How to turn a string into an XML file? [duplicate] - java

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Closed 10 years ago.
Possible Duplicate:
Writing to a XML file in Java
I have below XML text as a string.
<someNode>
<id>A124</id>
<status>404</status>
<message>No data</message>
</someNode>
I have above XML data as a String. Is it possible to convert the text into an XML file and archive the generated XML file?
Thanks!

DocumentBuilderFactory dbfac = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = dbfac.newDocumentBuilder();
Document doc = docBuilder.parse(new InputSource(new StringReader(theString)));

public class StringToXML {
public static void main(String[] args) {
String xmlString = "<?xml version=\"1.0\" encoding=\"utf-8\"?><soap:Envelope xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\"></soap:Envelope>";
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder;
try
{
builder = factory.newDocumentBuilder();
// Use String reader
Document document = builder.parse( new InputSource(
new StringReader( xmlString ) ) );
TransformerFactory tranFactory = TransformerFactory.newInstance();
Transformer aTransformer = tranFactory.newTransformer();
Source src = new DOMSource( document );
Result dest = new StreamResult( new File( "xmlFileName.xml" ) );
aTransformer.transform( src, dest );
} catch (Exception e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
}
This information is helpful.
Thanks,
Pavan

Its simple as that:
String text = "<your><xml>data</xml></your>";
Writer writer = new FileWriter("/tmp/filename.xml");
writer.write(text);
writer.flush();
writer.close();

You can, use the java.io.FileWriter to save your file.
String fileData = "<sample><xml>data</xml></sample>";
File outputFile = new File("someFile.xml");
BufferedWriter bw = null;
try{
bw = new BufferedWriter(new FileWriter(outputFile));
bw.write(fileData);
}
catch(IOException e)
{
e.printStackTrace();
}
finally
{
try{bw.close();}catch(Exception e){}
}
In case you need to manipulate the xml do like Kazekage Gaara said:
DocumentBuilderFactory dbfac = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = dbfac.newDocumentBuilder();
Document doc = docBuilder.parse(new InputSource(new StringReader(theString)));
And to save you can do as I said above. To transform the document back to string:
fileData = doc.toString();

I would recommend using commons-io. It has a single method that will do everything you need.
Code would look something like
FileUtils.writeStringToFile(new File("filename.xml"), xml);

Related

JAVA how to find and delete the structure of sentences?

I have a xml file, and its structure is like this.
<?xml version="1.0" encoding="MS949"?>
<pmd-cpd>
<duplication lines="123" tokens"123">
<file line="1" path="..">
<file line="1" path="..">
<codefragment><![CDATA[........]]></codefragment>
</duplication>
<duplication>
...
</duplication>
</pmd-cpd>
I want to delete 'codefragment' node, because my parser make an error 'invalid XML character(0x1). '
My parsing code is like this,
private void parseXML(File f){
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = null;
Document document = null;
try {
builder = factory.newDocumentBuilder();
document = builder.parse(f);
}catch(...)
The error happens in document = builder.parse(f); so I cannot use parser to delete the codefragment node.
This is why I want to delete these lines without the parser.
How can I delete this node without the parser...?
This is a followup answer to OP's self-answer, and the comment I made to that answer. Here's the recap, plus some extra:
Never do String += String in a loop. Use StringBuilder.
Read the XML in blocks, not lines.
Don't use String.replaceAll(). It has to recompile the regex every time, a regex you already have. Use Matcher.replaceAll().
Remember to close() the Reader. Better yet, use try-with-resources.
No need to save the clean XML back out, just use it directly.
Since XML is usually in UTF-8, read the file as UTF-8.
Don't print and ignore errors. Let caller handle errors.
private static void parseXML(File f) throws IOException, ParserConfigurationException, SAXException {
StringBuilder xml = new StringBuilder();
try (BufferedReader in = new BufferedReader(new InputStreamReader(new FileInputStream(f),
StandardCharsets.UTF_8))) {
Pattern badChars = Pattern.compile("[^\\u0009\\u000a\\u000d\\u0020-\\uD7FF\\uE000-\\uFFFD]+");
char[] cbuf = new char[1024];
for (int len; (len = in.read(cbuf)) != -1; )
xml.append(badChars.matcher(CharBuffer.wrap(cbuf, 0, len)).replaceAll(""));
}
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder domBuilder = domFactory.newDocumentBuilder();
Document document = domBuilder.parse(new InputSource(new StringReader(xml.toString())));
// insert code using DOM here
}
How I solved this problem was, to remove the bad characters such as x01, save as new XML file, and then parse the new file.
Because I could not even parse my old xml file, I could not remove the node with parser.
So removing invalid character and saving as a new file code was like this.
//save the xml string as a new file.
public static Document stringToDom(String xmlSource)
throws SAXException, ParserConfigurationException, IOException {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
return builder.parse(new InputSource(new StringReader(xmlSource)));
}
//get the file and remove bad characters in it
private static void cleanString(File fileName) {
try {
BufferedReader in = new BufferedReader(new FileReader(fileName));
String xmlLines, cleanXMLString="";
Pattern p = null;
Matcher m = null;
p = Pattern.compile("[^\\u0009\\u000a\\u000d\\u0020-\\uD7FF\\uE000-\\uFFFD]");
while (((xmlLines = in.readLine()) != null)){
m = p.matcher(xmlLines);
if (m.find()){
cleanXMLString = cleanXMLString + xmlLines.replaceAll("[^\\u0009\\u000a\\u000d\\u0020-\\uD7FF\\uE000-\\uFFFD]", "")+"\n";
}else
cleanXMLString = cleanXMLString + xmlLines+"\n";
}
Document doc = stringToDom(cleanXMLString);
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
DOMSource source = new DOMSource(doc);
StreamResult result = new StreamResult(new File("\\new\\"+fileName.getName()));
transformer.transform(source, result);
} catch (IOException | SAXException | ParserConfigurationException | TransformerException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
Maybe, that's not good method since it takes quite long time for even a small file(under 5MB).
But if your file is small, you can try this...

updated xml data not updated in the xml file

i have made a method for updating my xml in the xml file by a using a GUI..
but when I update it everything seem to be working fine and the console is printing out the correct things.
But when I open the xml file and press refrah nothing is updated.
What is my problem?
public void updateObjType(String newTxt, int x) throws ParserConfigurationException, SAXException, IOException, XPathExpressionException {
System.out.println("String value : " + newTxt);
System.out.println("Index value : " + x);
DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse("xmlFiles/CoreDatamodel.xml");
XPath xPath = XPathFactory.newInstance().newXPath();
// Go thru the Object_types in the XML file and get item x.
NodeList nodeList = (NodeList) xPath.compile("//OBJECT_TYPE/text()")
.evaluate(xmlDocument, XPathConstants.NODESET);
// Set new NodeValue
nodeList.item(x).setNodeValue(newTxt);
String value = nodeList.item(x).getTextContent();
System.out.println(value);
}
this is the output from the console :
Original data : IF150Data
Incoming String value : Data
Index value : 4
updated data : Data
I solved it by using a transformer.
Full solution :
// Update the object type name from the object type list.
public void updateObjType(String newTxt, int x)
throws ParserConfigurationException, SAXException, IOException,
XPathExpressionException {
File file = new File("xmlFiles/CoreDatamodel.xml");
System.out.println("Incoming String value : " + newTxt);
System.out.println("Index value : " + x);
DocumentBuilderFactory builderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(file);
XPath xPath = XPathFactory.newInstance().newXPath();
NodeList nodeList = (NodeList) xPath.compile("//OBJECT_TYPE/text()")
.evaluate(xmlDocument, XPathConstants.NODESET);
// Set new NodeValue
nodeList.item(x).setNodeValue(newTxt);
// Save the new updates
try {
save(file, xmlDocument);
} catch (Exception e) {
e.printStackTrace();
}
}
And then the method I added :
public void save(File file, Document doc) throws Exception {
TransformerFactory factory = TransformerFactory.newInstance();
Transformer transformer = factory.newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
StringWriter writer = new StringWriter();
StreamResult result = new StreamResult(writer);
DOMSource source = new DOMSource(doc);
transformer.transform(source, result);
String s = writer.toString();
System.out.println(s);
FileWriter fileWriter = new FileWriter(file);
BufferedWriter bufferedWriter = new BufferedWriter(fileWriter);
bufferedWriter.write(s);
bufferedWriter.flush();
bufferedWriter.close();
}

Cannot create XML Document from String

I am trying to create an org.w3c.dom.Document form an XML string. I am using this How to convert string to xml file in java as a basis. I am not getting an exception, the problem is that my document is always null. The XML is system generated and well formed. I wish to convert it to a Document object so that I can add new Nodes etc.
public static org.w3c.dom.Document stringToXML(String xmlSource) throws Exception {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
InputStream input = IOUtils.toInputStream(xmlSource); //uses Apache commons to obtain InputStream
BOMInputStream bomIn = new BOMInputStream(input); //create BOMInputStream from InputStream
InputSource is = new InputSource(bomIn); // InputSource with BOM removed
Document document = builder.parse(new InputSource(new StringReader(xmlSource)));
Document document2 = builder.parse(is);
System.out.println("Document=" + document.getDoctype()); // always null
System.out.println("Document2=" + document2.getDoctype()); // always null
return document;
}
I have tried these things: I created a BOMInputStream thinking that a BOM was causing the conversion to fail. I thought that this was my issue but passing the BOMInputStream to the InputSource doesn't make a difference. I have even tried passing a literal String of simple XML and nothing but null. The toString method returns [#document:null]
I am using Xpages, a JSF implementation that uses Java 6. Full name of Document class used to avoid confusion with Xpage related class of the same name.
Don't rely on what toString is telling you. It is providing diagnostic information that it thinks is useful about the current class, which is, in this case, nothing more then...
"["+getNodeName()+": "+getNodeValue()+"]";
Which isn't going to help you. Instead, you will need to try and transform the model back into a String, for example...
String text
= "<fruit>"
+ "<banana>yellow</banana>"
+ "<orange>orange</orange>"
+ "<pear>yellow</pear>"
+ "</fruit>";
InputStream is = null;
try {
is = new ByteArrayInputStream(text.getBytes());
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document document = builder.parse(is);
System.out.println("Document=" + document.toString()); // always null
Transformer tf = TransformerFactory.newInstance().newTransformer();
tf.setOutputProperty(OutputKeys.INDENT, "yes");
tf.setOutputProperty(OutputKeys.METHOD, "xml");
tf.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
ByteArrayOutputStream os = null;
try {
os = new ByteArrayOutputStream();
DOMSource domSource = new DOMSource(document);
StreamResult sr = new StreamResult(os);
tf.transform(domSource, sr);
System.out.println(new String(os.toByteArray()));
} finally {
try {
os.close();
} finally {
}
}
} catch (ParserConfigurationException | SAXException | IOException | TransformerConfigurationException exp) {
exp.printStackTrace();
} catch (TransformerException exp) {
exp.printStackTrace();
} finally {
try {
is.close();
} catch (Exception e) {
}
}
Which outputs...
Document=[#document: null]
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<fruit>
<banana>yellow</banana>
<orange>orange</orange>
<pear>yellow</pear>
</fruit>
You can try using this: http://www.wissel.net/blog/downloads/SHWL-8MRM36/$File/SimpleXMLDoc.java

How to load XML file from internet into a string

Currently I have a java application that loads XML from a local file into a string. My code looks like this
private String xmlFile = "D:\\mylocalcomputer\\extract-2339393.xml";
String fileStr = FileUtils.readFileToString(new File(xmlFile));
How can I get the contents of the XML file if it was located on the internet, at a URL like http://mydomain.com/xml/extract-2000.xml ?
try the sax interface
private String xmlURL = "http://mydomain.com/xml/extract-2000.xml";
XMLReader reader = XMLReaderFactory.createXMLReader();
reader.setContentHandler(handler);
reader.parse(new InputSource(new URL(xmlURL).openStream()));
For more information regarding SAX check this link
Check this code:
DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();
InputStream inputStream = new FileInputStream(new File("http://mydomain.com/xml/extract-2000.xml"));
org.w3c.dom.Document doc = documentBuilderFactory.newDocumentBuilder().parse(inputStream);
StringWriter stw = new StringWriter();
Transformer serializer = TransformerFactory.newInstance().newTransformer();
serializer.transform(new DOMSource(doc), new StreamResult(stw));
stw.toString();

How to convert String to DOM Document object in java?

I have a case like getting an XML and convert the XML elements to document object and getting the element values and attributes which i have been created already
Here is the piece of code i have tried to convert the string to DOM document object
String xmlString = " <r><e>d</e></r>";
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
org.w3c.dom.Document document = builder.parse(new InputSource(new StringReader(xmlString)));
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
DOMSource source = new DOMSource(document);
StreamResult result = new StreamResult(new StringWriter());
transformer.transform(source, result);
String str1 = result.getWriter().toString();
System.out.println(str1);
But this case is valid for only elements without attributes
what can we do if the
String xmlString = "<element attribname="value" attribname1="value1"> pcdata</element>"
we are using Double quotes for the attribute values"value". The compiler is showing error
Suggest me if there any xml encoder and decoder is there to handle this scenarios ??
you can try
DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
InputSource is = new InputSource();
is.setCharacterStream(new StringReader("<root><node1></node1></root>"));
Document doc = db.parse(is);
refer this http://www.java2s.com/Code/Java/XML/ParseanXMLstringUsingDOMandaStringReader.htm
Either escape the double quotes with \
String xmlString = "<element attribname=\"value\" attribname1=\"value1\"> pcdata</element>"
or use single quotes instead
String xmlString = "<element attribname='value' attribname1='value1'> pcdata</element>"
DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document document = db.parse(new ByteArrayInputStream(xmlString.getBytes("UTF-8"))); //remove the parameter UTF-8 if you don't want to specify the Encoding type.
this works well for me even though the XML structure is complex.
And please make sure your xmlString is valid for XML, notice the escape character should be added "\" at the front.
The main problem might not come from the attributes.
public static void main(String[] args) {
final String xmlStr = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>\n"+
"<Emp id=\"1\"><name>Pankaj</name><age>25</age>\n"+
"<role>Developer</role><gen>Male</gen></Emp>";
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder;
try
{
builder = factory.newDocumentBuilder();
Document doc = builder.parse( new InputSource( new StringReader( xmlStr )) );
} catch (Exception e) {
e.printStackTrace();
}
}

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