如何解组 <!DOCTYPE> 在JAXB中？

Question

I need to unmarshal !DOCTYPE section using JAXB so I can get the values of file1 and file2.

The xml file looks like this :

<?xml version="1.0" encoding="UTF-8"?>

  <!DOCTYPE doc [
     <!ENTITY file1 SYSTEM "conf/files/file1.xml">
      <!ENTITY file2 SYSTEM "conf/files/file2.xml">
    ]>

    <appels>
      <appel>
         &file1;
      </appel>
      <appel>
         &file1;
       </appel>
    </appels>

the java look like this :

@XmlRootElement(name = "appels")
public class Appels implements Serializable {

private List<Appel> appel;

}

I need to get file1 = conf/files/file1.xml and file2 = conf/files/file2.xml

is this possible? Thank you for the help

Answer 1

UPDATE

The following will resolve &file1; to the contents of the file conf/files/file1.xml .

<!ENTITY file1 SYSTEM "conf/files/file1.xml">

The following will resolve &file3; to the string conf/files/file3.xml .

<!ENTITY file3 "conf/files/file3.xml">

---

Input Files

input.xml

This is a slightly modified version of the XML document from your question. I changed one of the &file1; to &file2; and added &file3;

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE doc [
    <!ENTITY file1 SYSTEM "conf/files/file1.xml">
    <!ENTITY file2 SYSTEM "conf/files/file2.xml">
    <!ENTITY file3 "conf/files/file3.xml">
 ]>
<appels>
    <appel>
         &file1;
    </appel>
    <appel>
         &file2;
    </appel>
    <appel>
         &file3;
    </appel>
</appels>

conf/files/file1.xml

<foo>Hello World</foo>

conf/files/file2.xml

Instead of making this an XML file, I'm just going to add text.

Foo Bar

Equivalent XML File

As far as your parser is concerned your XML document looks like the following. Note that &file1; and &file2; resolved to the contents of their files and &file3; did not.

<?xml version="1.0" encoding="UTF-8"?>
<appels>
    <appel>
         <foo>Hello World</foo>
    </appel>
    <appel>
         Foo Bar
    </appel>
    <appel>
         conf/files/file3.xml
    </appel>
</appels>

Demo Code

Demo

The demo code below will convert the XML to objects, and then write it back to XML. The entities are resolved when the XML is unmarshalled into objects.

import java.io.File;
import javax.xml.bind.*;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Appels.class);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        File xml = new File("src/forum20637070/input.xml");
        Appels appels = (Appels) unmarshaller.unmarshal(xml);

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(appels, System.out);
    }

}

Output

Note how the first appel element marshalled. Since &file; resolved to an XML fragment, JAXB wasn't able to match it against the text property we had mapped.

<appels>
    <appel>
    </appel>
    <appel>
             Foo Bar
    </appel>
    <appel>
             conf/files/file3.xml
    </appel>
</appels>

Java Model

Below is the Java model I used for this example:

Appels

import java.io.Serializable;
import java.util.List;
import javax.xml.bind.annotation.*;

@XmlRootElement(name = "appels")
@XmlAccessorType(XmlAccessType.FIELD)
public class Appels implements Serializable {

    private List<Appel> appel;

}

Appel

import javax.xml.bind.annotation.*;

@XmlAccessorType(XmlAccessType.FIELD)
public class Appel {

    @XmlValue
    private String value;

}

Answer 2

在我看来，这就是你要找的东西： http ： //lauraliparulo.altervista.org/jaxb-part-6-entities-references-expansion-before-unmarshalling/