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How to convert XML file into String?

Hi,

When I run below code it is not reading file, the document object is null. What I want to do is to read a file and then convert it into sgtring.

File file = new File("C:\\Product\\abcd.xml");
       
        if (file.exists()) {
              
              System.out.println("File Exits   " + file.getAbsolutePath());
        }
       
    String content = "";
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    try {
        DocumentBuilder builder = factory.newDocumentBuilder();
       // Document document = builder.parse(new FileInputStream(file));
        Document document = builder.parse(new FileInputStream(file));


        TransformerFactory tranFactory = TransformerFactory.newInstance();
        Transformer aTransformer = tranFactory.newTransformer();
        aTransformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION,
                "yes");
        aTransformer.setOutputProperty(OutputKeys.INDENT, "yes");
        Source src = new DOMSource(document);
        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        Result dest = new StreamResult(stream);
        aTransformer.transform(src, dest);
        content = stream.toString();
       
        System.out.println("String is ....");
       
    } catch (ParserConfigurationException e) {
        //logger.error(e.getMessage(), e);
    } catch (SAXException e) {
        //logger.error(e.getMessage(), e);
    } catch (IOException e) {
       // logger.error(e.getMessage(), e);
    } catch (TransformerConfigurationException e) {
        //logger.error(e.getMessage(), e);
    } catch (TransformerException e) {
       // logger.error(e.getMessage(), e);
    }
    return content;
   }
0
AliAjoo
Asked:
AliAjoo
2 Solutions
 
mccarlIT Business Systems Analyst / Software DeveloperCommented:
There doesn't appear to be any issues with the code, other than the poor error handling.
When I run below code it is not reading file, the document object is null
How do you know that "document" is null? Are you running this in a debugger?


Ahh, or by the above do you actually mean that the return value of this method (ie. the "content" variable in the above code) is null? That is more likely the case, but for us (or even you) to figure it out, you will have to STOP ignoring any exceptions that might be thrown, as you are currently doing, and at the very least print the stack traces so that you can see what is going on.

Add the following line inside ALL the "catch" blocks in the above code, and run it again...
e.printStackTrace();

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AliAjooAuthor Commented:
thnaks for your resposne.

I am attaching a sample XML file. It is stored under C:\\Product\\student.xml. When I run the above program it shoud print xml contents?

Ali
student.xml
0
 
awking00Commented:
Are you using Eclipse? Do you have it?
0
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AliAjooAuthor Commented:
Yes, I am using eclipse and also the file is present in the directory
0
 
CEHJCommented:
http://technojeeves.com/joomla/index.php/free/137-serialize-xml-to-file-in-java

With variant

public static String saveToString(Document doc) {
        StringWriter out = null;

        try {
            DOMImplementationLS domImplementation = (DOMImplementationLS) doc.getImplementation();
            LSSerializer lsSerializer = domImplementation.createLSSerializer();
            lsSerializer.getDomConfig()
                        .setParameter("format-pretty-print", Boolean.TRUE);

            LSOutput output = domImplementation.createLSOutput();
            output.setEncoding(System.getProperty("file.encoding"));
            out = new StringWriter();
            output.setCharacterStream(out);
            lsSerializer.write(doc, output);

            return out.toString();
        } finally {
            if (out != null) {
                try {
                    out.close();
                } catch (IOException e) { /* ignore */
                }
            }
        }
    }

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AliAjooAuthor Commented:
It was really helpful
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