消除 XSLT 1.0 中节点数据的重复

Question

提前致谢，抱歉，您可能需要一段时间才能理解我的逻辑（我不是使用 XSLT 的专家）。

任何人都可以帮助消除标签 "PaySlip_Val" 的重复并将所有 "ListOfPaySlipData" 个孩子分组在一个 parent "ListOfPaySlipMonth".

下

下面是 XML 层级，

            <?xml version="1.0" encoding="UTF-8"?>
            <Data>
                <P>
                    <FstName>F1</FstName>
                    <LstName>L1</LstName>
                    <Type>
                        <Payslip>y</Payslip>
                        <Details>
                            <Year>2016</Year>
                            <Month>Jan</Month>
                            <Amount>$$$</Amount>
                        </Details>
                        <Details>
                            <Year>2016</Year>
                            <Month>Feb</Month>
                            <Amount>$$$</Amount>
                        </Details>
                    </Type>
                    <Type>
                        <Payslip>yes</Payslip>
                        <Details>
                            <Year>2016</Year>
                            <Month>Mar</Month>
                            <Amount>$$$</Amount>
                        </Details>
                        <Details>
                            <Year>2016</Year>
                            <Month>Apr</Month>
                            <Amount>$$$</Amount>
                        </Details>
                    </Type>
                    <Type>
                        <Payslip>n</Payslip>
                        <Details>
                            <Year>2016</Year>
                            <Month>May</Month>
                            <Amount>$$$</Amount>
                        </Details>
                    </Type>
                </P>
                <P>
                    <FstName>F2</FstName>
                    <LstName>L2</LstName>
                    <Type>
                        <Payslip>n</Payslip>
                        <Details>
                            <Year>2016</Year>
                            <Month>Feb</Month>
                            <Leaves>4</Leaves>
                        </Details>
                    </Type>
                </P>
            </Data>

我已在 XSLT 下应用，

                <?xml version="1.0" encoding="UTF-8"?>
            <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
                <xsl:variable name="smallcase" select="'abcdefghijklmnopqrstuvwxyz'" />
                <xsl:variable name="uppercase" select="'ABCDEFGHIJKLMNOPQRSTUVWXYZ'" />
                <xsl:template match="/">
                    <xsl:call-template name="Pay_Slip" /> 
                </xsl:template>
                <xsl:template name="Pay_Slip"> 
                        <xsl:element name="ListOfData">
                            <xsl:for-each select="Data/P">
                                <xsl:element name="First_Name">
                                        <xsl:value-of select="FstName" /> 
                                </xsl:element>
                                <xsl:element name="Last_Name">
                                    <xsl:value-of select="LstName" /> 
                                </xsl:element>
                                <xsl:element name="PaySlip">
                                    <xsl:for-each select="Type">
                                        <xsl:variable name="Type" select="Payslip" />
                                        <xsl:if test="$Type ='y' or $Type ='yes'">
                                                <xsl:element name="PaySlip_Val">
                                                        <xsl:value-of select="$Type" /> 
                                                </xsl:element>
                                        <xsl:element name="ListOfPaySlipMonth">
                                                <xsl:for-each select="Details">
                                                    <xsl:element name="ListOfPaySlipData">
                                                        <xsl:element name="Month">
                                                            <xsl:value-of select="Month" /> 
                                                        </xsl:element>
                                                        <xsl:element name="Salary">
                                                            <xsl:value-of select="Amount" /> 
                                                        </xsl:element>
                                                        <xsl:element name="Year">
                                                            <xsl:value-of select="Year" /> 
                                                        </xsl:element>
                                                    </xsl:element>
                                                </xsl:for-each> 
                                        </xsl:element>
                                        </xsl:if>
                                    </xsl:for-each>
                                </xsl:element>
                            </xsl:for-each>  
                        </xsl:element>
                </xsl:template>
            </xsl:stylesheet>

结果是：

                <?xml version="1.0" encoding="UTF-8"?>
            <ListOfData>
            <First_Name>F1</First_Name>
            <Last_Name>L1</Last_Name>
            <PaySlip>
            <PaySlip_Val>y</PaySlip_Val>
            <ListOfPaySlipMonth>
            <ListOfPaySlipData>
            <Month>Jan</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            <ListOfPaySlipData>
            <Month>Feb</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            </ListOfPaySlipMonth>
            <PaySlip_Val>yes</PaySlip_Val>
            <ListOfPaySlipMonth>
            <ListOfPaySlipData>
            <Month>Mar</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            <ListOfPaySlipData>
            <Month>Apr</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            </ListOfPaySlipMonth>
            </PaySlip>
            <First_Name>F2</First_Name>
            <Last_Name>L2</Last_Name>
            <PaySlip/>
            </ListOfData>

期望的结果：

            <ListOfData>
<ListOfP>
    <First_Name>F1</First_Name>
    <Last_Name>L1</Last_Name>
    <PaySlip>
        <PaySlip_Val>y</PaySlip_Val>
        <ListOfPaySlipMonth>
            <ListOfPaySlipData>
            <Month>Jan</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            <ListOfPaySlipData>
            <Month>Feb</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            <ListOfPaySlipData>
            <Month>Mar</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
            <ListOfPaySlipData>
            <Month>Apr</Month>
            <Salary>$$$</Salary>
            <Year>2016</Year>
            </ListOfPaySlipData>
        </ListOfPaySlipMonth>
    </PaySlip>
</ListOfP>
</ListOfData>

Answer 1

AFAICT，这 returns 预期结果：

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="*"/>

<!-- identity transform -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="/Data">
    <ListOfData>
        <xsl:apply-templates/>
    </ListOfData>
</xsl:template>

<xsl:template match="P">
    <First_Name><xsl:value-of select="FstName"/></First_Name>
    <Last_Name><xsl:value-of select="LstName"/></Last_Name>
    <xsl:variable name="payslips" select="Type[Payslip='yes' or Payslip='y']" />
    <PaySlip>
        <xsl:if test="$payslips">
            <PaySlip_Val>y</PaySlip_Val>
            <ListOfPaySlipMonth>
                <xsl:apply-templates select="Type[Payslip='yes' or Payslip='y']/Details"/>
            </ListOfPaySlipMonth>
        </xsl:if>
    </PaySlip>
</xsl:template>

<xsl:template match="Details">
    <ListOfPaySlipData>
        <xsl:copy-of select="Month"/>
        <Salary><xsl:value-of select="Amount"/></Salary>
        <xsl:copy-of select="Year"/>
    </ListOfPaySlipData>
</xsl:template>

</xsl:stylesheet>

编辑：

I have updated the desired result. The condition is that I need to get the First Name & Last name only if the "Payslip" tag has y or yes. In other cases, it should not fetch the data (F2 & L2 as per example).

这样试试，然后：

<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="*"/>

<!-- identity transform -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="/Data">
    <ListOfData>
        <xsl:apply-templates select="P[Type/Payslip='yes' or Type/Payslip='y']"/>
    </ListOfData>
</xsl:template>

<xsl:template match="P">
    <First_Name><xsl:value-of select="FstName"/></First_Name>
    <Last_Name><xsl:value-of select="LstName"/></Last_Name>
    <PaySlip>
        <PaySlip_Val>y</PaySlip_Val>
        <ListOfPaySlipMonth>
            <xsl:apply-templates select="Type[Payslip='yes' or Payslip='y']/Details"/>
        </ListOfPaySlipMonth>
    </PaySlip>
</xsl:template>

<xsl:template match="Details">
    <ListOfPaySlipData>
        <xsl:copy-of select="Month"/>
        <Salary><xsl:value-of select="Amount"/></Salary>
        <xsl:copy-of select="Year"/>
    </ListOfPaySlipData>
</xsl:template>

</xsl:stylesheet>

消除 XSLT 1.0 中节点数据的重复

Eliminating duplication of node data in XSLT 1.0

xslt

xpath

xslt-1.0

编辑：