Xslt转换/如何对实体进行分组_Xslt

Xslt转换/如何对实体进行分组

xslt

Xslt转换/如何对实体进行分组,xslt,Xslt,我想使用XSLT修改： <Office Code="1" OtherAttribute="5"> <Customer CustomerId="0010" CodeModifier="A"/> <Customer CustomerId="0011" CodeModifier="B"/> <Customer CustomerId="0012" CodeModifier="B"/> </Office> <Office

我想使用XSLT修改：

<Office Code="1" OtherAttribute="5">
   <Customer CustomerId="0010" CodeModifier="A"/>
   <Customer CustomerId="0011" CodeModifier="B"/>
   <Customer CustomerId="0012" CodeModifier="B"/>
</Office>
<Office Code="2" OtherAttribute="6">
   <Customer CustomerId="2010" CodeModifier="A"/>
   <Customer CustomerId="0011" CodeModifier="C"/>
</Office>

进入：

<Office Code="1A" OtherAttribute="5">
   <Customer CustomerId="0010"/>
</Office>
<Office Code="1B" OtherAttribute="5">
   <Customer CustomerId="0011"/>
   <Customer CustomerId="0012"/>
</Office>
<Office Code="2A" OtherAttribute="6">
   <Customer CustomerId="2010"/>
</Office>
<Office Code="2C" OtherAttribute="6">
   <Customer CustomerId="0011"/>
</Office>

我的目标：

将具有相同CodeModifier的每个客户实体分组为Office实体。如果有多个CodeModifier，我将添加Office实体。将修改Office中的Code属性（将客户端的CodeModifier连接到Office中）
（兼性但我认为很琐碎）抑制CodeModifier属性并保留所有其他属性

有人知道怎么做吗？

此样式表应用于您的输入示例：

<!-- a key to group Customers by their office code + modifier -->
<xsl:key name="kCustomerGroup" match="Customer" 
  use="concat(../@Code, @CodeModifier)" 
/>

<!-- identity template: copies everything that is not handled otherwise -->
<xsl:template match="node() | @*">
  <xsl:copy>
    <xsl:apply-templates select="node() | @*" />
  </xsl:copy>
</xsl:template>

<!-- apply templates directly to customers. modify as necessary -->
<xsl:template match="/">
  <xsl:apply-templates select="//Customer" mode="CustomerGroup" />
</xsl:template>

<xsl:template match="Customer" mode="CustomerGroup">
  <xsl:variable name="officeCode" select="concat(../@Code, @CodeModifier)" />
  <!-- if this Customer is first of his respective group... -->
  <xsl:if test="
    generate-id() 
    =
    generate-id(key('kCustomerGroup', $officeCode)[1])
  ">
    <!-- use for-each to switch the context node to the parent -->
    <xsl:for-each select="..">
      <xsl:copy>
        <xsl:apply-templates select="@*" />
        <!-- overwrite the @Code attribute -->
        <xsl:attribute name="Code">
          <xsl:value-of select="$officeCode" />
        </xsl:attribute>
        <!-- now handle the Customer group members -->
        <xsl:apply-templates select="key('kCustomerGroup', $officeCode)" />
      </xsl:copy>
    </xsl:for-each>
  </xsl:if>
</xsl:template>

<!-- remove unwanted attribute with empty template -->    
<xsl:template match="Customer/@CodeModifier" />

及

此样式表应用于您的输入示例：

<!-- a key to group Customers by their office code + modifier -->
<xsl:key name="kCustomerGroup" match="Customer" 
  use="concat(../@Code, @CodeModifier)" 
/>

<!-- identity template: copies everything that is not handled otherwise -->
<xsl:template match="node() | @*">
  <xsl:copy>
    <xsl:apply-templates select="node() | @*" />
  </xsl:copy>
</xsl:template>

<!-- apply templates directly to customers. modify as necessary -->
<xsl:template match="/">
  <xsl:apply-templates select="//Customer" mode="CustomerGroup" />
</xsl:template>

<xsl:template match="Customer" mode="CustomerGroup">
  <xsl:variable name="officeCode" select="concat(../@Code, @CodeModifier)" />
  <!-- if this Customer is first of his respective group... -->
  <xsl:if test="
    generate-id() 
    =
    generate-id(key('kCustomerGroup', $officeCode)[1])
  ">
    <!-- use for-each to switch the context node to the parent -->
    <xsl:for-each select="..">
      <xsl:copy>
        <xsl:apply-templates select="@*" />
        <!-- overwrite the @Code attribute -->
        <xsl:attribute name="Code">
          <xsl:value-of select="$officeCode" />
        </xsl:attribute>
        <!-- now handle the Customer group members -->
        <xsl:apply-templates select="key('kCustomerGroup', $officeCode)" />
      </xsl:copy>
    </xsl:for-each>
  </xsl:if>
</xsl:template>

<!-- remove unwanted attribute with empty template -->    
<xsl:template match="Customer/@CodeModifier" />

及

这里有另一种方法，只使用匹配的模板

在MSXSL 4.0下测试为XSLT 1.0

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output method="xml" indent="yes"/> <xsl:key name="kCustomerGroup" match="Customer" use="concat(../@Code, @CodeModifier)" /> <xsl:template match="Office"> <xsl:apply-templates select="Customer[generate-id() = generate-id(key('kCustomerGroup', concat(../@Code, @CodeModifier))[1])]" /> </xsl:template> <xsl:template match="Customer"> <Office Code="{concat(../@Code,@CodeModifier)}" OtherAttribute="{../@OtherAttribute}"> <xsl:apply-templates select="key('kCustomerGroup', concat(../@Code,@CodeModifier))" mode="copy"/> </Office> </xsl:template> <xsl:template match="Customer" mode="copy"> <xsl:copy> <xsl:copy-of select="@CustomerId"/> </xsl:copy> </xsl:template> </xsl:stylesheet>

输出至：

<?xml version="1.0" encoding="UTF-8"?> <Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011"/> <Customer CustomerId="0012"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011"/> </Office>

这里有另一种方法，只使用匹配的模板

在MSXSL 4.0下测试为XSLT 1.0

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output method="xml" indent="yes"/> <xsl:key name="kCustomerGroup" match="Customer" use="concat(../@Code, @CodeModifier)" /> <xsl:template match="Office"> <xsl:apply-templates select="Customer[generate-id() = generate-id(key('kCustomerGroup', concat(../@Code, @CodeModifier))[1])]" /> </xsl:template> <xsl:template match="Customer"> <Office Code="{concat(../@Code,@CodeModifier)}" OtherAttribute="{../@OtherAttribute}"> <xsl:apply-templates select="key('kCustomerGroup', concat(../@Code,@CodeModifier))" mode="copy"/> </Office> </xsl:template> <xsl:template match="Customer" mode="copy"> <xsl:copy> <xsl:copy-of select="@CustomerId"/> </xsl:copy> </xsl:template> </xsl:stylesheet>

输出至：

<?xml version="1.0" encoding="UTF-8"?> <Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011"/> <Customer CustomerId="0012"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011"/> </Office>

完整而简短的XSLT 2.0解决方案：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:template match="/*/*"> <xsl:for-each-group select="Customer" group-by="@CodeModifier"> <Office> <xsl:copy-of select="../@*"/> <xsl:attribute name="Code" select= "concat(../@Code, current-grouping-key())"/> <xsl:copy-of select="current-group()"/> </Office> </xsl:for-each-group> </xsl:template> </xsl:stylesheet>

<Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010" CodeModifier="A"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011" CodeModifier="B"/> <Customer CustomerId="0012" CodeModifier="B"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010" CodeModifier="A"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011" CodeModifier="C"/> </Office>

应用于以下XML文档时（基于提供的XML片段，并将其包装到单个顶部元素中，使其成为格式良好的XML文档）：

生成所需的正确结果：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:template match="/*/*"> <xsl:for-each-group select="Customer" group-by="@CodeModifier"> <Office> <xsl:copy-of select="../@*"/> <xsl:attribute name="Code" select= "concat(../@Code, current-grouping-key())"/> <xsl:copy-of select="current-group()"/> </Office> </xsl:for-each-group> </xsl:template> </xsl:stylesheet>

<Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010" CodeModifier="A"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011" CodeModifier="B"/> <Customer CustomerId="0012" CodeModifier="B"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010" CodeModifier="A"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011" CodeModifier="C"/> </Office>

完整而简短的XSLT 2.0解决方案：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:template match="/*/*"> <xsl:for-each-group select="Customer" group-by="@CodeModifier"> <Office> <xsl:copy-of select="../@*"/> <xsl:attribute name="Code" select= "concat(../@Code, current-grouping-key())"/> <xsl:copy-of select="current-group()"/> </Office> </xsl:for-each-group> </xsl:template> </xsl:stylesheet>

<Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010" CodeModifier="A"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011" CodeModifier="B"/> <Customer CustomerId="0012" CodeModifier="B"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010" CodeModifier="A"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011" CodeModifier="C"/> </Office>

应用于以下XML文档时（基于提供的XML片段，并将其包装到单个顶部元素中，使其成为格式良好的XML文档）：

生成所需的正确结果：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:template match="/*/*"> <xsl:for-each-group select="Customer" group-by="@CodeModifier"> <Office> <xsl:copy-of select="../@*"/> <xsl:attribute name="Code" select= "concat(../@Code, current-grouping-key())"/> <xsl:copy-of select="current-group()"/> </Office> </xsl:for-each-group> </xsl:template> </xsl:stylesheet>

<Office Code="1A" OtherAttribute="5"> <Customer CustomerId="0010" CodeModifier="A"/> </Office> <Office Code="1B" OtherAttribute="5"> <Customer CustomerId="0011" CodeModifier="B"/> <Customer CustomerId="0012" CodeModifier="B"/> </Office> <Office Code="2A" OtherAttribute="6"> <Customer CustomerId="2010" CodeModifier="A"/> </Office> <Office Code="2C" OtherAttribute="6"> <Customer CustomerId="0011" CodeModifier="C"/> </Office>

您使用的条件与我一样。只有您的条件在谓词中，而我的条件在
；-）如果你仔细观察的话，我不会使用循环来做任何工作。你使用的条件和我一样。只有您的条件在谓词中，而我的条件在
；-）如果你仔细看的话，我不会用循环做任何工作。好问题，+1。请参阅我的答案，了解完整、非常简短且简单的XSLT 2.0解决方案。：）好问题，+1。请参阅我的答案，了解完整、非常简短且简单的XSLT 2.0解决方案。：）