java解析xml

[yong_online]

<?xml version="1.0" encoding="GBK"?>
<test a="a" b="b" >
<temp a="4"
b="18"
c="0"
d="j2ee"
e =""
f =""
g =""
h="java"
i="java解析xml"
j=""
k="java解析xml"
l= ""
m="">
<file id= "7" n= "" x="0" y="4" z="0"></file>
</temp>
<temp a="12"
b="43"
c="1"
d="C#"
e =""
f =""
g =""
h=".net"
i="java解析xml"
j=""
k="java解析xml"
l= ""
m="">
<file id= "54" n= "" x="0" y="48" z="12"></file>
</temp>
</test>

我想获取父节点test中 a的属性值,b的属性值.
子节点temp中a的属性值,b的属性值,file节点中的 id值,z的值.
请问该如何实现.谢谢

[kokobox]

lz 用dom4j吧，看看他的api

使用很方便的

[灿烂星河]

你可以使用JDOM或DOM4J这样的第三方包。

[hmsuccess]

当然你也可以用xquery

[milo-higgs]

JDOM很好用的。也很简单。

[joejoe1991]

XPATH也很快吖。

import javax.xml.xpath.*;
import javax.xml.parsers.*;
import org.w3c.dom.*;
import java.io.*;

public class a {
    public static void main(String[] args) {
        try {
            XPathFactory fac = XPathFactory.newInstance();
            XPath pat = fac.newXPath();

            DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
            DocumentBuilder builder = factory.newDocumentBuilder();

            Document doc = builder.parse(new File("c:/a.xml"));

            NodeList list = (NodeList) pat.evaluate("/test/@a|/test/@b", doc,
                    XPathConstants.NODESET);
            print(list, "test");

            list = (NodeList) pat.evaluate("//temp/@a|//temp/@b", doc,
                                           XPathConstants.NODESET);
            print(list, "temp");

            list = (NodeList) pat.evaluate("//file/@id|//file/@z", doc,
                                           XPathConstants.NODESET);
            print(list, "file");

        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }

    public static void print(NodeList list, String ele) {
        System.out.println(ele + ":");
        for (int i = 0; i < list.getLength(); i++) {

            System.out.println("\t" + list.item(i).getNodeName() + ":" +
                               list.item(i).getNodeValue());
        }

    }
}

[kevinchj]

jdom比较简单的。而且对于这种少量数据的，效率也不错！

[milo-higgs]

[Quote=引用 4 楼 joejoe1991 的回复:]
XPATH也很快吖。

Java codeimport javax.xml.xpath.*;
import javax.xml.parsers.*;
import org.w3c.dom.*;
import java.io.*;

public class a {
public static void main(String[] args) {
try {
XPathFactory fac = XPathFactory.newInstance();
XPath pat = fac.newXPath();

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
…
[/Quote]

XPATH很强大，不过比jdom要多学习一点东西拉

[qusic]

学习··

[guohaijiao]

dom4j

[zhao_believe]

dom4j 关注 .学习

[灿烂星河]

还是JDOM好用点，呵呵

[LexChen]

这样简单的东西，建议楼主还是从JDK里面的DOM和SAX入手，打点基础知识，再用JDOM之类的东东，就学习而言来的要好！