如何基於獨特的元素或ID

Question

我如何基於「CorporationID」和他們組的元素進行排序的元素，並將它們組排序..

我有一個XML有效載荷像下面的一個...

 
<corporationActions> 
    <corporationAction> 
    <ActionID>5530974</ActionID> 
    <CorporationID>1044294</CorporationID> 
    <ActionDate>2009-05-03</ActionDate> 
    <ActionType>Articles of Organization2</ActionType> 
    <ActionNotes></ActionNotes> 
    <DocumentNumber>20110258775-68</DocumentNumber> 
    <NumberofPages>4</NumberofPages> 
    </corporationAction> 
    <corporationAction> 
    <ActionID>5530975</ActionID> 
    <CorporationID>1044294</CorporationID> 
    <ActionDate>2009-05-03</ActionDate> 
    <ActionType>Miscellaneous</ActionType> 
    <ActionNotes></ActionNotes> 
    <DocumentNumber>20110258777-80</DocumentNumber> 
    <NumberofPages>2</NumberofPages> 
    </corporationAction> 
</corporationActions> 

我需要首先基於元素進行排序CorporationID，然後將它們分組，如果CorporationID s的重複..像下面paylaod

 
<corporationActions> 
    <corporationAction> 
     <CorporationID>1044294</CorporationID> 
     <corporationActionDetails> 
      <ActionID>5530974</ActionID> 
      <ActionDate>2009-05-03</ActionDate> 
      <ActionType>Articles of Organization2</ActionType> 
     </corporationActionDetails> 
     <corporationActionDetails> 
      <ActionID>5530975</ActionID> 
      <ActionDate>2009-05-03</ActionDate> 
      <ActionType>Miscellaneous</ActionType> 
     </corporationActionDetails> 
    </corporationAction> 
</corporationActions> 

感謝&問候， anvv夏爾馬

Answer 1

這是分組（Muenchian法）溶液中的XSLT 1.0，即大多采用模板和推式中沒有<xsl:for-each>：

<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="kcorpActByID" match="corporationAction" 
    use="CorporationID"/> 

<xsl:template match="node()|@*"> 
    <xsl:copy> 
     <xsl:apply-templates select="node()|@*"> 
     <xsl:sort select="CorporationID" data-type="number"/> 
     </xsl:apply-templates> 
    </xsl:copy> 
</xsl:template> 

<xsl:template match= 
"corporationAction|CorporationID 
|ActionNotes|DocumentNumber|NumberofPages"/> 

<xsl:template match= 
    "corporationAction 
    [generate-id() 
    = 
    generate-id(key('kcorpActByID', CorporationID)[1]) 
    ]"> 
    <corporationAction> 
    <xsl:copy-of select="CorporationID"/> 
    <xsl:apply-templates mode="copy" 
    select="key('kcorpActByID',CorporationID)" /> 
    </corporationAction> 
</xsl:template> 

<xsl:template match="corporationAction" mode="copy"> 
    <corporationActionDetails> 
    <xsl:apply-templates/> 
    </corporationActionDetails> 
</xsl:template> 
</xsl:stylesheet> 

當這變換應用到提供的XML文檔：

<corporationActions> 
    <corporationAction> 
    <ActionID>5530974</ActionID> 
    <CorporationID>1044294</CorporationID> 
    <ActionDate>2009-05-03</ActionDate> 
    <ActionType>Articles of Organization2</ActionType> 
    <ActionNotes></ActionNotes> 
    <DocumentNumber>20110258775-68</DocumentNumber> 
    <NumberofPages>4</NumberofPages> 
    </corporationAction> 
    <corporationAction> 
    <ActionID>5530975</ActionID> 
    <CorporationID>1044294</CorporationID> 
    <ActionDate>2009-05-03</ActionDate> 
    <ActionType>Miscellaneous</ActionType> 
    <ActionNotes></ActionNotes> 
    <DocumentNumber>20110258777-80</DocumentNumber> 
    <NumberofPages>2</NumberofPages> 
    </corporationAction> 
</corporationActions> 

想要的，正確的結果產生：

<corporationActions> 
    <corporationAction> 
     <CorporationID>1044294</CorporationID> 
     <corporationActionDetails> 
     <ActionID>5530974</ActionID> 
     <ActionDate>2009-05-03</ActionDate> 
     <ActionType>Articles of Organization2</ActionType> 
     </corporationActionDetails> 
     <corporationActionDetails> 
     <ActionID>5530975</ActionID> 
     <ActionDate>2009-05-03</ActionDate> 
     <ActionType>Miscellaneous</ActionType> 
     </corporationActionDetails> 
    </corporationAction> 
</corporationActions> 

請注意：增加的靈活性，由於其所使用的和壓倒一切身份模板。與之相比，我們使用<xsl:copy-of select="ActionID|ActionDate|ActionType"/>：<xsl:apply-templates/>，這可以確保當前節點的任何子節點都可以按照我們想要的任何方式進行處理（並進一步在單獨的模板中進行指定）。

二， XSLT 2.0解決方案：

<xsl:stylesheet version="2.0" 
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:output omit-xml-declaration="yes" indent="yes"/> 
    <xsl:strip-space elements="*"/> 

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

<xsl:template match="/*"> 
    <xsl:for-each-group select="corporationAction" 
     group-by="CorporationID"> 
    <xsl:sort select="number(CorporationID)"/> 

    <corporationAction> 
    <xsl:sequence select="CorporationID"/> 
    <xsl:apply-templates mode="copy" 
    select="current-group()" /> 
    </corporationAction> 
    </xsl:for-each-group> 
</xsl:template> 

<xsl:template match="corporationAction" mode="copy"> 
    <corporationActionDetails> 
    <xsl:apply-templates/> 
    </corporationActionDetails> 
</xsl:template> 

    <xsl:template match= 
"corporationAction|CorporationID 
|ActionNotes|DocumentNumber|NumberofPages"/> 
</xsl:stylesheet> 

請注意：使用<xsl:for-each-group>和current-group()功能簡化分組。

Answer 2

此解決方案使用Muenchian方法：

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> 
<xsl:output indent="yes"/> 

<xsl:key name="bycorporationid" match="corporationActions/corporationAction" use="CorporationID"/> 

<xsl:template match="corporationActions"> 
    <corporationActions> 
    <xsl:for-each select="corporationAction[count(. | key('bycorporationid', CorporationID)[1]) = 1]"> 

    <xsl:sort select="CorporationID" /> 

    <corporationAction> 
     <CorporationID><xsl:value-of select="CorporationID"/></CorporationID> 

     <xsl:for-each select="key('bycorporationid',CorporationID)"> 
     <corporationActionDetails> 
      <xsl:copy-of select="ActionID"/> 
      <xsl:copy-of select="ActionDate"/> 
      <xsl:copy-of select="ActionType"/> 
     </corporationActionDetails> 
     </xsl:for-each> 

    </corporationAction> 

    </xsl:for-each> 
    </corporationActions> 
</xsl:template> 

</xsl:stylesheet>