得分：7)
最简单的方法是：
只需添加此DLL file。
使用如下代码：
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.
答案 10 :(得分：6)
您可以使用using Newtonsoft.Json
var jRoot =
JsonConvert.DeserializeObject<dynamic>(Encoding.UTF8.GetString(resolvedEvent.Event.Data));
resolvedEvent.Event.Data是我对调用核心事件的反应。
答案 11 :(得分：6)
您可以扩展JavaScriptSerializer以递归方式将其创建的字典复制到expando对象，然后动态使用它们：
static class JavaScriptSerializerExtensions
{
public static dynamic DeserializeDynamic(this JavaScriptSerializer serializer, string value)
{
var dictionary = serializer.Deserialize<IDictionary<string, object>>(value);
return GetExpando(dictionary);
}
private static ExpandoObject GetExpando(IDictionary<string, object> dictionary)
{
var expando = (IDictionary<string, object>)new ExpandoObject();
foreach (var item in dictionary)
{
var innerDictionary = item.Value as IDictionary<string, object>;
if (innerDictionary != null)
{
expando.Add(item.Key, GetExpando(innerDictionary));
}
else
{
expando.Add(item.Key, item.Value);
}
}
return (ExpandoObject)expando;
}
}
然后你只需要为你定义扩展的命名空间使用一个using语句（考虑在System.Web.Script.Serialization中定义它们......另一个技巧是不使用命名空间，那么你就不要需要使用using语句，你可以像这样使用它们：
var serializer = new JavaScriptSerializer();
var value = serializer.DeserializeDynamic("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");
var name = (string)value.Name; // Jon Smith
var age = (int)value.Age; // 42
var address = value.Address;
var city = (string)address.City; // New York
var state = (string)address.State; // NY
答案 12 :(得分：5)
看一下我在CodeProject上写的文章，这篇文章准确地回答了这个问题：
在这里重新发布所有内容的方式太多了，甚至更少，因为该文章附带了密钥/所需的源文件。
答案 13 :(得分：5)
为此，我将使用JSON.NET对JSON流进行低级解析，然后从ExpandoObject类的实例中构建对象层次结构。
答案 14 :(得分：5)
我使用http://json2csharp.com/来获取表示JSON对象的类。
输入：
{
"name":"John",
"age":31,
"city":"New York",
"Childs":[
{
"name":"Jim",
"age":11
},
{
"name":"Tim",
"age":9
}
]
}
输出：
public class Child
{
public string name { get; set; }
public int age { get; set; }
}
public class Person
{
public string name { get; set; }
public int age { get; set; }
public string city { get; set; }
public List<Child> Childs { get; set; }
}
之后我使用Newtonsoft.Json填写课程：
using Newtonsoft.Json;
namespace GitRepositoryCreator.Common
{
class JObjects
{
public static string Get(object p_object)
{
return JsonConvert.SerializeObject(p_object);
}
internal static T Get<T>(string p_object)
{
return JsonConvert.DeserializeObject<T>(p_object);
}
}
}
您可以这样称呼它：
Person jsonClass = JObjects.Get<Person>(stringJson);
string stringJson = JObjects.Get(jsonClass);
<强> PS：
如果您的JSON变量名称不是有效的C＃名称（名称以$开头），您可以这样修复：
public class Exception
{
[JsonProperty(PropertyName = "$id")]
public string id { get; set; }
public object innerException { get; set; }
public string message { get; set; }
public string typeName { get; set; }
public string typeKey { get; set; }
public int errorCode { get; set; }
public int eventId { get; set; }
}
答案 15 :(得分：5)
我在我的代码中使用这样的，并且它正常工作
using System.Web.Script.Serialization;
JavaScriptSerializer oJS = new JavaScriptSerializer();
RootObject oRootObject = new RootObject();
oRootObject = oJS.Deserialize<RootObject>(Your JSon String);
答案 16 :(得分：4)
在JavaScript中使用DataSet（C＃）。一个使用DataSet输入创建JSON流的简单函数。创建类似（多表数据集）的JSON内容：
[[{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;
}
答案 17 :(得分：4)
JSON.NET中的反序列化可以使用JObject类进行动态处理，该类包含在该库中。我的JSON字符串代表这些类：
public class Foo {
public int Age {get;set;}
public Bar Bar {get;set;}
}
public class Bar {
public DateTime BDay {get;set;}
}
现在我们反序列化字符串而不引用上面的类：
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());
或者如果你想更深入：
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());
有关完整示例，请参阅post。
答案 18 :(得分：4)
试试这个：
var units = new { Name = "Phone", Color= "White" };
var jsonResponse = JsonConvert.DeserializeAnonymousType(json, units);
答案 19 :(得分：4)
有一个名为SimpleJson的C＃轻量级JSON库。
它支持.NET 3.5 +，Silverlight和Windows Phone 7。
它支持.NET 4.0的动态
它也可以作为NuGet包安装
Install-Package SimpleJson
答案 20 :(得分：4)
您想要的对象DynamicJSONObject包含在ASP.NET Web Pages包中的System.Web.Helpers.dll中，该包是WebMatrix的一部分。
答案 21 :(得分：3)
获取ExpandoObject：
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;
Container container = JsonConvert.Deserialize<Container>(jsonAsString, new ExpandoObjectConverter());
答案 22 :(得分：3)
使用Newtonsoft.Json创建动态对象确实很棒。
//json is your string containing the JSON value
dynamic data = JsonConvert.DeserializeObject<dynamic>(json);
现在，您可以像常规对象一样访问data对象。这是我们当前作为示例的JSON对象：
{ "ID":123,"Name":"Jack","Numbers":[1, 2, 3] }
这是反序列化后的访问方式：
data.ID //Retrieve the int
data.Name //Retrieve the string
data.Numbers[0] //Retrieve the first element in the array
答案 23 :(得分：2)
请添加 System.Web.Extensions 的参考，并在顶部添加此命名空间using System.Web.Script.Serialization;：
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();
}
请添加 System.Web.Extensions 的参考，并在顶部添加此命名空间using System.Web.Script.Serialization;：
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();
}
答案 24 :(得分：1)
另一种选择是“将JSON粘贴为类”，以便可以快速轻松地反序列化。
这是一个更好的解释n piccas ... ‘Paste JSON As Classes’ in ASP.NET and Web Tools 2012.2 RC
答案 25 :(得分：1)
使用Cinchoo ETL-可用于将JSON解析为动态对象的开源库：
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("$.*")
)
{
foreach (var rec in p)
{
Console.WriteLine("Action: " + rec.action);
Console.WriteLine("Timestamp: " + rec.timestamp);
Console.WriteLine("URL: " + rec.url);
Console.WriteLine("IP address: " + rec.ip);
}
}
输出：
Action: open
Timestamp: 2018-09-05 20:46:00
URL: http://www.google.com
IP address: 66.102.6.98
免责声明：我是这个图书馆的作者。
答案 26 :(得分：1)
我想在单元测试中以编程方式进行此操作，我确实可以将其键入。
我的解决方法是：
dict.ContainsKey("ExpectedProperty");
现在我可以断言
import pandas as pd
df = pd.read_json(data)
new_df = df.transpose()
答案 27 :(得分：0)
尝试一下！
JSON示例：
[{
"id": 140,
"group": 1,
"text": "xxx",
"creation_date": 123456,
"created_by": "xxx@gmail.co",
"tags": ["xxxxx"]
}, {
"id": 141,
"group": 1,
"text": "xxxx",
"creation_date": 123456,
"created_by": "xxx@gmail.com",
"tags": ["xxxxx"]
}]
C＃代码：
var jsonString = (File.ReadAllText(Path.Combine(Directory.GetCurrentDirectory(),"delete_result.json")));
var objects = JsonConvert.DeserializeObject<dynamic>(jsonString);
foreach(var o in objects)
{
Console.WriteLine($"{o.id.ToString()}");
}