Muenchian分組 - 在一個節點內的組，而不是在整個文檔中

Question

我想在我的XSLT中使用Muenchian分組來分組匹配的節點，但我只想在父節點內分組，而不是跨整個源XML文檔。

鑑於XSLT和XML如下（道歉的我的樣本代碼的長度）：

XSLT

<?xml version="1.0" encoding="utf-8"?> 
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> 
<xsl:output method="html" indent="yes"/> 

<xsl:key name="contacts-by-surname" match="contact" use="surname" /> 
<xsl:template match="records"> 
    <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[1]) = 1]"> 
    <xsl:sort select="surname" /> 
    <xsl:value-of select="surname" />,<br /> 
    <xsl:for-each select="key('contacts-by-surname', surname)"> 
    <xsl:sort select="forename" /> 
    <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br /> 
    </xsl:for-each> 
    </xsl:for-each> 
</xsl:template> 
</xsl:stylesheet> 

XML

<root> 
<records> 
    <contact id="0001"> 
    <title>Mr</title> 
    <forename>John</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0002"> 
    <title>Dr</title> 
    <forename>Amy</forename> 
    <surname>Jones</surname> 
    </contact> 
    <contact id="0003"> 
    <title>Mrs</title> 
    <forename>Mary</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0004"> 
    <title>Ms</title> 
    <forename>Anne</forename> 
    <surname>Jones</surname> 
    </contact> 
    <contact id="0005"> 
    <title>Mr</title> 
    <forename>Peter</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0006"> 
    <title>Dr</title> 
    <forename>Indy</forename> 
    <surname>Jones</surname> 
    </contact> 
</records> 
<records> 
    <contact id="0001"> 
    <title>Mr</title> 
    <forename>James</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0002"> 
    <title>Dr</title> 
    <forename>Mandy</forename> 
    <surname>Jones</surname> 
    </contact> 
    <contact id="0003"> 
    <title>Mrs</title> 
    <forename>Elizabeth</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0004"> 
    <title>Ms</title> 
    <forename>Sally</forename> 
    <surname>Jones</surname> 
    </contact> 
    <contact id="0005"> 
    <title>Mr</title> 
    <forename>George</forename> 
    <surname>Smith</surname> 
    </contact> 
    <contact id="0006"> 
    <title>Dr</title> 
    <forename>Harry</forename> 
    <surname>Jones</surname> 
    </contact> 
</records> 
</root> 

RESULT

Jones, 
Amy (Dr) 
Anne (Ms) 
Harry (Dr) 
Indy (Dr) 
Mandy (Dr) 
Sally (Ms) 

Smith, 
Elizabeth (Mrs) 
George (Mr) 
James (Mr) 
John (Mr) 
Mary (Mrs) 
Peter (Mr) 

我怎麼組的每個<records>內實現這一結果：

Jones, 
Amy (Dr) 
Anne (Ms) 
Indy (Dr) 

Smith, 
John (Mr) 
Mary (Mrs) 
Peter (Mr) 

Jones, 
Harry (Dr) 
Mandy (Dr) 
Sally (Ms) 

Smith, 
Elizabeth (Mrs) 
George (Mr) 
James (Mr) 

Answer 1

我花了一些時間......我正要放棄，但仍繼續:)

的缺點關鍵功能是生成的密鑰將始終用於整個xml。因此，您應該連接密鑰中的其他信息以使其更加具體。在例如在下面，我將連接記錄節點的位置，以便爲每個記錄獲得不同姓氏的密鑰。

這裏的XSLT：

<?xml version="1.0" encoding="utf-8"?> 
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> 
    <xsl:output method="html" indent="yes"/> 
    <xsl:key name="distinct-surname" match="contact" use="concat(generate-id(..), '|', surname)"/> 
    <xsl:template match="records"> 
    <xsl:for-each select="contact[generate-id() = generate-id(key('distinct-surname', concat(generate-id(..), '|', surname))[1])]"> 
     <xsl:sort select="surname" /> 
     <xsl:value-of select="surname" />,<br /> 
     <xsl:for-each select="key('distinct-surname', concat(generate-id(..), '|', surname))"> 
     <xsl:sort select="forename" /> 
     <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br /> 
     </xsl:for-each> 
    </xsl:for-each> 
    </xsl:template> 
</xsl:stylesheet> 

這是結果：

Jones, 
Amy (Dr) 
Anne (Ms) 
Indy (Dr) 
Smith, 
John (Mr) 
Mary (Mrs) 
Peter (Mr) 
Jones, 
Harry (Dr) 
Mandy (Dr) 
Sally (Ms) 
Smith, 
Elizabeth (Mrs) 
George (Mr) 
James (Mr) 

請注意，結果在forenames排序了。如果你不希望它forenames排序，你需要刪除線<xsl:sort select="forename" />

Answer 2

有更簡單的方法，通過添加謂詞，確保比參與測試MUENCH接觸是的電流記錄孩子。

<xsl:key name="contacts-by-surname" match="contact" use="surname" /> 
<xsl:template match="records"> 
    <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[generate-id(parent::records) = generate-id(current())][1]) = 1]"> 
    <xsl:sort select="surname" /> 
    <xsl:value-of select="surname" />,<br /> 
    <xsl:for-each select="key('contacts-by-surname', surname)[generate-id(parent::records) = generate-id(current()/parent::records)]"> 
    <xsl:sort select="forename" /> 
    <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br /> 
    </xsl:for-each> 
    </xsl:for-each> 
</xsl:template>