handle many XML files in a directory (java)

Question

I have managed a code to handle a file.
Now I want to use the same code to handle all the XML files which are located in a directory. Can someone tell me how can I declare the path and how to look for a loop. Thanks in advance

import org.xml.sax.SAXException;
import org.w3c.dom.*;
import javax.xml.parsers.*;
import java.io.IOException;

public class XmlReadWrite3 {
    public static void main(String[] args) {
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        try {
            DocumentBuilder db = dbf.newDocumentBuilder();
            Document doc = db.parse("C:/Users/Desktop/1381.xml");

            Element langs = doc.getDocumentElement();



            Element filename= getElement(langs, "Filename");
            Element beschreibung = getElement(langs, "Beschreibung");
            Element name = getElement(langs, "Name");
            Element ide = getElement(langs, "IDe");

            System.out.println("Filename:      " + filename.getTextContent() + "\n" + "Beschreibung:  "
                    + beschreibung.getTextContent() + "\n" + "Ersteller:     " + name.getTextContent() + "\n"
                    + "Pnummer: " + ide.getTextContent() + "\n\n");
        }catch (ParserConfigurationException pce) {
            pce.printStackTrace();
        } catch (SAXException se) {
            se.printStackTrace();
        } catch (IOException ioe) {
            ioe.printStackTrace();
        }
   }

   private static  Element getElement(Element langs, String tag){
       return (Element) langs.getElementsByTagName(tag).item(0);
   }
}

Possible duplicate of [Read all files in a folder](https://stackoverflow.com/questions/1844688/read-all-files-in-a-folder) — UnholySheep, Aug 24 '17 at 07:30

score 0 · Accepted Answer · answered Aug 24 '17 at 07:46

Hi you can use the Path and File classes to loop through a directory:

import org.xml.sax.SAXException;
import org.w3c.dom.*;

import javax.xml.parsers.*;
import java.io.IOException;
import java.nio.file.DirectoryStream;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

class XmlReadWrite3 {
public static void main(String[] args) {

    // here you enter the path to your directory.
    // for example: Path workDir = Paths.get("c:\\workspace\\xml-files")

    Path workDir = Paths.get("path/to/dir"); // enter the path to your xml-dir
    // the if checks whether the directory truly exists
    if (!Files.notExists(workDir)) {
        // this part stores all files withn the directory in a list
        try (DirectoryStream<Path> directoryStream = Files.newDirectoryStream(workDir)) {
            for (Path path : directoryStream) {


                DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
                try {
                    DocumentBuilder db = dbf.newDocumentBuilder();
                    Document doc = db.parse(path.toString());

                    Element langs = doc.getDocumentElement();


                    Element filename = getElement(langs, "Filename");
                    Element beschreibung = getElement(langs, "Beschreibung");
                    Element name = getElement(langs, "Name");
                    Element ide = getElement(langs, "IDe");

                    System.out.println("Filename:      " + filename.getTextContent() + "\n" + "Beschreibung:  "
                            + beschreibung.getTextContent() + "\n" + "Ersteller:     " + name.getTextContent() + "\n"
                                 + "Pnummer: " + ide.getTextContent() + "\n\n");
                    } catch (ParserConfigurationException pce) {
                        pce.printStackTrace();
                    } catch (SAXException se) {
                        se.printStackTrace();
                        } catch (IOException ioe) {
                            ioe.printStackTrace();
                         }
                 }
            } catch (Exception e) {
                System.out.println(e.getMessage())
                          }
        }
    }

    private static Element getElement(Element langs, String tag) {
        return (Element) langs.getElementsByTagName(tag).item(0);
    }
}

handle many XML files in a directory (java)

1 Answers1