I want to convert an output stream into String object
JavaJaxbJava Problem Overview
I want to convert an OutputStream
into a String
object. I am having an OutputStream
object returned after marshalling the JAXB object.
Java Solutions
Solution 1 - Java
not very familiar with jaxb, from what i was able to find you can convert into a string using
public String asString(JAXBContext pContext,
Object pObject)
throws
JAXBException {
java.io.StringWriter sw = new StringWriter();
Marshaller marshaller = pContext.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_ENCODING, "UTF-8");
marshaller.marshal(pObject, sw);
return sw.toString();
}
but I'm not sure about a stirng object. still searching.
** EDIT
> Marshalling a non-element
>
> Another common use case is where you
> have an object that doesn't have
> @XmlRootElement on it. JAXB allows you
> to marshal it like this:
>
> marshaller.marshal( new JAXBElement(
> new
> QName("","rootTag"),Point.class,new
> Point(...)));
>
> This puts the
> new
> QName("","rootTag"),String.class,"foo
> bar"));
>
> But unfortunately it cannot be used to
> marshal objects like List or Map, as
> they aren't handled as the first-class
> citizen in the JAXB world.
found HERE
Solution 2 - Java
StringWriter sw = new StringWriter();
com.integra.xml.Integracao integracao = new Integracao();
integracao.add(...);
try {
JAXBContext context = JAXBContext.newInstance("com.integra.xml");
Marshaller marshaller = context.createMarshaller();
marshaller.marshal(integracao, sw );
System.out.println(sw.toString());
} catch (JAXBException e) {
e.printStackTrace();
}
Solution 3 - Java
public String readFile(String pathname) throws IOException {
File file = new File(pathname);
StringBuilder fileContents = new StringBuilder((int) file.length());
Scanner scanner = new Scanner(file);
String lineSeparator = System.getProperty("line.separator");
try {
while (scanner.hasNextLine()) {
fileContents.append(scanner.nextLine() + lineSeparator);
}
return fileContents.toString();
}
finally {
scanner.close();
}
}
Method to handle the XML and convert it to a string.
JAXBContext jc = JAXBContext.newInstance(ClassMatchingStartofXMLTags.class);
Unmarshaller unmarshaller = jc.createUnmarshaller();
//store filepath to be used
String filePath = "YourXMLFile.xml";
File xmlFile = new File(filePath);
//Set up xml Marshaller
ClassMatchingStartofXMLTags xmlMarshaller = (ClassMatchingStartofXMLTags) unmarshaller.unmarshal(xmlFileEdit);
Marshaller marshaller = jc.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
// Use marshall to output the contents of the xml file
System.out.println("Marshalled XML\n");
marshaller.marshal(xmlMarshaller, System.out);
//Use Readfile method to convert the XML to a string and output the results
System.out.println("\nXML to String");
String strFile = ReadFile(xmlFile)
System.out.println(strFile);
Method within your class to get the XML and Marshall it Edit The above method will both output the marshalled xml and also the xml as a string.
Solution 4 - Java
Yes, there is a way to do it: just pass the String writer as input to it. So that it the xml created will be written to it
public void saveSettings() throws IOException {
FileOutputStream os = null;
//Declare a StringWriter to which the output has to go
StringWriter sw = new StringWriter();
try {
Answer ans1=new Answer(101,"java is a programming language","ravi");
Answer ans2=new Answer(102,"java is a platform","john");
ArrayList<Answer> list=new ArrayList<Answer>();
list.add(ans1);
list.add(ans2);
settings=new Question(1,"What is java?",list);
os = new FileOutputStream(FILE_NAME);
this.marshaller.marshal(settings, new StreamResult(sw));
System.out.println(sw.toString());
new File(FILE_NAME).delete();
} finally {
if (os != null) {
os.close();
}
}
}