Newtonsoft.Json，填充字典失败

本文介绍了Newtonsoft.Json，填充字典失败的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ var serializeSettings = new JsonSerializerSettings 我将序列化一个字典到json by Newtonsoft.json { TypeNameHandling = TypeNameHandling.All， TypeNameAssemblyFormat = FormatterAssemblyStyle.Full， Formatting = Formatting.Indented }; var serializedObject = JsonConvert.SerializeObject（dic，serializeSettings）; 此代码生成一个这样的json： {$ type：System.Collections.Generic.Dictionary`2 [[System.Guid，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089]，[System.String，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089]]，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089，9648af76 -7986-4b34-8b2c-97b2345769ef：测试} 我尝试通过此代码将json反序列化为字典： var newDic = new Dictionay< Guid，string>（）; var deserializeSettings = new JsonSerializerSettings { TypeNameHandling = TypeNameHandling.All， TypeNameAssemblyFormat = FormatterAssemblyStyle.Full， Formatting = Formatting.Indented } JsonConvert.PopulateObject（serializedObject，newDic，deserializeSettings）; 但是发生这种异常：我写这样的GuidConverter并使用它。但不工作 public class GuidConverter：JsonConverter { public override bool CanConvert（Type objectType） { return objectType.IsAssignableFrom（typeof（Guid））; public override object ReadJson（JsonReader reader，Type objectType，object existingValue，JsonSerializer serializer） { try { return serializer.Deserialize<&的Guid GT;（读取器）; } catch { return Guid.Empty; } } public override void WriteJson（JsonWriter writer，object value，JsonSerializer serializer） { serializer.Serialize（writer，value）; } } 编辑： p> 我发现我的问题。将代码更改为序列化json到Dictionary<字符串，字符串>现在生成的字典中的第一个项目是： Kay：$ type值： System.Collections.Generic.Dictionary`2 [[System.Guid，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089]，[System.String，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089]]，mscorlib，Version = 4.0.0.0，Culture = neutral，PublicKeyToken = b77a5c561934e089 为什么？ 解决方案 问题是指定 TypeNameHandling = TypeNameHandling.All 当序列化你的字典。这将导致元数据$ type属性作为字典中的第一个对象发布：当使用 DeserializeObject ，当构造相应的c＃对象时，该令牌通常由Json.NET使用。但是您正在使用 PopulateObject 在预先分配的字典上。因此，元数据属性在构建期间不会消耗，而Json.NET会尝试将其添加到字典中，并失败。 解决方案设置 MetadataPropertyHandling = MetadataPropertyHandling.ReadAhead 在 deserializeSettings 。这样做会导致$ type属性无条件地被使用或忽略（如适用）： var deserializeSettings = new JsonSerializerSettings { TypeNameHandling = TypeNameHandling.All， TypeNameAssemblyFormat = FormatterAssemblyStyle.Full， Formatting = Formatting.Indented， MetadataPropertyHandling = MetadataPropertyHandling.ReadAhead }; JsonConvert.PopulateObject（serializedObject，newDic，deserializeSettings）; 请注意，从发行说明，使用此设置的内存使用情况和速度。 或，如果您无条件地在JSON中需要元数据类型信息您可以使用 TypeNameHandling = TypeNameHandling.Auto 序列化，并且只会发布多态类型的类型信息，您的 Dictionary< Guid，string> 不是。 I serialize a dictionary to json by Newtonsoft.json and bellow code :var serializeSettings = new JsonSerializerSettings { TypeNameHandling = TypeNameHandling.All, TypeNameAssemblyFormat = FormatterAssemblyStyle.Full, Formatting = Formatting.Indented }; var serializedObject = JsonConvert.SerializeObject(dic, serializeSettings);this code generate a json like this :{ "$type": "System.Collections.Generic.Dictionary`2[[System.Guid, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089],[System.String, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]], mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089", "9648af76-7986-4b34-8b2c-97b2345769ef": "Test"}I try to deserialize json to dictionary by this code :var newDic = new Dictionay<Guid,string>();var deserializeSettings = new JsonSerializerSettings{ TypeNameHandling = TypeNameHandling.All, TypeNameAssemblyFormat = FormatterAssemblyStyle.Full, Formatting = Formatting.Indented}JsonConvert.PopulateObject(serializedObject, newDic, deserializeSettings);But this exception occurs :I write GuidConverter like this and use it. but not workpublic class GuidConverter : JsonConverter{ public override bool CanConvert(Type objectType) { return objectType.IsAssignableFrom(typeof(Guid)); } public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer) { try { return serializer.Deserialize<Guid>(reader); } catch { return Guid.Empty; } } public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer) { serializer.Serialize(writer, value); }}EDIT:I found my problem. Change code to deserialize json to Dictionary< string, string > and now First item in generated dictionary is :Kay: "$type"Value : "System.Collections.Generic.Dictionary`2[[System.Guid, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089],[System.String, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]], mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089"Why?? 解决方案 The problem is that you have specified TypeNameHandling = TypeNameHandling.All when serializing your dictionary. This causes a metadata "$type" property to be emitted as the first object in the dictionary:When deserializing using DeserializeObject, this token is normally consumed by Json.NET when the corresponding c# object is constructed. But you are using PopulateObject on a pre-allocated dictionary. Thus the metadata property is not consumed during construction and instead Json.NET tries to add it to the dictionary, and fails.The solution is to set MetadataPropertyHandling = MetadataPropertyHandling.ReadAhead in deserializeSettings. Doing so will cause the "$type" property to be consumed or ignored (as appropriate) unconditionally:var deserializeSettings = new JsonSerializerSettings{ TypeNameHandling = TypeNameHandling.All, TypeNameAssemblyFormat = FormatterAssemblyStyle.Full, Formatting = Formatting.Indented, MetadataPropertyHandling = MetadataPropertyHandling.ReadAhead};JsonConvert.PopulateObject(serializedObject, newDic, deserializeSettings);Please note that, from the release notes, there is a slight cost in memory usage and speed from using this setting.Alternatively, if you don't unconditionally need metadata type information in your JSON, you could serialize with TypeNameHandling = TypeNameHandling.Auto and only emit type information for polymorphic types, which your Dictionary<Guid, string> is not. 这篇关于Newtonsoft.Json，填充字典失败的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！