Question

我的目标是进行Web调用并将返回的JSON转换为POJO。我正在尝试使用泽西+杰克逊，但在跑步时会遇到异常。

我的maven pom文件包含以下依赖项 -

<dependency>
    <groupId>org.glassfish.jersey.core</groupId>
    <artifactId>jersey-client</artifactId>
    <version>2.6</version>
</dependency>
<dependency>
    <groupId>org.glassfish.jersey.media</groupId>
    <artifactId>jersey-media-json-jackson</artifactId>
    <version>2.6</version>
</dependency>

我用来获取一些数据的代码如下 -

Client client = ClientBuilder.newBuilder()
            .register(JacksonFeature.class)
            .build();
ClientResponse response = client.target(url).request(MediaType.APPLICATION_JSON).get(ClientResponse.class);

但是以下异常是throw -

javax.ws.rs.ProcessingException: Error reading entity from input stream.
at org.glassfish.jersey.message.internal.InboundMessageContext.readEntity(InboundMessageContext.java:868)
at org.glassfish.jersey.message.internal.InboundMessageContext.readEntity(InboundMessageContext.java:785)
at org.glassfish.jersey.client.ClientResponse.readEntity(ClientResponse.java:335)
...
...
Caused by: org.codehaus.jackson.map.JsonMappingException: Can not find a deserializer for non-concrete Map type [map type; class javax.ws.rs.core.MultivaluedMap, [simple type, class java.lang.String] -> [collection type; class java.util.List, contains [simple type, class java.lang.String]]]
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createAndCache2(StdDeserializerProvider.java:315)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createAndCacheValueDeserializer(StdDeserializerProvider.java:290)
at org.codehaus.jackson.map.deser.StdDeserializerProvider.findValueDeserializer(StdDeserializerProvider.java:159)
at org.codehaus.jackson.map.deser.std.StdDeserializer.findDeserializer(StdDeserializer.java:620)
at org.codehaus.jackson.map.deser.BeanDeserializer.resolve(BeanDeserializer.java:379)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._resolveDeserializer(StdDeserializerProvider.java:407)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createAndCache2(StdDeserializerProvider.java:352)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createAndCacheValueDeserializer(StdDeserializerProvider.java:290)
at org.codehaus.jackson.map.deser.StdDeserializerProvider.findValueDeserializer(StdDeserializerProvider.java:159)
at org.codehaus.jackson.map.deser.StdDeserializerProvider.findTypedValueDeserializer(StdDeserializerProvider.java:180)
at org.codehaus.jackson.map.ObjectMapper._findRootDeserializer(ObjectMapper.java:2829)
at org.codehaus.jackson.map.ObjectMapper._readValue(ObjectMapper.java:2699)
at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1315)
at org.codehaus.jackson.jaxrs.JacksonJsonProvider.readFrom(JacksonJsonProvider.java:419)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor$TerminalReaderInterceptor.invokeReadFrom(ReaderInterceptorExecutor.java:257)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor$TerminalReaderInterceptor.aroundReadFrom(ReaderInterceptorExecutor.java:229)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor.proceed(ReaderInterceptorExecutor.java:149)
at org.glassfish.jersey.message.internal.MessageBodyFactory.readFrom(MessageBodyFactory.java:1124)
at org.glassfish.jersey.message.internal.InboundMessageContext.readEntity(InboundMessageContext.java:853)
... 90 more
Caused by: java.lang.IllegalArgumentException: Can not find a deserializer for non-concrete Map type [map type; class javax.ws.rs.core.MultivaluedMap, [simple type, class java.lang.String] -> [collection type; class java.util.List, contains [simple type, class java.lang.String]]]
at org.codehaus.jackson.map.deser.BasicDeserializerFactory.createMapDeserializer(BasicDeserializerFactory.java:424)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createDeserializer(StdDeserializerProvider.java:380)
at org.codehaus.jackson.map.deser.StdDeserializerProvider._createAndCache2(StdDeserializerProvider.java:310)
... 108 more

我是否缺少一些设置才能使其正常工作？我通过curl和浏览器尝试了url，它按预期返回JSON。

Answer 1

您需要的是响应，而不是ClientResponse：

javax.ws.rs.core.Response jsonResponse = client.target(url).request(MediaType.APPLICATION_JSON).get();

然后你可以看到你的回复中有什么（调试是你的朋友）。某种类型的地图有可能吗？如果是，你可以通过这样做来阅读它。

Map<SomeClassOfYours> entitiesFromResponse = jsonResponse.readEntity(new GenericType<Map<SomeClassOfYours>>() {});

如果您在响应中放置了一个普通实体，则可以执行以下操作：

SomeClassOfYours entityFromResponse = jsonResponse.readEntity(SomeClassOfYours.class);

编辑：为了实现这个目的，您还需要定义SomeClassOfYours并在其中放置相应的字段，构造函数，getter和setter。

Edit2：如果有疑问，您可以随时将jsonResponse读作String.class并将其放入String变量中。

Answer 2

Response jsonResponse = getClient().target(URI).request().get();
T result = jsonResponse.readEntity(type);

泽西岛2.6与杰克逊JSON反序列化

2 个答案: