我遇到了同样的问题,所以我自己写了这个。该解决方案与其他答案不同,因为它可以反序列化到多个级别。
只需将 JSON 字符串发送到 deserializeToDictionary 函数,它将返回非强类型 Dictionary<string, object> 对象。
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;
}
例如:这将返回 Facebook JSON 响应的 Dictionary<string, object> 对象。
测试
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:家乡进一步脱毒成Dictionary<string, object>
对象。
更新
如果 JSON 字符串上没有数组,我的旧答案效果很好。如果一个元素是一个数组,这进一步反序列化为List<object>。
只需将 JSON 字符串发送到 deserializeToDictionaryOrList 函数,它将返回非强类型 Dictionary<string, object> 对象或 @987654328 @。
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;
}
}