Is there a way to get the path of a XML-Document from a xPath- or Document-Object in the xPath-API ?
That´s how the Objects are initalized:
FileInputStream file = new FileInputStream(new File("C:\ExampleFile.xml"));
DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(file);
XPath xPath = XPathFactory.newInstance().newXPath();
So the question is:
Could the Objects xmlDocument or xpath somehow return "C:\ExampleFile.xml" ?
Using the Document object xmlDocument you can return the path of the file with:
xmlDocument.getDocumentURI();
Rather than creating a FileInputStream from the File and passing that to the parse method
FileInputStream file = new FileInputStream(new File("C:\\ExampleFile.xml"));
use the version of parse that takes a File directly.
File file = new File("C:\\ExampleFile.xml");
// rest of your code is unchanged - parse(file) is now the
// java.io.File version rather than the InputStream version
When you pass just a stream the parser has no way of knowing that that stream was created from a file, as far as the parser is concerned that could be a stream you received from a web server, or a ByteArrayInputStream, or some other non-file source. If you pass the File directly to parse then the parser will handle opening and closing the streams itself, and will be able to provide a meaningful URI to downstream components, and you'll get a sensible result from xmlDocument.getDocumentURI().
As an aside, if you want XPath to work reliably then you need to enable namespaces by calling builderFactory.setNamespaceAware(true) before you call newDocumentBuilder(). Even if your XML doesn't actually use any namespaces, you still need to parse with a namespace-aware DOM parser.
Related
I have never had to download an XML file in Java before and parse it after. I'm looking to download and parse this file http://api.irishrail.ie/realtime/realtime.asmx/getStationDataByNameXML?StationDesc=Bayside
All I want to do is read the train times. I've been reading about parsing XML but I'm not really getting anywhere with it. I just keep reading about parsers like stax, after that I'm a bit lost.
Can anyone give me some basic advice of what I need to do?
You can use JAXB for this and any other XML processing needs. Start here.
You can use DOM parser and new Java Architecture for XML Binding JAXB it will help you in marshalling (for converting an xml to Object) and unmarshalling(for converting an Object to xml)
link for the example
http://www.vogella.com/tutorials/JAXB/article.html
You can use the DOM Parser to create a Document object from the XML file.
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new File(filename));
The DOM Parser creates a traversable tree from your XML data.
You can then pick out the data you need from the DOM tree using XPath.
try this library
http://x-stream.github.io/download.html
it is simple and fast to use to write and read xml from java.
I'm trying to transform an XML file into a document like this:
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document document = db.parse("C:/xml/41111208890622000144550010000000011000003066-nfe.xml");
Document document = db.parse(new InputSource(new StringReader("C:/xml/41111208890622000144550010000000011000003066-nfe.xml")));
but it is giving the error message:
Exception in thread "main" org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 1;
someone knows what to do?
You're currently creating a reader containing the string
"C:/xml/41111208890622000144550010000000011000003066-nfe.xml"
and asking the DocumentBuilder to parse that as if it were XML, when it's clearly not. (I'm referring to the second parse call, which I suspect is the one in your actual code. The code you've provided wouldn't compile as you've declared document twice.)
You can create a FileInputStream or perhaps an InputStreamReader wrapped around it:
String filename = "C:/xml/41111208890622000144550010000000011000003066-nfe.xml";
try (FileInputStream input = new FileInputStream(filename))
{
Document document = db.parse(new InputSource(input));
}
(I prefer to use a stream directly, and let the parser detect the encoding.)
Now this call:
Document document = db.parse("C:/xml/...");
would nearly work and may actually work, using DocumentBuilder.parse(String) - it depends on whether parse is happy to handle a filename as a URI. (I've seen some XML APIs that are fine with that, and some that aren't.) If it doesn't work, try using the file:// scheme:
Document document = db.parse("file://C:/xml/...");
I'm building an android app that communicates to a web server and am struggling with the following scenario:
Given ONE line of XML in a String eg:
"<test one="1" two="2" />"
I would like to extract the values into a HashMap so that:
map.get("one") = "1"
map.get("two") = "2"
I already can do this with a full XML document using the SAX Parser, this complains when i try to just give it the above string with a MalformedUrlException: Protocol not found
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder;
Document doc = null;
builder = factory.newDocumentBuilder();
doc = builder.parse("<test one="1" two="2" />"); //here
I realize some regex could do this but Id really rather do it properly.
The same behaviour can be found at http://metacpan.org/pod/XML::Simple#XMLin which is what the web server uses.
Can anyone help? Thanks :D
DocumentBuilder.parse(String) treats the string as a URL. Try this instead:
Document doc = builder.parse(new InputSource(new StringReader(text)));
(where text contains the XML, of course).
I have a JAVA application where I am sending some xml requests and receiving xml responses. I first receive response in string and then write a file and storing this file into file system. Then while parsing the xml response file I am accessing this from file system and use some of the data for further business logic.
File file = new File("log\\XMLMessage\\LastXMLResponse.xml");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(file);
Now I am thinking to distribute this JAVA application via Java Web Start (JWS) application and as I know I cannot keep this file into jar file since there will be modification in this file on regular basis.
What do you suggest me to do? Can I parse the String directly (no need to store the response into file)?
Document doc = db.parse(xmlMessage);
or where can I keep this file? I don't want to show this file to the user of my application.
Take the String, make a StringReader from String, make a InputSource from StringReader, then call parse on your DocumentBuilder.
Yes, you can parse the string directly,no need to store it in a file.
Try this:
String xml = "<xml></xml>";
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new ByteArrayInputStream(xml.getBytes()));
System.out.println(doc);
I need to remove XML declaration from dom4j document type
I am creating document by
doc = (Document) DocumentHelper.parseText(someXMLstringWithoutXMLDeclaration);
String parsed to Document doc by DocumenHelper contains no XML declaration (it comes from XML => XSL => XML transformation)
I think that DocumentHelper is adding declaration to a document body ?
Is there any way to remove XML declaration from the body of
doc
The simpler solution I choose is
doc.getRootElement().asXML();
I'm not sure where exactly this the declaration is a problem in your code.
I had this once when I wanted to write an xml file without declaration (using dom4j).
So if this is your use case: "omit declaration" is what you're looking for.
http://dom4j.sourceforge.net/dom4j-1.6.1/apidocs/org/dom4j/io/OutputFormat.html
Google says this can be set as a property as well, not sure what it does though.
You need to interact with the root element instead of the document.
For example, using the default, compact OutputFormat mentioned by PhilW:
Document doc = (Document) DocumentHelper.parseText(someXMLstringWithoutXMLDeclaration);
final Writer writer = new StringWriter();
new XMLWriter(writer).write(doc.getRootElement());
String out = writer.toString();