使用XSLT,我试图找出如何使用来自另一组节点的数据合并/更新一组节点中的数据。节点具有相同的模式,但父节点不同。需要基于共享父属性合并数据。在下面的示例中,数据正从Principal复制到Driver。任何人都可以帮助我吗?
输入文件:
<Info>
<Principal id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
</PersonInfo>
<PrincipalInfo></PrincipalInfo>
</Principal>
<Policy>
<Driver id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<Surname>Smith</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>S</MaritalStatus>
<Occupation>Manager</Occupation>
</PersonInfo>
</Driver>
<PolicyInfo></PolicyInfo>
</Policy>
</Info>
期望的结果:
<Info>
<Principal id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
</PersonInfo>
<PrincipalInfo></PrincipalInfo>
</Principal>
<Policy>
<Driver id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
<Occupation>Manager</Occupation>
</PersonInfo>
</Driver>
<PolicyInfo></PolicyInfo>
</Policy>
</Info>
答案 0 :(得分:2)
这是一个完整的解决方案:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kPrincipalById" match="Principal"
use="@id"/>
<xsl:key name="kPrincipalChild" match="Principal/*/*"
use="concat(../../@id, name())"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Driver/*">
<xsl:variable name="vPrincipal"
select="key('kPrincipalById', ../@id)"/>
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates select=
"$vPrincipal/*[name()=name(current())]/*"/>
<xsl:apply-templates select=
"*[not(key('kPrincipalChild',
concat(../../@id,name())
)
)
]"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
在提供的XML文档上应用此转换时:
<Info>
<Principal id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
</PersonInfo>
<PrincipalInfo></PrincipalInfo>
</Principal>
<Policy>
<Driver id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<Surname>Smith</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>S</MaritalStatus>
<Occupation>Manager</Occupation>
</PersonInfo>
</Driver>
<PolicyInfo></PolicyInfo>
</Policy>
</Info>
产生了想要的正确结果:
<Info>
<Principal id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
</PersonInfo>
<PrincipalInfo/>
</Principal>
<Policy>
<Driver id="Insured">
<PersonName>
<GivenName>Jane</GivenName>
<OtherGivenName>A</OtherGivenName>
<Surname>Doe</Surname>
</PersonName>
<PersonInfo>
<BirthDate>01-01-1980</BirthDate>
<MaritalStatus>M</MaritalStatus>
<Occupation>Manager</Occupation>
</PersonInfo>
</Driver>
<PolicyInfo/>
</Policy>
</Info>
<强>解释强>:
identity rule /模板按原样复制每个节点。身份规则的使用和覆盖是最基本和最强大的XSLT设计模式。
只有一个额外的模板会覆盖Driver
的儿童元素的身份规则。它复制(并有效地替换Driver
的相同名称的孙子元素与Principal
的相应的祖父元素。然后它仍然处理(复制)Driver
的那些没有相应的子孙元素Principal
为了方便地访问Principal
及其子孙 - 通过id和id ++ name(),有两个keys已定义并已使用