我有一个简单的JSON键/值列表，通过POST发送回ASP.NET。例子：
{ "key1": "value1", "key2": "value2"}
我没有试图反序列化为强类型的.NET对象
我只需要一个普通的旧字典（Of String，String），或者一些等价的东西（哈希表，字典（OfString，Object），老式的StringDictionary——见鬼，一个二维字符串数组对我来说很有用。
我可以使用ASP.NET 3.5中可用的任何东西，也可以使用流行的Json.NET（我已经在使用它对客户端进行序列化）。
显然，这两个JSON库都没有这种开箱即用的能力——它们完全专注于通过强契约实现基于反射的反序列化。
有什么想法吗？
限制：
我不想实现我自己的JSON解析器尚无法使用ASP.NET 4.0希望远离旧的、不推荐使用的JSON ASP.NET类
当前回答
我也有同样的问题，所以我自己写了这个。此解决方案与其他解决方案不同，因为它可以反序列化到多个级别。
只需将JSON字符串发送到反序列化ToDictionary函数，它将返回非强类型Dictionary＜string，object＞object。
旧代码
private Dictionary<string, object> deserializeToDictionary(string jo)
{
var values = JsonConvert.DeserializeObject<Dictionary<string, object>>(jo);
var values2 = new Dictionary<string, object>();
foreach (KeyValuePair<string, object> d in values)
{
// if (d.Value.GetType().FullName.Contains("Newtonsoft.Json.Linq.JObject"))
if (d.Value is JObject)
{
values2.Add(d.Key, deserializeToDictionary(d.Value.ToString()));
}
else
{
values2.Add(d.Key, d.Value);
}
}
return values2;
}
示例：这将返回FacebookJSON响应的Dictionary<string，object>object。
Test
private void button1_Click(object sender, EventArgs e)
{
string responsestring = "{\"id\":\"721055828\",\"name\":\"Dasun Sameera Weerasinghe\",\"first_name\":\"Dasun\",\"middle_name\":\"Sameera\",\"last_name\":\"Weerasinghe\",\"username\":\"dasun\",\"gender\":\"male\",\"locale\":\"en_US\", hometown: {id: \"108388329191258\", name: \"Moratuwa, Sri Lanka\",}}";
Dictionary<string, object> values = deserializeToDictionary(responsestring);
}
注意：home进一步反序列化为Dictionary＜string，object＞对象
使现代化
如果JSON字符串上没有数组，我的旧答案很有用。如果元素是数组，则这一个将进一步反序列化为List<object>。
只需将JSON字符串发送到反序列化ToDictionaryOrList函数，它将返回非强类型Dictionary＜string，object＞object或List＜object＞。
private static object deserializeToDictionaryOrList(string jo,bool isArray=false)
{
if (!isArray)
{
isArray = jo.Substring(0, 1) == "[";
}
if (!isArray)
{
var values = JsonConvert.DeserializeObject<Dictionary<string, object>>(jo);
var values2 = new Dictionary<string, object>();
foreach (KeyValuePair<string, object> d in values)
{
if (d.Value is JObject)
{
values2.Add(d.Key, deserializeToDictionary(d.Value.ToString()));
}
else if (d.Value is JArray)
{
values2.Add(d.Key, deserializeToDictionary(d.Value.ToString(), true));
}
else
{
values2.Add(d.Key, d.Value);
}
}
return values2;
}else
{
var values = JsonConvert.DeserializeObject<List<object>>(jo);
var values2 = new List<object>();
foreach (var d in values)
{
if (d is JObject)
{
values2.Add(deserializeToDictionary(d.ToString()));
}
else if (d is JArray)
{
values2.Add(deserializeToDictionary(d.ToString(), true));
}
else
{
values2.Add(d);
}
}
return values2;
}
}
其他回答
如果您追求的是一种轻量级的、无添加引用的方法，那么我刚才写的这段代码可能会奏效（尽管我不能100%保证健壮性）。
using System;
using System.Collections.Generic;
using System.Text;
using System.Text.RegularExpressions;
public Dictionary<string, object> ParseJSON(string json)
{
int end;
return ParseJSON(json, 0, out end);
}
private Dictionary<string, object> ParseJSON(string json, int start, out int end)
{
Dictionary<string, object> dict = new Dictionary<string, object>();
bool escbegin = false;
bool escend = false;
bool inquotes = false;
string key = null;
int cend;
StringBuilder sb = new StringBuilder();
Dictionary<string, object> child = null;
List<object> arraylist = null;
Regex regex = new Regex(@"\\u([0-9a-z]{4})", RegexOptions.IgnoreCase);
int autoKey = 0;
for (int i = start; i < json.Length; i++)
{
char c = json[i];
if (c == '\\') escbegin = !escbegin;
if (!escbegin)
{
if (c == '"')
{
inquotes = !inquotes;
if (!inquotes && arraylist != null)
{
arraylist.Add(DecodeString(regex, sb.ToString()));
sb.Length = 0;
}
continue;
}
if (!inquotes)
{
switch (c)
{
case '{':
if (i != start)
{
child = ParseJSON(json, i, out cend);
if (arraylist != null) arraylist.Add(child);
else
{
dict.Add(key, child);
key = null;
}
i = cend;
}
continue;
case '}':
end = i;
if (key != null)
{
if (arraylist != null) dict.Add(key, arraylist);
else dict.Add(key, DecodeString(regex, sb.ToString()));
}
return dict;
case '[':
arraylist = new List<object>();
continue;
case ']':
if (key == null)
{
key = "array" + autoKey.ToString();
autoKey++;
}
if (arraylist != null && sb.Length > 0)
{
arraylist.Add(sb.ToString());
sb.Length = 0;
}
dict.Add(key, arraylist);
arraylist = null;
key = null;
continue;
case ',':
if (arraylist == null && key != null)
{
dict.Add(key, DecodeString(regex, sb.ToString()));
key = null;
sb.Length = 0;
}
if (arraylist != null && sb.Length > 0)
{
arraylist.Add(sb.ToString());
sb.Length = 0;
}
continue;
case ':':
key = DecodeString(regex, sb.ToString());
sb.Length = 0;
continue;
}
}
}
sb.Append(c);
if (escend) escbegin = false;
if (escbegin) escend = true;
else escend = false;
}
end = json.Length - 1;
return dict; //theoretically shouldn't ever get here
}
private string DecodeString(Regex regex, string str)
{
return Regex.Unescape(regex.Replace(str, match => char.ConvertFromUtf32(Int32.Parse(match.Groups[1].Value, System.Globalization.NumberStyles.HexNumber))));
}
[我意识到这违反了OP限制#1，但从技术上讲，你没有写，我写了]
马克·伦德尔（Mark Rendle）将此作为评论发布，我想将其作为答案发布，因为这是迄今为止唯一一个能够从Google reCaptcha响应中返回成功和错误代码json结果的解决方案。
string jsonReponseString= wClient.DownloadString(requestUrl);
IDictionary<string, object> dict = new JavaScriptSerializer().DeserializeObject(jsonReponseString) as IDictionary<string, object>;
再次感谢，马克！
您可以使用Tiny JSON
string json = "{\"key1\":\"value1\", \"key2\":\"value2\"}";
IDictionary<string, string> dict = Tiny.Json.Decode<Dictionary<string, string>>(json);
我在这里添加了jSnake04和Dasun提交的代码。我添加了从JArray实例创建对象列表的代码。它具有双向递归，但由于它在固定的有限树模型上运行，所以除非数据量很大，否则不会有堆栈溢出的风险。
/// <summary>
/// Deserialize the given JSON string data (<paramref name="data"/>) into a
/// dictionary.
/// </summary>
/// <param name="data">JSON string.</param>
/// <returns>Deserialized dictionary.</returns>
private IDictionary<string, object> DeserializeData(string data)
{
var values = JsonConvert.DeserializeObject<Dictionary<string, object>>(data);
return DeserializeData(values);
}
/// <summary>
/// Deserialize the given JSON object (<paramref name="data"/>) into a dictionary.
/// </summary>
/// <param name="data">JSON object.</param>
/// <returns>Deserialized dictionary.</returns>
private IDictionary<string, object> DeserializeData(JObject data)
{
var dict = data.ToObject<Dictionary<String, Object>>();
return DeserializeData(dict);
}
/// <summary>
/// Deserialize any elements of the given data dictionary (<paramref name="data"/>)
/// that are JSON object or JSON arrays into dictionaries or lists respectively.
/// </summary>
/// <param name="data">Data dictionary.</param>
/// <returns>Deserialized dictionary.</returns>
private IDictionary<string, object> DeserializeData(IDictionary<string, object> data)
{
foreach (var key in data.Keys.ToArray())
{
var value = data[key];
if (value is JObject)
data[key] = DeserializeData(value as JObject);
if (value is JArray)
data[key] = DeserializeData(value as JArray);
}
return data;
}
/// <summary>
/// Deserialize the given JSON array (<paramref name="data"/>) into a list.
/// </summary>
/// <param name="data">Data dictionary.</param>
/// <returns>Deserialized list.</returns>
private IList<Object> DeserializeData(JArray data)
{
var list = data.ToObject<List<Object>>();
for (int i = 0; i < list.Count; i++)
{
var value = list[i];
if (value is JObject)
list[i] = DeserializeData(value as JObject);
if (value is JArray)
list[i] = DeserializeData(value as JArray);
}
return list;
}
对于那些在互联网上搜索并偶然发现这篇文章的人，我写了一篇关于如何使用JavaScriptSerializer类的博客文章。
阅读更多。。。http://procbits.com/2011/04/21/quick-json-serializationdeserialization-in-c/
下面是一个示例：
var json = "{\"id\":\"13\", \"value\": true}";
var jss = new JavaScriptSerializer();
var table = jss.Deserialize<dynamic>(json);
Console.WriteLine(table["id"]);
Console.WriteLine(table["value"]);
