JsonConvert.DeserializeObject<> (字符串)为$ id属性返回空值 [英] JsonConvert.DeserializeObject<> (string) returns null value for $id property
问题描述
我正在使用System.Net.WebClient.DownloadString下载JSON.我得到一个有效的答复:
I'm downloading the JSON using System.Net.WebClient.DownloadString. I'm getting a valid response:
{
"FormDefinition": [
{
"$id":"4",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Punchworks Form"
},
{
"$id":"6",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Punchworks Form test second"
},
{
"$id":"46",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"any_Name"
},
{
"$id":"47",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Punchworks Form test second"
},
{
"$id":"49",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Testing Name ??´????? ???? ACEeišuu { [ ( ~ ! @ # "
},
{
"$id":"50",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"something new"
},
{
"$id":"56",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Testing Name руÌÑÑкий 汉è¯æ¼¢èªž ĄČĘėįšųū { [ ( ~ ! @ # "
},
{
"$id":"57",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Test Name"
},
{
"$id":"58",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 12:59:29 PM"
},
{
"$id":"59",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:01:18 PM"
},
{
"$id":"60",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:40:44 PM"
},
{
"$id":"61",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:43:46 PM"
},
{
"$id":"62",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:48:21 PM"
},
{
"$id":"63",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:57:00 PM"
},
{
"$id":"64",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:57:53 PM"
},
{
"$id":"65",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Unique Name - 12/16/2013 1:58:46 PM"
},
{
"$id":"79",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Testing Name1211"
},
{
"$id":"80",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Testing Name1211"
},
{
"$id":"81",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"any_nami"
},
{
"$id":"90",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Test_something3"
},
{
"$id":"91",
"Class":558,
"ClassDisplayLabel":"Punchworks",
"Name":"Test_something4"
}]
}
这是我的模特:
public class FormDefinitionList
{
[JsonProperty("FormDefinition")]
public List<FormDefinition> FormDefinitions { get; set; }
}
public class FormDefinition
{
[JsonProperty ("$id")]
public string Id { get; set; }
[JsonProperty ("Class")]
public int Class { get; set; }
[JsonProperty ("ClassName")]
public string ClassName { get; set; }
[JsonProperty ("ClassDisplayLabel")]
public string ClassDisplayLabel { get; set; }
[JsonProperty ("Definition")]
public string Definition { get; set; }
[JsonProperty ("Name")]
public string Name { get; set; }
}
当我这样做时，一切都会正常工作
Everything works when I do:
string response = "json as above";
FormDefinitionList root = JsonConvert.DeserializeObject<FormDefinitionList> (response);
，除了Id($ id)属性始终为空.最初，我试图弄清楚我从服务器取回的美元符号是否不同，但事实并非如此.我不确定从这里去哪里，有什么想法吗?
except that the Id ($id) property is always null. At first I tried to figure out if the dollar sign symbol I was getting back from the server was different, but that doesn't appear to be the case. I'm not sure where to go from here, so any ideas?
谢谢.
注意:如果我尝试使用JavaScriptSerializer之类的东西进行反序列化，则它可以完美工作，因此，我很确定我的模型或JSON.net都存在问题.不过可能是错误的.
NOTE: If I try to deserialize with something like JavaScriptSerializer, it works perfectly, so I'm fairly sure it's something wrong with my model or with JSON.net. Could be wrong though.
推荐答案
Json.Net通常将$id和$ref用作元数据，以在JSON中保留对象引用.因此，当看到$id时，它假定该属性不是实际JSON属性集的一部分，而是一个内部标识符.因此，即使您包含一个指示该属性的[JsonProperty]属性，它也不会填充对象的Id属性.
Json.Net normally uses $id along with $ref as metadata to preserve object references in JSON. So when it sees $id it assumes that property is not part of the actual JSON property set, but an internal identifier. Thus it does not populate the Id property on your object, even though you included a [JsonProperty] attribute indicating that it should.
更新
从Json.Net版本 6.0.4 开始，有一个新设置，您可以通过该设置指示解串器将这些元数据"属性视为普通属性，而不是使用它们.您需要做的就是将MetadataPropertyHandling设置设置为Ignore，然后照常反序列化.
As of Json.Net version 6.0.4, there is a new setting by which you can instruct the deserializer to treat these "metadata" properties as normal properties instead of consuming them. All you need to do is set the MetadataPropertyHandling setting to Ignore and then deserialize as usual.
var settings = new JsonSerializerSettings();
settings.MetadataPropertyHandling = MetadataPropertyHandling.Ignore;
var obj = JsonConvert.DeserializeObject<FormDefinitionList>(json, settings);
在6.0.4版之前，需要一种解决方法来解决此问题.该答案的其余部分讨论了可能的解决方法.如果您使用的是6.0.4或更高版本，则不需要解决方法，并且可以立即停止阅读.
Prior to version 6.0.4, a workaround was needed to solve this issue. The rest of this answer discusses the possible workarounds. If you are using 6.0.4 or later, you do not need the workaround and can stop reading now.
我看到的最简单的解决方法是，在反序列化之前，用@ c8>(包括引号)对JSON进行"$id"的字符串替换，如@Carlos Coelho所建议的.由于您必须对每个响应都执行此操作，因此，如果您采用这种方式，我建议您使用一种简单的辅助方法来避免代码重复，例如:
The simplest workaround I can see is to do a string replace of "$id" with "id" (including the quotes) on the JSON prior to deserializing it, as @Carlos Coelho suggested. Since you would have to do this with each response, if you go this route I would recommend making a simple helper method to avoid code duplication, e.g.:
public static T Deserialize<T>(string json)
{
return JsonConvert.DeserializeObject<T>(json.Replace("\"$id\"", "\"id\""));
}
但是，由于您在评论中说您不太热衷于使用字符串替换的想法，因此我研究了其他选项.我确实找到了另一种可能适合您的替代方法-自定义JsonConverter.转换器背后的想法是，它将尝试使用Json.Net的内置反序列化机制来创建和填充对象(sans ID)，然后从JSON手动检索$id属性，并使用其填充
However, since you said in your comments that you are not so keen on the idea of using a string replace, I looked into other options. I did find one other alternative that might work for you-- a custom JsonConverter. The idea behind the converter is that it would try to use Json.Net's built-in deserialization mechanisms to create and populate the object (sans ID), then manually retrieve the $id property from the JSON and use it to populate the Id property on the object via reflection.
这是转换器的代码:
public class DollarIdPreservingConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(FormDefinition);
}
public override object ReadJson(JsonReader reader, Type objectType,
object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
object o = jo.ToObject(objectType);
JToken id = jo["$id"];
if (id != null)
{
PropertyInfo prop = objectType.GetProperty("Id");
if (prop != null && prop.CanWrite &&
prop.PropertyType == typeof(string))
{
prop.SetValue(o, id.ToString(), null);
}
}
return o;
}
public override void WriteJson(JsonWriter writer, object value,
JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
我试图编写该转换器，使其适用于具有$id的任何对象-您只需要相应地更改CanConvert方法，以便对于所有需要使用的类型返回true除了FormDefinition.
I tried to write the converter such that it would work for any object that has the $id -- you just need to change the CanConvert method accordingly so that it returns true for all the types that you need to use it for in addition to FormDefinition.
要使用该转换器，只需将其实例传递给DeserializeObject<T>即可，如下所示:
To use the converter, you just need to pass an instance of it to DeserializeObject<T> like this:
FormDefinitionList root = JsonConvert.DeserializeObject<FormDefinitionList>(
json, new DollarIdPreservingConverter());
重要说明:您可能会想用JsonConverter属性修饰类，而不是将转换器传递到DeserializeObject调用中，但是不要这样做-这会导致转换器进入递归循环，直到堆栈溢出. (有一种方法可以使转换器使用该属性，但是您必须重写ReadJson方法以手动创建目标对象并填充其属性，而不是调用jo.ToObject(objectType).这是可行的，但要注意一点更加混乱.)
Important note: you might be tempted to decorate your classes with a JsonConverter attribute instead of passing the converter into the DeserializeObject call, but don't do this-- it will cause the converter to go into a recursive loop until the stack overflows. (There is a way to get the converter to work with the attribute, but you would have to rewrite the ReadJson method to manually create the target object and populate its properties instead of calling jo.ToObject(objectType). It is doable, but a little more messy.)
让我知道这是否对您有用.
Let me know if this works for you.
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