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java - 如何从xml文件中获取属性?

转载 作者:行者123 更新时间:2023-12-01 15:24:07 25 4
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我的 Xml 文件如下所示:

    <?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE pointList SYSTEM "point.dtd">
<pointList>
<point unit="mm">
<x>2</x>
<y>3</y>
</point>

<point unit="cm">
<x>9</x>
<y>3</y>
</point>

<point unit="px">
<x>4</x>
<y>7</y>
</point>

</pointList>

当我尝试获取标记点的属性时:

import java.io.File;
import java.io.IOException;


import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;


public class TryXml {
public static void main (String [] args) {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = null;
Document doc = null;
try {
builder = factory.newDocumentBuilder();
}
catch (ParserConfigurationException e){
System.out.println(e.getMessage());
}
File f = new File("p1.xml");
try {
doc=builder.parse(f);
}
catch (SAXException e){
e.printStackTrace();
}
catch (IOException e){
e.printStackTrace();
}

Element root = doc.getDocumentElement();
System.out.println(root.getTagName());

System.out.println("****************");


NodeList nList = root.getChildNodes();
for (int i=0;i<nList.getLength();i++) {
if(nList.item(i) instanceof Element)

System.out.println(nList.item(i).getAttributes());
}

}
}

我得到的只是类似地址的内容:

com.sun.org.apache.xerces.internal.dom.AttributeMap@3214512e
com.sun.org.apache.xerces.internal.dom.AttributeMap@53ddbcb1
com.sun.org.apache.xerces.internal.dom.AttributeMap@28f337b

任何人都可以给我一个关于如何获取 point 的属性以及其他内部标签的提示吗?

最佳答案

您可以将 println 替换为以下内容:

System.out.println(((Element) nList.item(i)).getAttribute("unit"));

这应该为您提供当前元素的“unit”属性。

关于java - 如何从xml文件中获取属性?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10431450/

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