使用java泛型的JAXB解組

Question

我想從使用JAXB的遺留系統解開XML文檔。我有一個XML結構如下：

<response> 
    <id>000000</id> 
    <results> 
     <result> 
<!-- Request specific xml content --> 
      <year>2003</year> 
      <title>Lorem Ipsum</title> 
      <items> 
       <item>I1</item> 
       <item>I2</item> 
      </items> 
     </result> 
     <result> 
      <year>2007</year> 
      <title>Dolor sit amet</title> 
      <items> 
       <item>K1</item> 
       <item>K2</item> 
      </items> 
     </result> 
    </results> 
</response> 

通過<result>標籤中指定將根據我的要求更改部分內的標籤。由於內容可能會改變我決定用結果項仿製藥，我已經準備了我的Java bean與註解如下：

JAXBContext context = JAXBContext.newInstance(XResponse.class, X1Result.class); 
Unmarshaller um = context.createUnmarshaller(); 
XResponse<X1Result> response = (XResponse<X1Result>) um.unmarshal(xmlContent); 

List<X1Result> results = unmarshal.getResults(); 
for (X1Result object : results) { 
    System.out.println(object.getClass()); 
} 

：

// imports here 
@XmlRootElement(name="response") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class XResponse<T>{ 
    private String id; 

    @XmlElementWrapper(name="results") 
    @XmlElement(name="result") 
    private List<T> results; 

// setters and getters 
} 

... 

@XmlRootElement(name="result") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class X1Result{ 
    private String year; 
    private String title; 
    @XmlElementWrapper(name="items") 
    @XmlElement(name="item") 
    private List<String> items; 

// setters and getters 
} 
... 

我試圖通過下面的代碼解組XML文檔我在解組期間遇到問題，無法將列表項目轉換爲X1Result類。相反，它使用org.apache.xerces.dom.ElementNSImpl。

我該怎麼做才能讓JAXB Unmarshaller使用X1Result類？

在此先感謝

Answer 1

我認爲你應該使用繼承而不是泛型。給定一個這樣的XML：

<?xml version="1.0" encoding="UTF-8"?> 
<response xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> 
    <id>000000</id> 
    <results> 
     <result xsi:type="X1Result"> 
      <year>2003</year> 
      <title>Lorem Ipsum</title> 
      <items> 
       <item>I1</item> 
       <item>I2</item> 
      </items> 
     </result> 
     <result xsi:type="X1Result"> 
      <year>2007</year> 
      <title>Dolor sit amet</title> 
      <items> 
       <item>K1</item> 
       <item>K2</item> 
      </items> 
     </result> 
    </results> 
</response> 

您可以動態綁定您的<result>條目。你有一個頂層類型：

@XmlAccessorType(XmlAccessType.FIELD) 
@XmlType(name = "XResult") 
@XmlSeeAlso({ 
    X1Result.class 
})public abstract class XResult { 

} 

而且你必須實現類：

@XmlAccessorType(XmlAccessType.FIELD) 
@XmlType(name = "X1Result") 
public class X1Result extends XResult { 
    @XmlElement(name = "year") 
    private String year; 

    @XmlElement(name = "title") 
    private String title; 

    @XmlElementWrapper(name = "items") 
    @XmlElement(name = "item") 
    private List<String> items; 
    ... 
} 

使用您XResponse類的頂級類型：

@XmlRootElement(name = "response") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class XResponse { 
    @XmlElement(name = "id") 
    private String id; 

    @XmlElementWrapper(name = "results") 
    @XmlElement(name = "result") 
    private List<XResult> results; 
    ... 
} 

你也可以解組使用頂級類型：

context = JAXBContext.newInstance(XResponse.class, XResult.class); 
Unmarshaller unmarshaller = context.createUnmarshaller(); 
XResponse response = (XResponse) unmarshaller.unmarshal(new File("testfile.xml")); 

List<XResult> results = response.getResults(); 
for (XResult object : results) { 
    System.out.println(object.getClass()); 
}