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Java 修改XML文件_Java_Xml - Fatal编程技术网

Java 修改XML文件

Java 修改XML文件,java,xml,Java,Xml,例如,我有一个XML文件: <http auto-config="true"> <intercept-url access="ROLE_ADMIN,ROLE_USER" pattern="/"/> <intercept-url access="ROLE_ADMIN,ROLE_USER" pattern="/index*"/> <intercept-url access="ROLE_ADMIN" patter

例如,我有一个XML文件:

<http auto-config="true">
        <intercept-url access="ROLE_ADMIN,ROLE_USER" pattern="/"/>
        <intercept-url access="ROLE_ADMIN,ROLE_USER" pattern="/index*"/>
        <intercept-url access="ROLE_ADMIN" pattern="/page1"/>
        <intercept-url access="ROLE_ADMIN" pattern="/page2"/>
        <intercept-url access="ROLE_ADMIN" pattern="/page33"/>

</http>

我想更改此行元素“access”的文本:

<intercept-url access="ROLE_ADMIN" pattern="/page33"/>

因此,我创建了以下代码:

import java.io.File;
import java.io.IOException;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
 import org.w3c.dom.Document;
 import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;

import com.sun.xml.internal.ws.org.objectweb.asm.Attribute;

public class ChangeXMLFile {

    public static void main(String argv[]) {

       try {
        String filepath = "D:\\Myfile.xml";
        DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
        Document doc = docBuilder.parse(filepath);


        Node company = doc.getFirstChild();

        Node staff = doc.getElementsByTagName("intercept-url").item(0);

        NodeList list = staff.getChildNodes();

        for (int i = 0; i < list.getLength(); i++) {

                   Node node = list.item(i);


           if ("pattern".equals(node.getNodeName())) {
                if(("/page33").equals(node.getTextContent())){


                    NamedNodeMap attr = staff.getAttributes();
                    Node nodeAttr = attr.getNamedItem("access");
                    nodeAttr.setTextContent("ROLE_ANONYM");
                }

           }




        }


        TransformerFactory transformerFactory = TransformerFactory.newInstance();
        Transformer transformer = transformerFactory.newTransformer();
        DOMSource source = new DOMSource(doc);
        StreamResult result = new StreamResult(new File(filepath));
        transformer.transform(source, result);

        System.out.println("Done");

       } catch (ParserConfigurationException pce) {
        pce.printStackTrace();
       } catch (TransformerException tfe) {
        tfe.printStackTrace();
       } catch (IOException ioe) {
        ioe.printStackTrace();
       } catch (SAXException sae) {
        sae.printStackTrace();
       }
    }
}
导入java.io.File;
导入java.io.IOException;
导入javax.xml.parsers.DocumentBuilder;
导入javax.xml.parsers.DocumentBuilderFactory;
导入javax.xml.parsers.parserConfiguration异常;
导入javax.xml.transform.Transformer;
导入javax.xml.transform.TransformerException;
导入javax.xml.transform.TransformerFactory;
导入javax.xml.transform.dom.DOMSource;
导入javax.xml.transform.stream.StreamResult;
导入org.w3c.dom.Document;
导入org.w3c.dom.NamedNodeMap;
导入org.w3c.dom.Node;
导入org.w3c.dom.NodeList;
导入org.xml.sax.SAXException;
导入com.sun.xml.internal.ws.org.objectweb.asm.Attribute;
公共类ChangeXMLFile{
公共静态void main(字符串argv[]){
试一试{
String filepath=“D:\\Myfile.xml”;
DocumentBuilderFactory docFactory=DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder=docFactory.newDocumentBuilder();
Document doc=docBuilder.parse(文件路径);
节点公司=doc.getFirstChild();
节点staff=doc.getElementsByTagName(“截取url”)。项(0);
NodeList list=staff.getChildNodes();
对于(int i=0;i

但是我没有零钱。如何让程序读取xml文件的每一行,并根据“模式”的内容文本更改“访问”属性

    Node staff = doc.getElementsByTagName("intercept-url").item(0);

    NodeList list = staff.getChildNodes();
列表为空:staff是您的第一个“拦截url”节点,因此当您尝试获取其子节点时,结果列表为空

请尝试以下方式:

    NodeList staffList = doc.getElementsByTagName("intercept-url");

你的层次结构都错了。请尝试以下方法:

        DocumentBuilderFactory docFactory = DocumentBuilderFactory
                .newInstance();
        DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
        Document doc = docBuilder.parse(...;

        NodeList list = doc.getElementsByTagName("intercept-url");

        for (int i = 0; i < list.getLength(); i++) {
            Element node = (Element) list.item(i);
            Attr pattern = node.getAttributeNode("pattern");
            if (pattern != null && pattern.getValue().equals("/page33")) {
                node.setAttribute("access", "ROLE_ANONYM");
            }
        }

        TransformerFactory transformerFactory = TransformerFactory
                .newInstance();
        Transformer transformer = transformerFactory.newTransformer();
        DOMSource source = new DOMSource(doc);
        StreamResult result = new StreamResult(...);
        transformer.transform(source, result);
DocumentBuilderFactory docFactory=DocumentBuilderFactory
.newInstance();
DocumentBuilder docBuilder=docFactory.newDocumentBuilder();
documentdoc=docBuilder.parse(。。。;
NodeList list=doc.getElementsByTagName(“截取url”);
对于(int i=0;i
heyyy我是第一个得到同样好答案的人(甚至更早),为什么你的答案被验证了?我提出了一个声明:)这就是生活,残酷而不公平:-)但是你忽略了属性节点不是任何人的孩子这一部分,因此,它不会与现有的属性逻辑一起工作。好吧,这是真的,我甚至没有试图理解为什么以及是否可以通过NamedNodeMap使用这些属性。主要问题很明显,我就到此为止。