XML（带命名空间）到对象解组

Question

我从 Web 服务调用中得到了以下响应，我尝试使用 JAXB 对其进行解组以将其映射到 Java 类。这样做时我遇到了解组异常。

<?xml version="1.0" encoding="UTF-8"?>
<ns0:QueryByLNResponse xmlns:ns0="UIS_CTMPeople_WS" xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <ns0:getListValues>
        <ns0:First_Name>Pradeep</ns0:First_Name>
        <ns0:Internet_E-mail/>
        <ns0:ManagersName/>
        <ns0:Person_ID>PPL1</ns0:Person_ID>
        <ns0:Last_Name>Srinivasa Reddy</ns0:Last_Name>
        <ns0:Full_Name>Pradeep M Srinivasa Reddy</ns0:Full_Name>
    </ns0:getListValues>
    <ns0:getListValues>
        <ns0:First_Name>Geeth </ns0:First_Name>
        <ns0:Internet_E-mail>bas@yahoo.com</ns0:Internet_E-mail>
        <ns0:ManagersName/>
        <ns0:Person_ID>PPL2</ns0:Person_ID>
        <ns0:Last_Name>Srinivasan</ns0:Last_Name>
        <ns0:Full_Name>Geeth  Srinivasan</ns0:Full_Name>
    </ns0:getListValues>
</ns0:QueryByLNResponse>

我尝试使用解组上述代码

public static Object xmlToObject(String xml, Class... objClass) throws Exception {
    JAXBContext jc = JAXBContext.newInstance(objClass);
    final Unmarshaller unmarshaller = jc.createUnmarshaller();
    return unmarshaller.unmarshal(new StringReader(xml.toString()));
}

它抛出以下错误

javax.xml.bind.UnmarshalException: unexpected element (uri:"UIS_CTMPeople_WS", local:"QueryByLNeResponse"). Expected elements are (none)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.handleEvent(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportUnexpectedChildElement(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext$DefaultRootLoader.childElement(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext._startElement(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.startElement(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.SAXConnector.startElement(Unknown Source)
    at org.apache.xerces.parsers.AbstractSAXParser.startElement(Unknown Source)
    at org.apache.xerces.impl.XMLNSDocumentScannerImpl.scanStartElement(Unknown Source)
    at org.apache.xerces.impl.XMLNSDocumentScannerImpl$NSContentDispatcher.scanRootElementHook(Unknown Source)
    at org.apache.xerces.impl.XMLDocumentFragmentScannerImpl$FragmentContentDispatcher.dispatch(Unknown Source)
    at org.apache.xerces.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
    at org.apache.xerces.parsers.XML11Configuration.parse(Unknown Source)
    at org.apache.xerces.parsers.XML11Configuration.parse(Unknown Source)
    at org.apache.xerces.parsers.XMLParser.parse(Unknown Source)
    at org.apache.xerces.parsers.AbstractSAXParser.parse(Unknown Source)
    at org.apache.xerces.jaxp.SAXParserImpl$JAXPSAXParser.parse(Unknown Source)
    at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(Unknown Source)

我如何使用 JAXB ( xml to object ) 解组它。

Answer 1

以下是一些应该有帮助的项目：

命名空间

您应该@XmlSchema在package-info类上使用注释来指定命名空间限定。下面是一个示例，您需要更改包名称以匹配您的模型。

包信息.java

@XmlSchema(
    namespace = "UIS_CTMPeople_WS",
    elementFormDefault = XmlNsForm.QUALIFIED)
package example;

import javax.xml.bind.annotation.XmlNsForm;
import javax.xml.bind.annotation.XmlSchema;

想要查询更多的信息

• http://blog.bdoughan.com/2010/08/jaxb-namespaces.html

根元素

您似乎没有使用@XmlRootElement( 或@XmlElementDecl)映射任何类。我希望你有如下内容：

QueryByLNResponse

package example;

@XmlRootElement(name="QueryByLNResponse")
public class QueryByLNResponse {
}

或者，您可以使用带Class参数的解组方法之一来指定要解组到的类：

return unmarshaller.unmarshal(xml, QueryByLNResponse.class)

想要查询更多的信息

• http://blog.bdoughan.com/2012/07/jaxb-and-root-elements.html

表现

在相同的代码中，JAXBContext每次执行解组时都会创建一个新代码。 JAXBContext是一个线程安全对象，可以创建一次并重用以提高性能。