我正在尝试使用XSLT将XML从一种形式转换为另一种形式。
我已将XML输入为
<Employee>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>ABC</COMPANY>
<DESIGNATION>Developer</DESIGNATION>
</JOBS>
<JOBS>
<COMPANY>DEF</COMPANY>
<DESIGNATION>Tester</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>GHT</COMPANY>
<DESIGNATION>QA Lead</DESIGNATION>
</JOBS>
<JOBS>
<COMPANY>Levor</COMPANY>
<DESIGNATION>Project Lead</DESIGNATION>
</JOBS>
</Emp>
</Employee>
我想转换为(期望输出)
<Employee>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>ABC</COMPANY>
<DESIGNATION>Developer</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>DEF</COMPANY>
<DESIGNATION>Tester</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>GHT</COMPANY>
<DESIGNATION>QA Lead</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>Levor</COMPANY>
<DESIGNATION>Project Lead</DESIGNATION>
</JOBS>
</Emp>
</Employee>
我已经在下面的XSLT中尝试过了,但不同的跟踪都不起作用,你能帮我纠正这个XSLT以获得所需的输出吗
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="Employee">
<xsl:element name="Employee">
<xsl:for-each select="Employee">
<xsl:sort select="Sequence"/>
<xsl:element name="Emp">
<xsl:copy-of select="EMPID"/>
<xsl:copy-of select="NAME"/>
<xsl:copy-of select="Sequence"/>
<xsl:for-each select="child::JOBS">
<xsl:copy-of select="COMPANY"/>
<xsl:copy-of select="DESIGNATION"/>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
请帮帮我?
已添加:
感谢您的快速响应,但我忘了在这里添加一个案例,JOBS聚合不会适用于下面的所有员工
<Employee>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>ABC</COMPANY>
<DESIGNATION>Developer</DESIGNATION>
</JOBS>
<JOBS>
<COMPANY>DEF</COMPANY>
<DESIGNATION>Tester</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>789</EMPID>
<NAME>Anderson</NAME>
<Sequence>3</Sequence>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>GHT</COMPANY>
<DESIGNATION>QA Lead</DESIGNATION>
</JOBS>
<JOBS>
<COMPANY>Levor</COMPANY>
<DESIGNATION>Project Lead</DESIGNATION>
</JOBS>
</Emp>
</Employee>
所以这里Anderson并没有emp,结果应该是这样的
<Employee>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>ABC</COMPANY>
<DESIGNATION>Developer</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>123</EMPID>
<NAME>James</NAME>
<Sequence>1</Sequence>
<JOBS>
<COMPANY>DEF</COMPANY>
<DESIGNATION>Tester</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>GHT</COMPANY>
<DESIGNATION>QA Lead</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>456</EMPID>
<NAME>Clark</NAME>
<Sequence>2</Sequence>
<JOBS>
<COMPANY>Levor</COMPANY>
<DESIGNATION>Project Lead</DESIGNATION>
</JOBS>
</Emp>
<Emp>
<EMPID>789</EMPID>
<NAME>Anderson</NAME>
<Sequence>3</Sequence>
</Emp>
</Employee>
你能帮我一下吗?
根据澄清进行编辑:
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Employee">
<xsl:copy>
<xsl:apply-templates select="Emp">
<xsl:sort select="Sequence"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Emp[not(JOBS)]">
<xsl:copy-of select="."/>
</xsl:template>
<xsl:template match="Emp">
<xsl:apply-templates select="JOBS"/>
</xsl:template>
<xsl:template match="JOBS">
<Emp>
<xsl:copy-of select="../EMPID | ../NAME | ../Sequence | ."/>
</Emp>
</xsl:template>
</xsl:stylesheet>