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Home/ Questions/Q 8366205
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T12:51:29+00:00 2026-06-09T12:51:29+00:00

I want to upload an xml file via JSF Tomahawk and then parse it.

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I want to upload an xml file via JSF Tomahawk and then parse it.

I have created the following form:

<!DOCTYPE html>
<html lang="en"
    xmlns="http://www.w3.org/1999/xhtml"
    xmlns:f="http://java.sun.com/jsf/core"
    xmlns:h="http://java.sun.com/jsf/html"
    xmlns:t="http://myfaces.apache.org/tomahawk"
    xmlns:ui="http://java.sun.com/jsf/facelets">
    <h:head>
       <title>Import</title>
    </h:head>
    <h:body>
       <h1>Import</h1>
       <h:form enctype="multipart/form-data">
           XML Datei mit den Kursdaten: 
           <t:inputFileUpload value="#{installationBean.uploadedFile}" />
           <h:commandButton value="submit" action="#{installationBean.submit}" />
           <h:messages />
       </h:form>
   </h:body> 

This is the bean:

package installation;

import java.io.IOException;
import javax.ejb.EJB;
import javax.ejb.Stateless;
import javax.ejb.LocalBean;
import javax.faces.bean.ManagedBean;
import javax.faces.bean.ViewScoped;
import javax.faces.context.FacesContext;
import org.apache.commons.io.FilenameUtils;
import org.apache.myfaces.custom.fileupload.UploadedFile;


@Stateless
@LocalBean
@ManagedBean
@ViewScoped
public class InstallationBean {

    private UploadedFile uploadedFile;


    public UploadedFile getUploadedFile() {
       return uploadedFile;
    }

    public void setUploadedFile(UploadedFile value) {
        uploadedFile = value;
    }


    public void submit()throws IOException {
        String fileName = FilenameUtils.getName(uploadedFile.getName());
        String contentType = uploadedFile.getContentType();
        byte[] bytes = uploadedFile.getBytes();

        // Parse xml
    }
}

The file that i upload is an xml file.
How can I parse the xml?
In my code above I have got a byte array after uploading the file.
How can I convert this to an xml file and parse it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T12:51:30+00:00Added an answer on June 9, 2026 at 12:51 pm

    You could use the following to parse this byte[] into a org.w3c.dom.Document.

       InputStream is = new ByteArrayInputStream(byteArray);  
       DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();  
       DocumentBuilder builder = factory.newDocumentBuilder();  
       Document xml = builder.parse(is);
    
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