blocks|key|510|text|使用JsonIgnore+-这将导致Json.Net在序列化和反序列化时完全忽略该属性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|511|另外，检查这个link。|512|entityMap|0|LINK|mutability|MUTABLE|url|http://james.newtonking.com/archive/2009/10/23/efficient-json-with-json-net-reducing-serialized-json-size^0|2|A|0|7|4|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Use <code>JsonIgnore</code> - this will cause the property to be completely ignored by Json.Net, both for serializing and deserializing.

Also, check this <a href="http://james.newtonking.com/archive/2009/10/23/efficient-json-with-json-net-reducing-serialized-json-size" rel="nofollow">link</a>.

blocks|key|550|text|+var+json+=+"{+"PropOne":+{+"Text":+"Data"+}+"PropTwo":+"Data2"+}";

+JObject+o+=+JObject.Parse(json);
+var+val+=+o.PropTwo;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|551|使用JSON+Linq提供程序，您不需要将对象反序列化为已知类型。|unstyled|552|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> var json = "{ "PropOne": { "Text": "Data" } "PropTwo": "Data2" }";

 JObject o = JObject.Parse(json);
 var val = o.PropTwo;
</code></pre>

Using JSON Linq provider you do not need to deserialize the object into a known type.

blocks|key|217607|text|JSON不是太大，所以我将采纳Matt+Johnson的建议并对整个JSON进行反序列化。多亏了jcwrequests+answer，我才能使用这个方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|217608|var+jObject+=+JObject.Parse(json);
var+jToken+=+jObject.GetValue("PropTwo");
PropTwoClass+value+=+jToken.ToObject(typeof(PropTwoClass));|code-block|syntax|javascript|217609|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The JSON isn't too large, so I'll take Matt Johnson's suggestion and deserialize the whole thing. Thanks to jcwrequests answer, I was able to use this method:

<pre><code>var jObject = JObject.Parse(json);
var jToken = jObject.GetValue("PropTwo");
PropTwoClass value = jToken.ToObject(typeof(PropTwoClass));
</code></pre>

blocks|key|649|text|Omar's+answer的一种更简单的解决方案是使用包装器。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|650|class+Wrapper
{
++++public+PropOneClass+PropOne;
}

JsonConvert.Deserialize<Wrapper>(json).PropOne|code-block|syntax|javascript|651|我的测试发现它快了大约30%25。|652|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/19439146/33236^0|0|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

A simpler solution to <a href="https://stackoverflow.com/a/19439146/33236">Omar's answer</a> would be to have a wrapper.

<pre><code>class Wrapper
{
 public PropOneClass PropOne;
}

JsonConvert.Deserialize&lt;Wrapper&gt;(json).PropOne
</code></pre>

My tests found it to be about 30% faster.

blocks|key|709|text|Matt's+answer是到目前为止最快的解决方案，尽管它有一个bug。这是我修复它的尝试。此方法将仅返回根级别的匹配属性。在计算开始和结束令牌时，仍然有一种幼稚的方法，尽管对于有效的JSON，它可能会起作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|710|马特，你可以把这个复制到你的答案里。|711|public+T+GetFirstInstance<T>(string+propertyName,+string+json)
{
++++using+(var+stringReader+=+new+StringReader(json))
++++using+(var+jsonReader+=+new+JsonTextReader(stringReader))
++++{
++++++++int+level+=+0;

++++++++while+(jsonReader.Read())
++++++++{
++++++++++++switch+(jsonReader.TokenType)
++++++++++++{
++++++++++++++++case+JsonToken.PropertyName:
++++++++++++++++++++if+(level+!=+1)
++++++++++++++++++++++++break;
++++++++++++++++++++if+((string)jsonReader.Value+==+propertyName)
++++++++++++++++++++{
++++++++++++++++++++++++jsonReader.Read();

++++++++++++++++++++++++return+(T)jsonReader.Value;+
++++++++++++++++++++}
++++++++++++++++++++break;

++++++++++++++++case+JsonToken.StartArray:
++++++++++++++++case+JsonToken.StartConstructor:
++++++++++++++++case+JsonToken.StartObject:
++++++++++++++++++++level%2B%2B;
++++++++++++++++++++break;

++++++++++++++++case+JsonToken.EndArray:
++++++++++++++++case+JsonToken.EndConstructor:
++++++++++++++++case+JsonToken.EndObject:
++++++++++++++++++++level--;
++++++++++++++++++++break;
++++++++++++}

++++++++}
++++++++return+default(T);
++++}
}|code-block|syntax|javascript|712|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/19438910/33236^0|0|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<a href="https://stackoverflow.com/a/19438910/33236">Matt's answer</a> is by far the fastest solution though it has a bug.
This is my attempt in fixing that.
This method will only return a matching property at the root level.
There is still a naive approach in counting start and end tokens though for valid JSON it will probably work.

Matt, feel free to copy this into your answer.

<pre><code>public T GetFirstInstance&lt;T&gt;(string propertyName, string json)
{
 using (var stringReader = new StringReader(json))
 using (var jsonReader = new JsonTextReader(stringReader))
 {
 int level = 0;

 while (jsonReader.Read())
 {
 switch (jsonReader.TokenType)
 {
 case JsonToken.PropertyName:
 if (level != 1)
 break;
 if ((string)jsonReader.Value == propertyName)
 {
 jsonReader.Read();

 return (T)jsonReader.Value; 
 }
 break;

 case JsonToken.StartArray:
 case JsonToken.StartConstructor:
 case JsonToken.StartObject:
 level++;
 break;

 case JsonToken.EndArray:
 case JsonToken.EndConstructor:
 case JsonToken.EndObject:
 level--;
 break;
 }

 }
 return default(T);
 }
}
</code></pre>

I have the following <code>JSON</code> text:

<pre><code>{
 "PropOne": {
 "Text": "Data"
 }
 "PropTwo": "Data2"
} 
</code></pre>

I want to deserialize <code>PropOne</code> into type <code>PropOneClass</code> without the overhead of deserializing any other properties on the object. Can this be done using JSON.NET?

JSON.NET deserialize a specific property

JSON

我有以下JSON文本：{    "PropOne": {        "Text": "Data"    }    "PropTwo": "Data2"}    我希望将PropOne反序列化为类型PropOneClass，而不需要对对象上的任何其他属性进行反序列化。使用JSON.NET可以做到这一点吗？

问JSON.NET反序列化特定属性
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