blocks|key|513567|text|您需要为使用的每个泛型类型创建一个TypeReference对象，并将其用于反序列化。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|513568|mapper.readValue(jsonString,+new+TypeReference<Data<String>>()+{});|code-block|syntax|javascript|513569|entityMap^0|H|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You need to create a <code>TypeReference</code> object for each generic type you use and use that for deserialization. For example - 

<pre><code>mapper.readValue(jsonString, new TypeReference&lt;Data&lt;String&gt;&gt;() {});
</code></pre>

blocks|key|297067|text|您不能这样做:您必须指定完全解析的类型，如Data<MyType>。T只是一个变量，as是没有意义的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|297068|但是，如果您的意思是T将是已知的，而不是静态的，那么您需要动态地创建TypeReference的等价物。引用的其他问题可能已经提到了这一点，但它应该类似于：|297069|public+Data<T>+read(InputStream+json,+Class<T>+contentClass)+{
+++JavaType+type+=+mapper.getTypeFactory().constructParametricType(Data.class,+contentClass);
+++return+mapper.readValue(json,+type);
}|code-block|syntax|javascript|297070|entityMap^0|L|C|Y|1|0|A|1|Y|D|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

You can't do that: you must specify fully resolved type, like <code>Data&lt;MyType&gt;</code>. <code>T</code> is just a variable, and as is meaningless.

But if you mean that <code>T</code> will be known, just not statically, you need to create equivalent of <code>TypeReference</code> dynamically. Other questions referenced may already mention this, but it should look something like:

<pre><code>public Data&lt;T&gt; read(InputStream json, Class&lt;T&gt; contentClass) {
 JavaType type = mapper.getTypeFactory().constructParametricType(Data.class, contentClass);
 return mapper.readValue(json, type);
}
</code></pre>

blocks|key|513721|text|你要做的第一件事是序列化，然后你就可以反序列化了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|513722|因此，在进行序列化时，应该使用@JsonTypeInfo让jackson将类信息写入到json数据中。你可以这样做：|offset|length|style|CODE|513723|Class+Data+<T>+{
++++int+found;
++++@JsonTypeInfo(use=JsonTypeInfo.Id.CLASS,+include=JsonTypeInfo.As.PROPERTY,+property="@class")
++++Class<T>+hits
}|code-block|syntax|javascript|513724|然后，当你反序列化时，你会发现jackson已经将你的数据反序列化为一个类，你的变量在运行时实际命中了这个类。|513725|entityMap^0|0|F|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

First thing you do is serialize, then you can do deserialize.
 
so when you do serialize, you should use <code>@JsonTypeInfo</code> to let jackson write class information into your json data. What you can do is like this:

<pre><code>Class Data &lt;T&gt; {
 int found;
 @JsonTypeInfo(use=JsonTypeInfo.Id.CLASS, include=JsonTypeInfo.As.PROPERTY, property="@class")
 Class&lt;T&gt; hits
}
</code></pre>

Then when you deserialize, you will find jackson has deserialize your data into a class which your variable hits actually is at runtime.

blocks|key|297353|text|从Jackson+2.5开始，解决这个问题的一种优雅方法是使用TypeFactory.constructParametricType(Class+parametrized,+Class...+parameterClasses)方法，该方法允许通过指定参数化的类及其参数化类型来直接定义Jackson+JavaType。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|297354|假设你想反序列化到Data<String>，你可以这样做：|297355|//+the+json+variable+may+be+a+String,+an+InputStream+and+so+for...
JavaType+type+=+mapper.getTypeFactory().constructParametricType(Data.class,+String.class);
Data<String>+data+=+mapper.readValue(json,+type);|code-block|syntax|javascript|297356|请注意，如果类声明了多个参数化类型，实际上并不会很难：|297357|class+Data+<T,+U>+{
++++int+found;
++++Class<T>+hits;
++++List+list;
}|297358|我们可以这样做：|297359|JavaType+type+=+mapper.getTypeFactory().constructParametricType(Data.class,+String.class,+Integer);
Data<String,+Integer>+data+=+mapper.readValue(json,+type);|297360|entityMap|0|LINK|mutability|MUTABLE|url|https://fasterxml.github.io/jackson-databind/javadoc/2.8/com/fasterxml/jackson/databind/type/TypeFactory.html#constructParametricType(java.lang.Class,%2520java.lang.Class...)^0|46|8|V|2A|0|0|9|C|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

From Jackson 2.5, an elegant way to solve that is using the 
<a href="https://fasterxml.github.io/jackson-databind/javadoc/2.8/com/fasterxml/jackson/databind/type/TypeFactory.html#constructParametricType(java.lang.Class,%20java.lang.Class...)" rel="noreferrer">TypeFactory.constructParametricType(Class parametrized, Class... parameterClasses)</a> method that allows to define straigthly a Jackson <code>JavaType</code> by specifying the parameterized class and its parameterized types. 

Supposing you want to deserialize to <code>Data&lt;String&gt;</code>, you can do :

<pre><code>// the json variable may be a String, an InputStream and so for...
JavaType type = mapper.getTypeFactory().constructParametricType(Data.class, String.class);
Data&lt;String&gt; data = mapper.readValue(json, type);
</code></pre>

Note that if the class declared multiple parameterized types, it would not be really harder : 

<pre><code>class Data &lt;T, U&gt; {
 int found;
 Class&lt;T&gt; hits;
 List&lt;U&gt; list;
}
</code></pre>

We could do :

<pre><code>JavaType type = mapper.getTypeFactory().constructParametricType(Data.class, String.class, Integer);
Data&lt;String, Integer&gt; data = mapper.readValue(json, type);
</code></pre>

blocks|key|513806|text|对于类Data<>|type|unstyled|depth|inlineStyleRanges|entityRanges|data|513807|ObjectMapper+mapper+=+new+ObjectMapper();
JavaType+type+=+mapper.getTypeFactory().constructParametrizedType(Data.class,+Data.class,+Parameter.class);
Data<Parameter>+dataParam+=+mapper.readValue(jsonString,type)|code-block|syntax|javascript|513808|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|$]]

For class Data&lt;> 

<pre><code>ObjectMapper mapper = new ObjectMapper();
JavaType type = mapper.getTypeFactory().constructParametrizedType(Data.class, Data.class, Parameter.class);
Data&lt;Parameter&gt; dataParam = mapper.readValue(jsonString,type)
</code></pre>

blocks|key|513764|text|只需在Util类中编写一个静态方法即可。我正在从文件中读取Json。您还可以将字符串提供给readValue|type|unstyled|depth|inlineStyleRanges|entityRanges|data|513765|public+static+<T>+T+convertJsonToPOJO(String+filePath,+Class<?>+target)+throws+JsonParseException,+JsonMappingException,+IOException,+ClassNotFoundException+{
++++++++ObjectMapper+objectMapper+=+new+ObjectMapper();
++++++++return+objectMapper.readValue(new+File(filePath),+objectMapper+.getTypeFactory().constructCollectionType(List.class,+Class.forName(target.getName())));
}|code-block|syntax|javascript|513766|用法：|513767|List<TaskBean>+list+=++Util.<List<TaskBean>>convertJsonToPOJO("E:/J2eeWorkspaces/az_workspace_svn/az-client-service/dir1/dir2/filename.json",+TaskBean.class);|513768|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just write a static method in Util class. I am reading a Json from a file. you can give String also to readValue

<pre><code>public static &lt;T&gt; T convertJsonToPOJO(String filePath, Class&lt;?&gt; target) throws JsonParseException, JsonMappingException, IOException, ClassNotFoundException {
 ObjectMapper objectMapper = new ObjectMapper();
 return objectMapper.readValue(new File(filePath), objectMapper .getTypeFactory().constructCollectionType(List.class, Class.forName(target.getName())));
}
</code></pre>

Usage:

<pre><code>List&lt;TaskBean&gt; list = Util.&lt;List&lt;TaskBean&gt;&gt;convertJsonToPOJO("E:/J2eeWorkspaces/az_workspace_svn/az-client-service/dir1/dir2/filename.json", TaskBean.class);
</code></pre>

blocks|key|513511|text|您可以将其包装在另一个知道您的泛型类型的类中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|513512|例如，|513513|class+Wrapper+{
+private+Data<Something>+data;
}
mapper.readValue(jsonString,+Wrapper.class);|code-block|syntax|javascript|513514|这里的东西是一个具体的类型。你需要一个包装器具体化的类型。否则Jackson不知道要创建什么对象。|513515|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can wrap it in another class which knows the type of your generic type.

Eg,

<pre><code>class Wrapper {
 private Data&lt;Something&gt; data;
}
mapper.readValue(jsonString, Wrapper.class);
</code></pre>

Here Something is a concrete type. You need a wrapper per reified type. Otherwise Jackson does not know what objects to create.

blocks|key|513988|text|public+class+Data<T>+extends+JsonDeserializer+implements+ContextualDeserializer+{
++++private+Class<T>+cls;
++++public+JsonDeserializer+createContextual(DeserializationContext+ctx,+BeanProperty+prop)+throws+JsonMappingException+{
++++++++cls+=+(Class<T>)+ctx.getContextualType().getRawClass();
++++++++return+this;
++++}
++++...
+}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|513989|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class Data&lt;T&gt; extends JsonDeserializer implements ContextualDeserializer {
 private Class&lt;T&gt; cls;
 public JsonDeserializer createContextual(DeserializationContext ctx, BeanProperty prop) throws JsonMappingException {
 cls = (Class&lt;T&gt;) ctx.getContextualType().getRawClass();
 return this;
 }
 ...
 }
</code></pre>

blocks|key|297278|text|如果你正在使用scala，并且在编译时知道泛型类型，但又不想在你的所有api中手动传递TypeReference，你可以使用以下代码(使用jackson+2.9.5)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|297279|def+read[T](entityStream:+InputStream)(implicit+typeTag:+WeakTypeTag[T]):+T+=+{

++++//nathang:+all+of+this+*crazy*+scala+reflection+allows+us+to+handle+List[Seq[Map[Int,Value]]]]+without+passing
++++//+new+TypeReference[List[Seq[Map[Int,Value]]]]](){}+to+the+function

++++def+recursiveFindGenericClasses(t:+Type):+JavaType+=+{
++++++val+current+=+typeTag.mirror.runtimeClass(t)

++++++if+(t.typeArgs.isEmpty)+{
++++++++val+noSubtypes+=+Seq.empty[Class[_]]
++++++++factory.constructParametricType(current,+noSubtypes:_*)
++++++}

++++++else+{
++++++++val+genericSubtypes:+Seq[JavaType]+=+t.typeArgs.map(recursiveFindGenericClasses)
++++++++factory.constructParametricType(current,+genericSubtypes:_*)
++++++}

++++}

++++val+javaType+=+recursiveFindGenericClasses(typeTag.tpe)

++++json.readValue[T](entityStream,+javaType)
++}|code-block|syntax|javascript|297280|它可以像这样使用：|297281|read[List[Map[Int,+SomethingToSerialize]]](inputStream)|297282|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

if you're using scala and know the generic type at compile time, but don't want to manually pass TypeReference everywhere in all your api l ayers, you can use the following code (with jackson 2.9.5):

<pre><code>def read[T](entityStream: InputStream)(implicit typeTag: WeakTypeTag[T]): T = {

 //nathang: all of this *crazy* scala reflection allows us to handle List[Seq[Map[Int,Value]]]] without passing
 // new TypeReference[List[Seq[Map[Int,Value]]]]](){} to the function

 def recursiveFindGenericClasses(t: Type): JavaType = {
 val current = typeTag.mirror.runtimeClass(t)

 if (t.typeArgs.isEmpty) {
 val noSubtypes = Seq.empty[Class[_]]
 factory.constructParametricType(current, noSubtypes:_*)
 }

 else {
 val genericSubtypes: Seq[JavaType] = t.typeArgs.map(recursiveFindGenericClasses)
 factory.constructParametricType(current, genericSubtypes:_*)
 }

 }

 val javaType = recursiveFindGenericClasses(typeTag.tpe)

 json.readValue[T](entityStream, javaType)
 }
</code></pre>

which can be used like this:

<pre><code>read[List[Map[Int, SomethingToSerialize]]](inputStream)
</code></pre>

I have a json string, which I should deSerialize to the following class

<pre><code>class Data &lt;T&gt; {
 int found;
 Class&lt;T&gt; hits
}
</code></pre>

How do I do it?
This is the usual way

<pre><code>mapper.readValue(jsonString, Data.class);
</code></pre>

But how do I mention what T stands for?

Jackson - Deserialize using generic class

我有一个json字符串，我应该将它deSerialize到下面的类class Data <T> { int found; Class<T> hits}我该怎么做呢？这是通常的方式mapper.readValue(jsonString, Data.class);但是我怎么提到T代表什么呢？

问Jackson -使用泛型类进行反序列化
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回答 9

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