使用jaxb:version=“2.1”生成java Spring wsdl

Question

我试图生成一个wsdl，其中包含xsd文件中使用的jaxb绑定版本。我想要的最终结果是：

<wsdl:definitions xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
       jaxb:version="2.1" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:tns="http://schemas.yyy.co.za/zzz" targetNamespace="http://schemas.yyy.co.za/zzz">

而不是：

<wsdl:definitions xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:tns="http://schemas.yyy.co.za/zzz" targetNamespace="http://schemas.yyy.co.za/zzz">

我将Spring2.5.6与SpringWeb服务组件1.5.5一起使用。我怀疑问题出在我的Spring上下文设置中，它看起来像：

<bean id="ClientServices" class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition">
    <property name="portTypeName" value="ClientServicesPort" />
    <property name="locationUri" value="http://localhost/client/services/ClientServices" />
    <property name="schemaCollection" ref="ClientServicesSchemaCollection" />
    <property name="requestSuffix" value="RequestMessage"/>
    <property name="responseSuffix" value="ResponseMessage" />
</bean>

<bean id="ClientServicesSchemaCollection" class="org.springframework.xml.xsd.commons.CommonsXsdSchemaCollection">
    <description>
        This bean wrap all the xsds and inlines them as a one.
    </description>
    <property name="inline" value="true" />
    <property name="xsds">
        <list>
            <value>classpath*:xsd/sample.xsd</value>
        </list>
    </property>
</bean>

部分样本。xsd:

<?xml version="1.0" encoding="UTF-8"?> <xs:schema xmlns="http://schemas.yyy.co.za/zzz" xmlns:jaxb="http://java.sun.com/xml/ns/jaxb" jaxb:version="2.1"

提前谢谢