Encoding XML entities in string using Apache XMLBeans

Question

I'd like to use Apache XMLBeans to escape a String for embedding it in an XML document, e.g. encode all XML entities.

XmlString does provide this functionality, but insists on wrapping the output in xml-fragment tags, which I'd like to get rid of.

However, I'm not interested in sugestions

to use anything other than XMLBeans (like org.apache.commons.lang.StringEscapeUtils)
to remove the enclosing tag after escaping (e.g. using a regex)

Here's a test case. Can you help me fix it?

import org.apache.xmlbeans.*;

public class Test {
  @Test public void test(){
    String input = "You & me";
    String expected = "You &amp; me";
    String actual = escape(input);
    Assert.assertEquals(expected, actual);
    // Fails with: ComparisonFailure: expected:<[You &amp; me]> 
    //             but was:<[<xml-fragment>You &amp; me</xml-fragment>]>
  }

  private String escape(String str){
    XmlString value = XmlString.Factory.newInstance();
    value.setStringValue(input);
    XmlOptions opts = new XmlOptions();
    // do I need to set one of the 54 available options?
    // see http://xmlbeans.apache.org/docs/2.4.0/reference/org/apache/xmlbeans/XmlOptions.html
    return value.xmlText(opts);
  }
}

I'd be pretty surprised if XmlBeans didn't automatically escape what must be escaped. Why do you want to escape manually? — JB Nizet, Jul 08 '11 at 11:40
JB Nizet: I'm using XMLBeans to implement parts of a custom XML marshalling framework. I didn't mean to imply that XmlBeans wouldn't automatically escape what must be escaped. — otto.poellath, Jul 08 '11 at 17:07
Why can't you use e.g. StringEscapeUtils? This may reveal some requirement that's not clear. — Ed Staub, Jul 11 '11 at 00:25
Ed: I could indeed use StringEscapeUtils, and that's what I'm currently using. But given that the required functionality is already in XMLBeans, I'd like to get rid of the Commons Lang dependency. — otto.poellath, Jul 11 '11 at 12:41

decal · Answer 1 · 2017-12-04T17:11:11.950

Use the setSaveOuter() option before passing the XmlOptions to the xmlText() method. Like this.

XmlOptions opts = new XmlOptions();
opts.setSaveOuter();
return value.xmlText(opts);

I would like to clarify that if you just create a new element out of thin air it seems like the element name will not be serialized. For example If I have a Model element that has a Property element on it the following will not display the property tag, it will display the xml fragment.

Property property = Property.Factory.newInstance(); 

XmlOptions opts = new XmlOptions();
opts.setSaveOuter();
return property.xmlText(opts);

To have the Property Element show up I must do the following.

ModelDocument modelDoc = ModelDocument.Factory.newInstance();
ModelType model = modelDoc.addNewModel();
PropertyType propertyType = model.addNewProperty();
Property property = Property.Factory.newInstance(); 

XmlOptions opts = new XmlOptions();
opts.setSaveOuter();
return property.xmlText(opts);

I have tried and this is not working.I even tried [this link](http://osdir.com/ml/user-xmlbeans.apache.org/2009-12/msg00020.html). Any help is appreciated. — dileepVikram, Oct 08 '14 at 06:24
+1 on this answer, worked for me - not sure why it didnt work for you dileepvikram - I just added this before i created a JMS TextMessage object `XmlOptions opts = new XmlOptions();` `opts.setSaveOuter();` and then just added the "opts" here: `sendMessage(_OrderStateChangeEventTypeImpl.xmlText(opts));` And it worked like a charm — JGlass, Nov 28 '17 at 21:06

score 0 · Answer 2 · answered Dec 12 '14 at 17:29

0

your expected string is incorrect. by virtue of calling xmlText, it will return an XML, hence the string must be wrapped in xml-fragment element.

answered Dec 12 '14 at 17:29

Jerry Sy

1

Consider supporting your answer with the relevant code. – carlodurso Dec 12 '14 at 17:51

score 0 · Answer 3 · answered Mar 22 '16 at 17:19

create xml reader as a normal xml string then remove the xml-fragment, i have tried like this,

if(request != null && request.contains("<xml")){
XMLInputFactory xif = XMLInputFactory.newFactory();
StringReader reader = new StringReader(request);
StreamSource xml = new StreamSource(reader);
XMLStreamReader xsr = xif.createXMLStreamReader(xml);
xsr.nextTag();          
request = xsr.getElementText();
System.out.println("updated request is \n"+request);
}

then

JAXBContext jaxbContext = JAXBContext.newInstance(YourClass.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
StringReader reader = new StringReader(request);

Encoding XML entities in string using Apache XMLBeans

3 Answers3