CODEWITHSUNDEEP
javaxmlxsdjaxb
Oxobo
Oxobo

Reputation: 673

How to create UUID in my Java code using JAXB to pass XSD pattern restriction

I'm going to create ID field in my Java code using UUID. I need to create xml from my Book class and validate it based on below XSD.

My XSD looks like this

<xsd:complexType name="Book" >
    <xsd:sequence>
        <xsd:element name="Publisher" type="ns:PublisherType"/>
        <xsd:element name="MessageId" type="ns:GUID"/>
        <xsd:element name="Author" type="xsd:string"/>
        <xsd:element name="Title" type="xsd:string"/>
        <xsd:any processContents="lax" minOccurs="0" maxOccurs="unbounded"
                 namespace="##other"/>
    </xsd:sequence>
</xsd:complexType>

<xsd:simpleType name="GUID">
    <xsd:restriction base="xsd:string">
        <xsd:pattern value="[a-fA-F0-9]{8}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{12}"/>
    </xsd:restriction>
</xsd:simpleType>

My Java class looks like this

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement
@XmlType(name="Book", propOrder = {"publisher", "messageId", "author", "title"
})
@Getter
@Setter
public class Book {

    private Publisher publisher;
    private GUID messageId;    
    private String author;  
    private String title;
}

How should I implement my GUID class to return a UUID.randomUUID()or any other way to pass the XSD ?

Upvotes: 0

Views: 785

Answers (1)

deFreitas
deFreitas

Reputation: 4448

Take a look at this example, basically you have to annotate the method which return the xml value with @XmlValue

public class GUID {

    private final UUID uuid;

    public GUID() {
        this.uuid = UUID.randomUUID();
    }

    @XmlValue
    public String getValue(){
        return uuid.toString();
    }
}

Upvotes: 1

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