XML模式验证，如何忽略元素内部的树

Question

考虑到此XML：

<myName myClass="grandfather" name="grandfather_name" transaction="new">
<attributes>
    <name>test name</name>
    <description>some stuff for desc</description>
    <version>the current version</version>
</attributes>
<myName myClass="father" name="father_name" transaction="new">
    <attributes>
    <name>test name</name>
    <description>some stuff for desc</description>
    <version>the current version</version>
    <wifeName>name of wife</wifeName>
    </attributes>
    <myName myClass="child_1" name="child_1_name" transaction="new">
        <attributes>
            <name>test name</name>
            <description>some stuff for desc</description>
            <version>the current version</version>
            <GirlFriendName>name of GirlFriend</GirlFriendName>
        </attributes>
    </myName>
    <myName myClass="child_2" name="child_2_name" transaction="new">
        <attributes>
            <name>test name</name>
            <description>some stuff for desc</description>
            <version>the current version</version>
            <GirlFriendName>name of GirlFriend</GirlFriendName>
        </attributes>
    </myName>
</myName>

我只想验证“祖父”及其属性，而忽略XML的其余部分。为此，我在模式中添加了这一行:type=“xs:anyType”

正在使用的XSD如下所示：

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

<xs:group name="AttributesGroup"> 
    <xs:all>
        <xs:element name="name" type="xs:string" minOccurs="1" maxOccurs="1" />
        <xs:element name="description" type="xs:string" minOccurs="0" maxOccurs="1" />
        <xs:element name="version" type="xs:string" minOccurs="0" maxOccurs="1" />
    </xs:all>
</xs:group>

<!-- BASE COMPLEX TYPES DEFINITIONS -->

<xs:complexType name="MyComplexType">
    <xs:sequence>
        <xs:element name="attributes" minOccurs="1" maxOccurs="unbounded">
            <xs:complexType>
                    <xs:group ref="AttributesGroup"/>
            </xs:complexType>
        </xs:element>

        <xs:element name="myName" type="xs:anyType" minOccurs="0" maxOccurs="unbounded"/>

    </xs:sequence>

    <xs:attribute name="myClass" type="xs:string" use="required" />
    <xs:attribute name="name" type="xs:string" />
    <xs:attribute name="transaction" type="xs:string" use="required" />
</xs:complexType>


<!-- ROOT ELEMENT -->
<xs:element name="myName" type="MyComplexType" />

我怎能无视“父亲”下面的一切？

Answer 1

您可以在MyComplexType内的本地定义的MyName中使用xsd:any；以下XSD将成功验证XML：

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"
           elementFormDefault="unqualified">
    <xs:group name="AttributesGroup"> 
        <xs:all>
            <xs:element name="name" type="xs:string" minOccurs="1" maxOccurs="1" />
            <xs:element name="description" type="xs:string"
                  minOccurs="0" maxOccurs="1" />
            <xs:element name="version" type="xs:string" minOccurs="0" maxOccurs="1" />
        </xs:all>
    </xs:group>
    <xs:complexType name="MyComplexType">
        <xs:sequence>
            <xs:element name="attributes" minOccurs="1" maxOccurs="unbounded">
                <xs:complexType>
                    <xs:group ref="AttributesGroup"/>
                </xs:complexType>
            </xs:element>
            <xs:element name="myName" minOccurs="1" maxOccurs="unbounded">
                <xs:complexType>
          <xs:sequence>
            <xs:any processContents="skip"
                    minOccurs="0" maxOccurs="unbounded"/>
          </xs:sequence>
          <xs:anyAttribute processContents="skip"/>
                </xs:complexType>
            </xs:element>
        </xs:sequence>
        <xs:attribute name="myClass" type="xs:string" use="required" />
        <xs:attribute name="name" type="xs:string" />
        <xs:attribute name="transaction" type="xs:string" use="required" />
    </xs:complexType>
    <xs:element name="myName" type="MyComplexType" />
</xs:schema>