CODEWITHSUNDEEP
javaxmlfiledompath
Dreamy
Dreamy

Reputation: 67

How to open a XML doc in Java

So, I'm trying to open a XML document with java using docBuilder.parse(filepath)

This is what my code looks like:

    public void openXMLfile(String filepath) {
        try { 
            DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
            Document doc = docBuilder.parse(filepath);
        }
        catch(Exception e) {
            e.printStackTrace();
        }
    }

I have written the absolute path of my file which is:

C:\\Users\\"my User account"\\Desktop, being "my user account", replaced by the actual name of my user account

In my main function, which looks like this:

    public static void main(String[] args) {

        App aplication = new App();

        String filepath = "C:\\Users\\"my User account"\\Desktop";
        aplicacao.openXMLfile(filepath);
    }

However, I get the following line of erros:

java.net.MalformedURLException: unknown protocol: c
    at java.net.URL.<init>(Unknown Source)
    at java.net.URL.<init>(Unknown Source)
    at java.net.URL.<init>(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(Unknown Source)
    at javax.xml.parsers.DocumentBuilder.parse(Unknown Source)
    at Trabalho.App.openXMLfile(App.java:39)
    at Trabalho.App.main(App.java:70)

Could anyone clarify me in what I might be doing wrong? May it be the filepath itself?

Thank you very much in advance!

Upvotes: 1

Views: 72

Answers (1)

Chris
Chris

Reputation: 1804

The parse method takes a URI for the file location, so something along the lines of file:///c:/user/

This question gives examples - How should a file: URI corresponding to a Windows path name look like? particularly Paulo's comment.

Examples also at https://en.m.wikipedia.org/wiki/File_URI_scheme

Upvotes: 2

Related Questions