CODEWITHSUNDEEP
javaxsdjaxb
denys
denys

Reputation: 2529

jaxb how to sort elements by attribute?

f.e. I have next xml scheme:

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema  xmlns:xs="http://www.w3.org/2001/XMLSchema"
        targetNamespace="http://www.example.com"
        xmlns="http://www.example.com"
        elementFormDefault="qualified">

    <xs:complexType name="TItem">
        <xs:simpleContent>
            <xs:extension base="xs:string">
                <xs:attribute name="id" type="xs:int" use="required"/>
            </xs:extension>
        </xs:simpleContent>
    </xs:complexType>

    <xs:element name="Resource">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="item" type="TItem" maxOccurs="unbounded" />
            </xs:sequence>
        </xs:complexType>
    </xs:element>

</xs:schema>

Based on that xsd jaxb generates java code using which I'm creating next xml:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Resource xmlns="http://www.example.com">
    <item id="10">asdf</item>
    <item id="2">some text</item>
    <item id="9">some item</item>
    <item id="14">vzxc</item>
</Resource>

So. I want to produce xml with sorted items by id. So it should be: 

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Resource xmlns="http://www.example.com">
    <item id="2">some text</item>
    <item id="9">some item</item>
    <item id="10">asdf</item>
    <item id="14">vzxc</item>
</Resource>

Is it possible with jaxb?

P.S. id is not unique - there could be a lot of items with id="22", etc.

Upvotes: 0

Views: 1636

Answers (1)

Ronson
Ronson

Reputation: 479

Try to change your item to a TreeSet in Resource

@XmlRootElement(name = "Resource")
public class Resource {
    @XmlElement(required = true)
    protected Collection<TItem> item;
    ....
    public Collection<TItem> getItem() {
        if (item == null) {
            item = new TreeSet<TItem>(cmp);//cmp is an instance of Comparator<TItem>
        }
    return this.item;
}

Upvotes: 2

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