具有HTTP状态500而不是HTTP状态400的JAX-RS响应

Question

有时，JAR-RS客户端发送错误的语法请求正文。服务器应以HTTP状态400（错误请求）进行响应，但它以HTTP状态500（内部服务器错误）进行响应。

码：

JAX-B模型类：

@XmlRootElement(namespace = "http://www.test.com/test")
@XmlAccessorType(value = XmlAccessType.FIELD)
public class TestModel {

    @XmlElement
    private String id;
}

JAX-RS资源类：

@Path("test")
public class TestResource {

    @POST
    @Consumes(MediaType.APPLICATION_JSON)
    public void create(TestModel testModel) {
        // some code
    }
}

CXF配置：

<jaxrs:server address="/rest" id="test" staticSubresourceResolution="true">
    <jaxrs:serviceBeans>
        <ref bean="testResource" /> 
    </jaxrs:serviceBeans>
    <jaxrs:providers>
        <bean class="com.fasterxml.jackson.jaxrs.json.JacksonJaxbJsonProvider" />
    </jaxrs:providers>
</jaxrs:server>

例：

要求正文：

{"id2": "test"}

这id2是错误的，因此客户端应获得HTTP状态400，但其应获得HTTP状态500。

服务器日志：

Caused by: com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "id2" (class test.TestModel), not marked as ignorable (one known property: "id"])
    at [Source: org.apache.cxf.transport.http.AbstractHTTPDestination$1@6f30793d; line: 1, column: 10] (through reference chain: test.TestModel["id2"])
    at com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:51)
    at com.fasterxml.jackson.databind.DeserializationContext.reportUnknownProperty(DeserializationContext.java:839)
    at com.fasterxml.jackson.databind.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:1045)
    at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownProperty(BeanDeserializerBase.java:1352)
    at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownVanilla(BeanDeserializerBase.java:1330)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:264)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:125)
    at com.fasterxml.jackson.databind.ObjectReader._bind(ObjectReader.java:1470)
    at com.fasterxml.jackson.databind.ObjectReader.readValue(ObjectReader.java:912)
    at com.fasterxml.jackson.jaxrs.base.ProviderBase.readFrom(ProviderBase.java:811)
    at org.apache.cxf.jaxrs.utils.JAXRSUtils.readFromMessageBodyReader(JAXRSUtils.java:1343)
    at org.apache.cxf.jaxrs.utils.JAXRSUtils.readFromMessageBody(JAXRSUtils.java:1294)
    at org.apache.cxf.jaxrs.utils.JAXRSUtils.processParameter(JAXRSUtils.java:826)
    at org.apache.cxf.jaxrs.utils.JAXRSUtils.processParameters(JAXRSUtils.java:789)
    at org.apache.cxf.jaxrs.interceptor.JAXRSInInterceptor.processRequest(JAXRSInInterceptor.java:212)
    ... 68 more

有没有一种方法可以配置Jackson和/或CXF为错误的语法请求正文返回HTTP状态400，而无需模式验证或bean验证？

Answer 1

问题在于，未映射到响应（通过ExceptionMappers）的异常会转换为一般服务器错误响应，因为运行时不知道如何处理异常。

该jackson-jaxrs-provider模块有ExceptionMappers 来处理 Jackson 基异常类JsonMappingException和JsonParseException. 映射器分别是JsonMappingExceptionMapper和。JsonParseExceptionMapper这些映射器会将异常映射到 400 响应，并将异常消息作为响应正文。如果您不喜欢这个响应体，您可以编写自己的映射器。