使用XPath从WSDL文件中的导入(“xsd:import”)模式获取XML元素声明

时间:2013-01-08 21:59:49

标签: xml xpath wsdl xsd

我有一个问题已经让我紧张了一个星期。

我使用Netbeans 7.2.1编写了一个虚拟Web服务作为无状态EJB:

@WebService(serviceName = "WSPersonLookup")
@Stateless()
public class WSPersonData {

  @PersistenceContext(unitName="PUWSPersonData")
  private EntityManager em;

 /**
  * Web service operation
  */
  @WebMethod(operationName = "getPersonData")
  public Person getPersonData(@WebParam(name = "personId") final String personId) {
    return em.find(Person.class, personId);
  }
}

类“Person”只是映射到此表的JPA实体(我使用PostgreSQL作为DBMS):

CREATE TABLE person
(
  id character varying(10) NOT NULL DEFAULT ''::character varying,
  nit text NOT NULL DEFAULT ''::text,
  name text NOT NULL DEFAULT ''::text,
  address text NOT NULL DEFAULT ''::text,
  phone integer NOT NULL DEFAULT 0,
  creation_date timestamp with time zone NOT NULL DEFAULT now(),
  CONSTRAINT pk_person PRIMARY KEY (id)
)

生成的WSDL是:

<?xml version="1.0" encoding="UTF-8"?>
<definitions xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd" xmlns:wsp="http://www.w3.org/ns/ws-policy" xmlns:wsp1_2="http://schemas.xmlsoap.org/ws/2004/09/policy" xmlns:wsam="http://www.w3.org/2007/05/addressing/metadata" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:tns="http://ws.i2b.com/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="http://schemas.xmlsoap.org/wsdl/" targetNamespace="http://ws.i2b.com/" name="WSPersonLookup">
  <types>
    <xsd:schema>
      <xsd:import namespace="http://ws.i2b.com/" schemaLocation="http://pc-i2b:8080/WSPersonLookup/WSPersonData?xsd=1" />
    </xsd:schema>
  </types>
  <message name="getPersonData">
    <part name="parameters" element="tns:getPersonData" />
  </message>
  <message name="getPersonDataResponse">
    <part name="parameters" element="tns:getPersonDataResponse" />
  </message>
  <portType name="WSPersonData">
    <operation name="getPersonData">
      <input wsam:Action="http://ws.i2b.com/WSPersonData/getPersonDataRequest" message="tns:getPersonData" />
      <output wsam:Action="http://ws.i2b.com/WSPersonData/getPersonDataResponse" message="tns:getPersonDataResponse" />
    </operation>
  </portType>
  <binding name="WSPersonDataPortBinding" type="tns:WSPersonData">
    <soap:binding transport="http://schemas.xmlsoap.org/soap/http" style="document" />
    <operation name="getPersonData">
      <soap:operation soapAction="" />
      <input>
        <soap:body use="literal" />
      </input>
      <output>
        <soap:body use="literal" />
      </output>
    </operation>
  </binding>
  <service name="WSPersonLookup">
    <port name="WSPersonDataPort" binding="tns:WSPersonDataPortBinding">
      <soap:address location="http://pc-i2b:8080/WSPersonLookup/WSPersonData" />
    </port>
  </service>
</definitions>

XSD声明:

<?xml version='1.0' encoding='UTF-8'?>
<xs:schema xmlns:tns="http://ws.i2b.com/" xmlns:xs="http://www.w3.org/2001/XMLSchema"
           version="1.0" targetNamespace="http://ws.i2b.com/">
  <xs:element name="getPersonData" type="tns:getPersonData"/>
  <xs:element name="getPersonDataResponse" type="tns:getPersonDataResponse"/>
  <xs:element name="person" type="tns:person"/>
  <xs:complexType name="getPersonData">
    <xs:sequence>
      <xs:element name="personId" type="xs:string" minOccurs="0"/>
    </xs:sequence>
  </xs:complexType>
  <xs:complexType name="getPersonDataResponse">
    <xs:sequence>
      <xs:element name="return" type="tns:person" minOccurs="0"/>
    </xs:sequence>
  </xs:complexType>
  <xs:complexType name="person">
    <xs:sequence>
      <xs:element name="address" type="xs:string" minOccurs="0"/>
      <xs:element name="creationDate" type="xs:dateTime" minOccurs="0"/>
      <xs:element name="id" type="xs:string" minOccurs="0"/>
      <xs:element name="name" type="xs:string" minOccurs="0"/>
      <xs:element name="nit" type="xs:string" minOccurs="0"/>
      <xs:element name="phone" type="xs:int"/>
    </xs:sequence>
  </xs:complexType>
</xs:schema>

问题是:我需要访问上面的模式声明,在WSDL中导入:

 <types>
    <xsd:schema>
      <xsd:import namespace="http://ws.i2b.com/" schemaLocation="http://pc-i2b:8080/WSPersonLookup/WSPersonData?xsd=1" />
    </xsd:schema>
  </types>

使用XPath。我不允许直接访问架构。

有办法吗?如果没有,是否有人知道解决方法?

亲切的问候。

1 个答案:

答案 0 :(得分:1)

您不能仅使用XPath来获取主XML文档(WSDL)引用的辅助XML文档(在本例中为模式)。

使用XSLT,您可以使用document()函数执行此操作:

<xsl:variable name="schema" select="document(/soap:definitions/soap:types/xsd:schema/xsd:import/@schemaLocation)"/>

或者,不是尝试获取导入的模式,而是使用类似/soap:definitions/soap:types/xsd:schema的XPath来获取WSDL中的模式片段。这是一个有效的架构,导入http://pc-i2b:8080/WSPersonLookup/WSPersonData?xsd=1的架构 - 您应该可以直接使用它。

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