How to modify xml node values using java? (Not by tag name)

Question

i'm trying to modify some parameters of a xml file from a java file. In all the posts i found people modify these parameters searching for it by tag name, but in my case it doesn't work.

An example of my xml file is:

<?xml version="1.0" encoding="UTF-8"?>
<comp>
    <params>
        <param name="id">1</param>
        <param name="type">aaaa</param>
        <param name="bus">123</param>
    </params>
</comp>

How i can access the parameters by "name"?

Finally my code in java is:

public void modifyXml(){
    String path = "C:\\Workspace\\Project\\data.xml";
    DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
    Document doc = docBuilder.parse(path);
    Node comp = doc.getFirstChild();
    Node params = doc.getElementsByTagName("params").item(0);
    loadParams(params);
}

private void loadParams(Node params) {
    NodeList list = params.getChildNodes();
    for (int i = 0; i < list.getLength(); i++) {
        Node url =   list.item(i);
        String nodeName = url.getNodeName();
        NamedNodeMap attributes = url.getAttributes();

        //System.out.println(attributes.item(i));
        if (nodeName.equalsIgnoreCase("param") && url.getFirstChild()!=null) {
            ServerProperties.getExternalParamsMap().put(attributes.getNamedItem("name").getNodeValue(), url.getFirstChild().getNodeValue());
            System.out.println((String) url.getFirstChild().getNodeName());
            System.out.println(url.getFirstChild().getNodeValue());
         }
    }
}

It prints me the node's values but not the parameter name. Parameters name appear like: "#text"

Thanks!

Regards!!

Eric P.

Don't write it by hand with DOM. Use XPath with these expressions: //param[@name="id"], //param[@name="type"], //param[@name="bus"], etc. — user207421
– user207421, Commented Aug 18, 2016 at 6:39

vtd-xml-author · Accepted Answer · 2016-08-18 06:57:01Z

1

You should use XPath to simplify coding... below is the code based on XPath and vtd-xml.

import com.ximpleware.*;

public class accessName {

    public static void main(String[] args) throws VTDException{
        VTDGen vg = new VTDGen();
        if (!vg.parseFile("input.xml", false))
            return;
        VTDNav vn = vg.getNav();
        AutoPilot ap = new AutoPilot(vn);
        ap.selectXPath("/param[@name='id']/text()");
        int i=0;
        while((i=ap.evalXPath())!=-1){
            System.out.println("text value "+vn.toString(i));
        }
    }
}

answered Aug 18, 2016 at 6:57

vtd-xml-author

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Eric · Accepted Answer · 2016-08-19 09:05:22Z

0

Now it works for me:

    String path = "C:\\Workspace\\Project\\data.xml";
    File file = new File(path);
    DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory
            .newInstance();
    javax.xml.parsers.DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();
    Document document = documentBuilder.parse(file);
    Node id = (Node) document.getElementsByTagName("param").item(0);
    id.setTextContent("xxxxxxx");
    System.out.println(usr.getTextContent());

This will modify the "id" parameter

answered Aug 19, 2016 at 9:05

Eric

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