如何验证具有重复子元素的xml作为Id唯一的标记?

时间:2016-05-18 22:04:06

标签: java xml validation xsd

我有一个以下格式的xml:

<data>
   <index id="Name">Mesut</index>
   <index id="Age">28</index>
</data>

现在这个xml的元素是索引相同的。由此生成的XSD如下:

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="data">
          <xs:complexType>
                 <xs:sequence>
                        <xs:element name="index" maxOccurs="unbounded" type="xs:string">
                               <xs:complexType>
                                      <xs:attribute name="id" type="xs:string"></xs:attribute>                                                            
                                </xs:complexType>                   
                        </xs:element>
                    </xs:sequence>
           </xs:complexType>
</xs:element>

上面xsd的问题在于,这个xsd无法验证一个xml,它具有与索引相同的标记,一次以String的形式出现2次,第二次以Integer的形式出现。因为xsd只能验证字符串。

现在我用来验证的代码如下:

    public static void main(String[] args){
    boolean b = true;
    File fXml = new File("C:\\Users\\Mesut\\Desktop\\XMLAndXSD\\MainXml.xml");
    File fXsd = new File("C:\\Users\\Mesut\\Desktop\\XMLAndXSD\\MainXml.xsd");
    SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
    try {           
        Schema schema = schemaFactory.newSchema(fXsd);
        schema.newValidator().validate(new StreamSource(fXml));
    } catch(SAXException sax) {         
        System.out.println("exception in sax");
        b = false;
        sax.printStackTrace();
    } catch(IOException io) {
        System.out.println("exception in io");
        b = false;
        io.printStackTrace();
    }       
    System.out.println(b);}

运行上面代码的例外情况如下:

enter code hereorg.xml.sax.SAXParseException; systemId: file:/C:/Users/Vikas/Desktop/XMLAndXSD/MainXml.xsd; lineNumber: 6; columnNumber: 93; src-element.3: Element 'index' has both a 'type' attribute and a 'anonymous type' child. Only one of these is allowed for an element.
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(ErrorHandlerWrapper.java:198)
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(ErrorHandlerWrapper.java:134)

2 个答案:

答案 0 :(得分:1)

使用extension标记创建一个新类型,扩展xs:string以包含属性id。在架构开始时:

<xs:complexType name="indexType">
   <xs:simpleContent>
     <xs:extension base="xs:string">
       <xs:attribute name="id" type="xs:string"/>
     </xs:extension>
   </xs:simpleContent>
</xs:complexType>

然后在“数据”中执行:

<xs:element name="index" maxOccurs="unbounded" type="indexType"/>

答案 1 :(得分:0)

在@ mascoj的答案的基础上,尝试这个架构:

<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema" targetNamespace="http://myns">
  <xs:element name="data">
    <xs:complexType>
      <xs:sequence>
        <xs:element name="index" maxOccurs="unbounded" minOccurs="0">
          <xs:complexType>
            <xs:simpleContent>
              <xs:extension base="xs:string">
                <xs:attribute type="xs:string" name="id" use="optional"/>
              </xs:extension>
            </xs:simpleContent>
          </xs:complexType>
        </xs:element>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
</xs:schema>

验证以下xml实例:

<f:data xmlns:f="http://myns">
   <f:index id="Name">Mesut</f:index>
   <f:index id="Age">28</f:index>
</f:data>