生成wsdl时,JAX-WS复制复杂类型

时间:2013-04-18 08:49:58

标签: java web-services jaxb wsdl jax-ws

我正在开发一个Web服务,其中有几种方法将相同的复杂数据类型作为输入。数据类型具有JAXB注释和setter和getter,Web服务类具有JAX-WS注释。

我的service.java文件的模板:

@WebService(serviceName = "ServiceWS")
public class SericeWS {

private static ServiceIF serviceImpl;

static {
    serviceImpl = new ServiceImpl();
}

public Result Method1(Credentials credentials) {
        @WebParam(name = "credentials") Credentials credentials) { 

    return serviceImpl.Method1(credentials);
}

    public Result Method2(Credentials credentials) {
        @WebParam(name = "credentials") Credentials credentials) { 

    return serviceImpl.Method2(credentials);
}

}

编辑:我的Credentials.java文件:

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
    "name",
    "password"
})
@XmlRootElement(name = "Credentials")
public class Credentials implements MyBean {

    @XmlElement(required = true)
    protected String name;
    @XmlElement(required = true)
    protected String password;

    /**
     * Gets the value of the name property.
     * 
     * @return The name property of the credentials
     *     
     */
    public String getName() {
        return name;
    }

    /**
     * Sets the value of the name property.
     * 
     * @param value The name property of the credentials
     *     
     */
    public void setName(String value) {
       this.name = value;
    }

    /**
     * Gets the value of the password property.
     * 
     * @return The password property of the credentials
     *     
     */
    public String getPassword() {
        return password;
    }

    /**
     * Sets the value of the password property.
     * 
     * @param value The password property of the credentials
     *       
     */
    public void setPassword(String password) {
        this.password = password;
    }  
}

该服务部署在Tomcat中,并且自动生成wsdl。当使用wsimport生成客户端存根时,我会获得Credentials类型(Credentials,Method1.Credentials和Method2.Credentials)的重复生成,即每个方法的不同(内部)类。

生成wsdl和schema时似乎存在问题:

<xs:schema xmlns:tns="http://service.my.package.com/"            
xmlns:xs="http://www.w3.org/2001/XMLSchema" version="1.0" 
targetNamespace="http://service.my.package.com/">
<xs:element name="Credentials">
 <xs:complexType>
  <xs:sequence>
   <xs:element name="name" type="xs:string"/>
   <xs:element name="password" type="xs:string"/>
  </xs:sequence>
 </xs:complexType>
</xs:element>  
....
<xs:complexType name="getLockBoxKeys">
 <xs:sequence>
  <xs:element name="credentials" minOccurs="0">
   <xs:complexType>
    <xs:sequence>
     <xs:element name="name" type="xs:string"/>
     <xs:element name="password" type="xs:string"/>
    </xs:sequence>
  </xs:complexType>
 </xs:element>    
 .....

如何才能完成所有这些工作,以便我只有一个凭据定义?我对Web服务,JAX-WS和JAXB都很陌生,所以我不确定我的注释是否合适。

非常感谢任何帮助。

2 个答案:

答案 0 :(得分:3)

我不会给出解释所有规则的完整答案,因为我不记得/了解所有规则。

但我认为,如果您将name元素添加到@XMLType注释中,您将获得所需内容(或者至少更进一步)。

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "Credentials", propOrder = {
    "name",
    "password"
})
@XmlRootElement(name = "Credentials")
public class Credentials {
顺便说一句,您原来的service.java文件似乎没有太干净地粘贴(我认为有一些不正确的括号)使得任何人都难以重新创建。

答案 1 :(得分:1)

在MyCredentials.java文件中,@ XmlRootElement(name =“Credentials”)的名称以大写字母C开头,而@WebParam(name =“credentials”)则不是。也许是因为这个?