使用JAX-RS处理JSON响应 [英] Processing JSON response using JAX-RS

问题描述

I have a json payload like this:

{
   "account":    {

      "sample_id": 1424876658095,
      "parameters":       [
                  {
            "name": "email",
            "value": "[e-mail]"
         },
                  {
            "name": "first_name",
            "value": "FIRSTNAME"
         },
                  {
            "name": "last_name",
            "value": "LASTNAME"
         }
      ]
   },
   "assests": [   {

      "tran_id": "1234567",

   }]
}

以上json有效负载是在休息API调用的响应中生成的。
我想在java中处理这个响应，生成如下内容：

The above json payload is getting generated in the response of a rest API call. I would like to process this response in java to generate something like this:

{
   "account":    {

      "sample_id": 1424876658095,
      "email_id": "[e-mail]",
      "first_name": "FIRSTNAME",
      "last_name": "LASTNAME",
   },
   "assets": [   {

      "tran_id": "1234567",

   }]
}

我正在使用REST API的JAX-RS规范，但我不是能够找到任何库来处理响应。

I am using the JAX-RS specification for the REST API, but I am not able to find any library to process the response.

推荐答案

如果你想利用JAX-R中的Jackson序列化，你需要实现自定义序列化器。

If you want to leverage the Jackson serialization within JAX-Rs, you need to implement as custom serializer.

有两个步骤可以做到：

• 创建自定义序列化程序

以下是基于bean的自定义Jackson序列化程序示例 AccountBean 和 ParameterBean ：

Here is a sample of a custom Jackson serializer for your needs based on beans AccountBean and ParameterBean:

public class AccountBeanSerializer extends JsonSerializer<AccountBean> {
    @Override
    public void serialize(AccountBean accountBean, JsonGenerator jgen,
            SerializerProvider provider) throws IOException,
            JsonProcessingException {
        jgen.writeStartObject();
        jgen.writeNumberField("sample_id", accountBean.getSampleId());
        List<ParameterBean> parameters = accountBean.getParameters();
        for (ParameterBean parameterBean : parameters) {
            jgen.writeStringField(parameterBean.getName(),
                    parameterBean.getValue());
        }
        jgen.writeEndObject();
    }
}

我假设您的回复类是否类似：

I assume the class for your response if something like that:

class ResponseBean
    field account = class AccountBean
       field sampleId (long)
       field parameters = class ParameterBean
    (...)

注册自定义序列化程序

然后，您需要在上下文解析程序中提供自定义Jackson配置。为此，您可以创建接口 ContextResolver 的实现，并使用 Provider 进行注释。

You need then to provide a custom Jackson configuration within a context resolver. For this, you can create an implementation of the interface ContextResolver annotated with Provider.

@Provider
public class CustomJacksonConfig implements ContextResolver<ObjectMapper> {
    private ObjectMapper objectMapper;

    public JacksonConfig() {
        this.objectMapper = new ObjectMapper();
        SimpleModule module = new SimpleModule("MyModule", new Version(1, 0, 0, null));
        module.addSerializer(AccountBean.class, new AccountSerializer());
        this.objectMapper.registerModule(module);
    }

    public ObjectMapper getContext(Class<?> objectType) {
        return objectMapper;
    }
}

以下链接可以帮助您：

• http://www.baeldung.com/jackson-custom-serialization


• https://books.google.fr/books?id=cFBptKRXrk4C&pg=PA81&lpg=PA81&dq=jaxrs+ContextResolver& ;源= BL&安培; OTS = FlOzDeookf&安培; SIG = iY0dM8l5A0svwxB-hsfFO94eIqc&安培; HL = FR&安培; SA = X&安培; EI = B5vwVKSmO8bbat2BguAB&安培; VED = 0CEkQ6AEwBQ＃v = onepage&安培; q = JAXRS％20ContextResolver&安培; F =假


• https://github.com/Fa sterXML / jackson-jaxrs-json-provider

• http://www.baeldung.com/jackson-custom-serialization
• https://books.google.fr/books?id=cFBptKRXrk4C&pg=PA81&lpg=PA81&dq=jaxrs+ContextResolver&source=bl&ots=FlOzDeookf&sig=iY0dM8l5A0svwxB-hsfFO94eIqc&hl=fr&sa=X&ei=B5vwVKSmO8bbat2BguAB&ved=0CEkQ6AEwBQ#v=onepage&q=jaxrs%20ContextResolver&f=false
• https://github.com/FasterXML/jackson-jaxrs-json-provider

希望这会有所帮助，
Thierry

Hope this helps, Thierry

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