Question

我想提供一个接口给我在wsdl文件中定义的一些复杂类型。我尝试了以下操作，结果为：XPath evaluation of "wsdl:definitions" results in empty target node

这是我的binding.xml文件：

 <jaxb:bindings node="wsdl:definitions/wsdl:types"> 
   <jaxb:bindings node="xs:complexType[@name='test']"> 
      <inheritance:implements>com.package.Interface</inheritance:implements> 
    </jaxb:bindings>
   </jaxb:bindings>
</jaxb:bindings>

我通过maven cxf-codegen-plugin生成Java文件：

<wsdlOption>
  <wsdl>src/main/resources/wsdl/Service.wsdl</wsdl>
    <frontEnd>jaxws21</frontEnd>
      <bindingFiles>
        <bindingFile>src/main/resources/wsdl/binding.xml</bindingFile>
      </bindingFiles>
</wsdlOption>

所以我不知道是否必须将wsdl文件作为模式导入，因此我尝试将其添加到binding.xml中，如下所示：

 <jaxb:bindings schemaLocation="wsdl/Service#type2">
    ...
 </jaxb:bindings>

结果是：

"file:/home/xxx/xxx/src/src/main/resources/wsdl/Service#type2" is not a part of this compilation. Is this a mistake for .,../binding.xml"

说实话，我不知道是否需要将wsdl作为schemaLocation以及如何精确地引用它。

wsdl的某些部分：

<?xml version="1.0" encoding="utf-8"?><wsdl:definitions name="Service" targetNamespace="http://xxx/Service" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:wsap="http://schemas.xmlsoap.org/ws/2004/08/addressing/policy" xmlns:wsa10="http://www.w3.org/2005/08/addressing" xmlns:msc="http://schemas.microsoft.com/ws/2005/12/wsdl/contract" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/" xmlns:wsx="http://schemas.xmlsoap.org/ws/2004/09/mex" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" >
....
 <wsdl:types>
   <xs:schema elementFormDefault="qualified" targetNamespace="http://zzzz/201/" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:tns="http://zzzzz/201/">
     <xs:complexType name="test">

请让我知道是否需要其他信息，谢谢！

WSDL JAXB绑定空目标节点

0 个答案: