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我正在尝试设置一个处理 content-type=application/x-www-form-urlencoded 请求的休息服务。我目前正在使用 Jaxb2Marshaller 来解组请求。然而，在解组时，它会抛出错误“[org.xml.sax.SAXParseException: Content is not allowed in prolog.]”。

我将 xml 请求检查为字符串。它的 url 编码形式为：%3C%3Fxml+version=%221.0%22+encoding%3D%22UTF-8%22+standalone%3D%22yes%22%3F%3E%3Cxrsi%。

似乎这个编码的 xml 字符串请求导致了错误。有没有比解组先解码请求的方法？

以下是我的上下文设置：

<bean id="multipartResolver" class="org.springframework.web.multipart.commons.CommonsMultipartResolver">
        <property name="maxUploadSize" value="100000000" />
    </bean>

    <bean id="xstreamMarshaller" class="org.springframework.oxm.xstream.XStreamMarshaller">
        <property name="autodetectAnnotations" value="true" />
        <!-- Set some properties to make the outputted xml look nicer -->
    </bean>

    <mvc:annotation-driven>
        <mvc:message-converters>
            <!-- Configure the XStream message converter -->
            <bean class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter">
                <property name="marshaller" ref="jaxbMarshaller2" />
                <property name="unmarshaller" ref="jaxbMarshaller2" />

                <property name="supportedMediaTypes">
                    <list>
                        <bean class="org.springframework.http.MediaType">
                            <constructor-arg index="0" value="application" />
                            <constructor-arg index="1" value="xml" />
                        </bean>
                        <bean class="org.springframework.http.MediaType">
                            <constructor-arg index="0" value="application" />
                            <constructor-arg index="1" value="x-www-form-urlencoded" />
                        </bean>
                    </list>
                </property>
            </bean>
        </mvc:message-converters>
    </mvc:annotation-driven>

    <bean id="jaxbMarshaller2" class="org.springframework.oxm.jaxb.Jaxb2Marshaller">

        <property name="classesToBeBound">
            <list>
                <value>com.auto.server.schema.ReceiveRequest</value>
                <value>com.auto.server.schema.ReceiveReply</value>

            </list>

        </property>
    </bean>

    <bean name="viewResolver" class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">

        <property name="ignoreAcceptHeader" value="true" />
        <property name="favorParameter" value="true" />
        <property name="favorPathExtension" value="true" />
        <!-- if no content type is specified, return json. -->
        <property name="defaultContentType" value="application/x-www-form-urlencoded" />

        <property name="mediaTypes">
            <map>
                <entry key="xml" value="application/x-www-form-urlencoded" />
            </map>
        </property>
        <property name="defaultViews">
            <list>
                <bean class="org.springframework.web.servlet.view.xml.MarshallingView">
                    <constructor-arg ref="jaxbMarshaller" />
                    <property name="modelKey" value="responseObject" />
                </bean>
            </list>
        </property>
    </bean>

    <!-- REST API controllers -->
    <context:component-scan base-package="com.auto.server.schema" />

Here is the controller part

@ResponseBody
    @RequestMapping(value = "/heartbeat", method = RequestMethod.POST, headers = { "content-type=application/x-www-form-urlencoded" }, consumes = "application/x-www-form-urlencoded;charset=UTF-8")
    public Object heartBeat(@RequestBody ReceiveRequest request) {

        ReceiveReply reply = new ReceiveReply();        
        return reply;
    }
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1 回答 1

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如果您总是要以 URL 编码样式获取数据，即content-type="application/x-www-form-urlencoded"

然后，当您通过以下代码收到它时，您可以在 bean 中简单地对其进行解码：

String decodedString = java.net.URLDecoder( inputString, "UTF-8" );
于 2013-11-08T14:40:03.103 回答