How to convert book's reference to XML?

Question

I have several book's reference be must convert to XML.
I want to create application in Java for this action.

Book's reference:

 Schulz V, Hansel R, Tyler VE. Rational phytotherapy: a physician's guide to herbal   
 medicine. 3rd ed., fully rev. and expand. Berlin: Springer; c1998. 306 p.

XML:

<element-citation publication-type="book" publication-format="print">
    <name>
        <surname>Schulz</surname>
        <given-names>V</given-names>
    </name>
    <name>
        <surname>Hansel</surname>
        <given-names>R</given-names>
    </name>
    <name>
        <surname>Tyler</surname>
        <given-names>VE</given-names>
    </name>
    <source>Rational phytotherapy: a physician's guide to herbal medicine</source>
    <edition>3rd ed., fully rev. and expand</edition>
    <publisher-loc>Berlin</publisher-loc>
    <publisher-name>Springer</publisher-name>
    <year>c1998</year>
    <size units="page">306 p</size>
</element-citation>

How to convert book's reference to XML format?
What do you suggest?

What is the structure of such a reference? Does it have fields and an order of these fields? How can the fields be recognized? — , Jul 05 '13 at 12:45
Personally I think that the near-unstructured nature of the input references will prove your biggest challenge, converting it to Java using eg JAXB is quite easy to do. — fvu, Jul 05 '13 at 12:49
Check out this answer: [Mapping XML to an object in Java](http://stackoverflow.com/a/16755094/814702). There are also other popular Object to XML binding frameworks, like [Castor](http://castor.codehaus.org/index.html), for example. — informatik01, Jul 05 '13 at 12:53

score 2 · Answer 1 · edited May 23 '17 at 11:56

2

For example, use JAXB.

Get an XSD for your desired XML format.
Generate java classes from the XSD - see how here.
Implement a simple program that will parse your input file and build a tree with the help of the generated classes. This may be trivial or very difficult depending on your input.
Serialize the result - see how here.

EDIT: As hinted by Joop Eggen you may also use annotations instead of steps 1-3. This makes things maybe even simpler. See how here.

edited May 23 '17 at 11:56

Community

1
1

answered Jul 05 '13 at 12:48

Lachezar Balev

11,498
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Step 3. is the difficult step unless we know more about the structure of a record. – Jul 05 '13 at 12:49
1

Instead of an XSD you can use annotations. Then it is very simple. – Joop Eggen Jul 05 '13 at 12:56
1

@Tichodroma Yeap, to be honest in seems like a free text to me. But if this is the problem the question seems to be wrong. It should not deal with XML at all :-) – Lachezar Balev Jul 05 '13 at 12:57
@Joop Eggen - correct, this would be easier, I'll extend my answer. – Lachezar Balev Jul 05 '13 at 12:58

score 0 · Accepted Answer · answered Jul 05 '13 at 13:23

As you might not be experienced in Java, the dull, simple solution (Java 7):

writing the XML as text;
parsing with String.split(regex) (Scanner would do too).

Mind, special characters < > & " ' in bookref text might need to be replaced by < > & " '.

String bookRef = "Schulz V, Hansel R, Tyler VE. Rational phytotherapy: a physician's guide to herbal "
        + "medicine. 3rd ed., fully rev. and expand. Berlin: Springer; c1998. 306 p.";

File file = new File("D:/dev/xml-part.txt");
final String TAB = "    ";
try (PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(new FileOutputStream(file), "UTF-8")))) {
    out.println(TAB + "<element-citation publication-type=\"book\" publication-format=\"print\">");

    String[] lines = bookRef.split("\\.\\s*");

    String names = lines[0];
    String[] nameArray = names.split(",\\s*");
    for (String name : nameArray) {
        String[] nameParts = name.split(" +", 2);
        out.println(TAB + TAB + "<name>");
        out.println(TAB + TAB + TAB + "<surname>" + nameParts[0] + "</surname>");
        out.println(TAB + TAB + TAB + "<given-name>" + nameParts[1] + "</given-name>");
        out.println(TAB + TAB + "</name>");
    }
    out.println(TAB + TAB + "<source>" + lines[1] + "</source>");
    ...

    out.println(TAB + "</element-citation>");
} catch (FileNotFoundException | UnsupportedEncodingException ex) {
    Logger.getLogger(Test.class.getName()).log(Level.SEVERE, null, ex);
}

How to convert book's reference to XML?

2 Answers2