6

我在根元素上创建了key/keyref以便基于指定元素创建文档范围的唯一性。

因此,通过所有实例的.//foo/@name每次出现必须是唯一的;同样对于. 这似乎工作正常。这些分别由和引用,也在根节点处定义。@namefoo.//bar/@name.//foo-ref/@name-ref.//bar-ref/@name-ref

但是,我收集到无法创建可选密钥,这带来了一些问题。foo从语义上讲,根据符合文件的性质,在or的每个实例上都不需要密钥bar。的实例foo-ref/@name-ref显然需要以现有的 为目标foo/@name,但是没有 a 的情况在语义上并不是无效foo@name

有什么解决方法吗?我不喜欢消费者必须为每个元素定义一个键的想法,而合理地只有少数人需要它们。

这是示例架构(当然,我没有部署一些foobar架构,但结构是相同的;这只是我一直在玩弄的一个测试架构

<xs:schema elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
    <xs:complexType name="ref">
        <xs:attribute name="name-ref" type="xs:string" use="required" />
    </xs:complexType>
    <xs:complexType name="obj">
        <xs:attribute name="name" type="xs:string" use="optional" />
    </xs:complexType>
    <xs:complexType name="foo">
        <xs:complexContent>
            <xs:extension base="obj">
                <xs:sequence>
                    <xs:choice maxOccurs="unbounded">
                        <xs:element name="foo" type="foo" />
                        <xs:element name="bar" type="bar" />
                        <xs:element name="foo-ref" type="foo-ref" />
                        <xs:element name="bar-ref" type="bar-ref" />
                    </xs:choice>
                </xs:sequence>
            </xs:extension>
        </xs:complexContent>
    </xs:complexType>
    <xs:complexType name="foo-ref">
        <xs:complexContent>
            <xs:extension base="ref" />
        </xs:complexContent>
    </xs:complexType>
    <xs:complexType name="bar">
        <xs:complexContent>
            <xs:extension base="obj">
                <xs:sequence>
                    <xs:choice maxOccurs="unbounded">
                        <xs:element name="bar" type="bar" />
                        <xs:element name="qux" type="qux" />
                        <xs:element name="bar-ref" type="bar-ref" />
                    </xs:choice>
                </xs:sequence>
            </xs:extension>
        </xs:complexContent>
    </xs:complexType>
    <xs:complexType name="bar-ref">
        <xs:complexContent>
            <xs:extension base="ref" />
        </xs:complexContent>
    </xs:complexType>
    <xs:complexType name="qux">
        <xs:simpleContent>
            <xs:extension base="xs:string" />
        </xs:simpleContent>
    </xs:complexType>
    <xs:element name="root">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="foo" type="foo" maxOccurs="unbounded" />
            </xs:sequence>
        </xs:complexType>
        <xs:key name="foo">
            <xs:selector xpath=".//foo" />
            <xs:field xpath="@name" />
        </xs:key>
        <xs:key name="bar">
            <xs:selector xpath=".//bar" />
            <xs:field xpath="@name" />
        </xs:key>
        <xs:keyref name="foo-ref" refer="foo">
            <xs:selector xpath=".//foo-ref" />
            <xs:field xpath="@name-ref" />
        </xs:keyref>
        <xs:keyref name="bar-ref" refer="bar">
            <xs:selector xpath=".//bar-ref" />
            <xs:field xpath="@name-ref" />
        </xs:keyref>
    </xs:element>
</xs:schema>

附录

感谢@PetruGardea,只需跟进我的修改。所以unique可以被 a 引用keyref,谁知道呢?(显然不是我

<xs:element name="root">
    <xs:complexType>
        <xs:sequence>
            <xs:element name="foo" type="foo" maxOccurs="unbounded" />
        </xs:sequence>
    </xs:complexType>
    <xs:keyref name="foo-ref" refer="foo">
        <xs:selector xpath=".//foo-ref" />
        <xs:field xpath="@name-ref" />
    </xs:keyref>
    <xs:keyref name="bar-ref" refer="bar">
        <xs:selector xpath=".//bar-ref" />
        <xs:field xpath="@name-ref" />
    </xs:keyref>
    <!--
        the use of xs:unique here, in lieu of xs:key allows for
        nullable "keys", retaining referential integrity with the
        above defined keyrefs. awesome possum.
    -->
    <xs:unique name="foo">
        <xs:selector xpath=".//foo" />
        <xs:field xpath="@name" />
    </xs:unique>
    <xs:unique name="bar">
        <xs:selector xpath=".//bar" />
        <xs:field xpath="@name" />
    </xs:unique>
</xs:element>
4

1 回答 1

4

使用xsd:unique;与键不同,它的匹配值要么是唯一的,要么是 nil(nil 或不存在)。

例子:

<?xml version="1.0" encoding="utf-8" ?>
<!-- XML Schema generated by QTAssistant/XSD Module (http://www.paschidev.com) -->
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <xsd:element name="root">
        <xsd:complexType>
            <xsd:sequence>
                <xsd:element name="uk" maxOccurs="unbounded">
                    <xsd:complexType>
                        <xsd:attribute name="name" type="xsd:string"/>
                    </xsd:complexType>
                </xsd:element>
                <xsd:element name="fk" maxOccurs="unbounded">
                    <xsd:complexType>
                        <xsd:attribute name="name" type="xsd:string"/>
                    </xsd:complexType>
                </xsd:element>
            </xsd:sequence>
        </xsd:complexType>
        <xsd:unique name="uq">
            <xsd:selector xpath="uk"/>
            <xsd:field xpath="@name"/>
        </xsd:unique>
        <xsd:keyref name="fk" refer="uq">
            <xsd:selector xpath="fk"/>
            <xsd:field xpath="@name"/>
        </xsd:keyref>
    </xsd:element>
</xsd:schema>

示例(有效)XML:

<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<!-- Sample XML generated by QTAssistant (http://www.paschidev.com) -->
<root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <uk name="name1"/>
    <uk />
    <fk/>
    <fk name="name1"/>
</root>
于 2013-04-05T21:45:11.860 回答