我创建了一个XML文件和DTD,可以在HERE找到。
我已经编写了一个代码,但它一直工作到一个级别,然后它无法正常工作。我还创建了一些对象来存储xml文件的值。但是我只能遍历到xml的sheet
标签,然后它才能正常工作。
Recon recon = new Recon();
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(configFile);
doc.getDocumentElement().normalize();
System.out.println("Root Element : " + doc.getDocumentElement().getNodeName());
String outputPath = doc.getDocumentElement().getAttribute("outputPath");
String withCompareFilePath = doc.getDocumentElement().getAttribute("withCompareFile");
String toCompareFilePath = doc.getDocumentElement().getAttribute("toCompareFile");
recon.setOutputPath(outputPath);
recon.setToCompareFile(new File(toCompareFilePath));
recon.setWithCompareFile(new File(withCompareFilePath));
NodeList sheetNodeList = doc.getElementsByTagName("sheet");
List<ReconSheet> reconSheets = new ArrayList<ReconSheet>();
for(int i = 0; i< sheetNodeList.getLength() ; i++) {
Node tempNode = sheetNodeList.item(i);
ReconSheet reconSheet = new ReconSheet();
NamedNodeMap attMap = tempNode.getAttributes();
Node sheetNode = attMap.getNamedItem("sheetNumber");
String sheetNumber = sheetNode.getNodeValue();
reconSheet.setSheetNumber(Integer.parseInt(sheetNumber));
NodeList list = tempNode.getChildNodes();
for(int j = 0; j< list.getLength(); j++) {
Node inNode = list.item(j);
System.out.println(inNode);
}
}
答案 0 :(得分:1)
注意:我是EclipseLink JAXB (MOXy)主管,是JAXB 2 (JSR-222)专家组的成员。
您可以使用JAXB实现将XML直接映射到域模型。 JAXB需要Java SE 5,Java SE 6中包含JAXB实现。
<强>侦察强>
您的Recon
课程类似于:
package forum7673323;
import java.io.File;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Recon {
@XmlAttribute
private String outputPath;
@XmlAttribute
@XmlJavaTypeAdapter(FileAdapter.class)
private File withCompareFile;
@XmlAttribute
@XmlJavaTypeAdapter(FileAdapter.class)
private File toCompareFile;
@XmlElement(name="sheet")
private List<ReconSheet> reconSheets;
}
<强> ReconSheet 强>
package forum7673323;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
@XmlAccessorType(XmlAccessType.FIELD)
public class ReconSheet {
@XmlAttribute
int sheetNumber;
}
<强> FileAdapter 强>
由于JAXB实现无法直接与java.io.File
对象交互,因此我们将使用JAXB适配器来处理此转换。使用Recon
注释在@XmlJavaTypeAdapter
类中指定了此适配器的使用:
package forum7673323;
import java.io.File;
import javax.xml.bind.annotation.adapters.XmlAdapter;
public class FileAdapter extends XmlAdapter <String, File>{
@Override
public String marshal(File file) throws Exception {
if(null == file) {
return null;
}
return file.getPath();
}
@Override
public File unmarshal(String path) throws Exception {
return new File(path);
}
}
<强>演示强>
package forum7673323;
import java.io.File;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.Marshaller;
import javax.xml.bind.Unmarshaller;
public class Demo {
public static void main(String[] args) throws Exception {
JAXBContext jc = JAXBContext.newInstance(Recon.class);
File xml = new File("src/forum7673323/input.xml");
Unmarshaller unmarshaller = jc.createUnmarshaller();
Recon recon = (Recon) unmarshaller.unmarshal(xml);
Marshaller marshaller = jc.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(recon, System.out);
}
}
<强>输出强>
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<recon toCompareFile="h:\work\two.xls" withCompareFile="h:\work\one.xls" outputPath="h:/work">
<sheet sheetNumber="1"/>
</recon>
答案 1 :(得分:0)
我可能错了,但不是getAttributes()
方法负责带来标签的属性而不是子元素?