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
获取一个ExpandoObject:
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;
Container container = JsonConvert.Deserialize<Container>(jsonAsString, new ExpandoObjectConverter());
其他回答
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.
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;
}
我做了一个使用Expando对象的DynamicJsonConverter的新版本。我使用了expando对象,因为我想使用Json.NET将动态序列化回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) })); }
}
}
}
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
.NET 4.0有一个内置库可以做到这一点:
using System.Web.Script.Serialization;
JavaScriptSerializer jss = new JavaScriptSerializer();
var d = jss.Deserialize<dynamic>(str);
这是最简单的方法。
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