XSLT 2，按屬性和元素分組多個

我似乎無法根據他們的MaritalStatus來管理分組情侶。
我已經設法將所有處於關係中的人（'DeFacto'，'已婚'）分組，但無法弄清楚如何將他們分組到不同的家庭。XSLT 2，按屬性和元素分組多個

基本上，如果兩個人做一對夫婦，他們將創建一個家庭。家庭名稱將屬於家庭中的第一人稱姓氏，NumberOfAdults將被硬編碼爲'2'，NumberOfDependants將被計算在內。

這對夫妻是基於鏈接的Party.Identifier和Party.MaritalStatus.RelatedEntityRef

我有什麼（簡體）：

<PartySegment> 
<Party Type="Guarantor" PrimaryApplicant="No"> 
    <Identifier>b8b0f908b08e</Identifier> 
    <Person Sex="Female" FirstHomeBuyer="No" CustomerOfLender="No"> 
     <PersonName> 
      <NameTitle Value="Lady"/> 
      <FirstName>Clemansa</FirstName> 
      <Surname>Sanchez</Surname> 
     </PersonName> 
     <MaritalStatus Status="DeFacto"> 
      <RelatedEntityRef>ea384b0bf3f5</RelatedEntityRef> 
     </MaritalStatus> 
     <NumberOfDependents>1</NumberOfDependents> 
    </Person> 
</Party> 
<Party Type="Applicant" PrimaryApplicant="Yes" ExistingCustomerID="1231"> 
    <Identifier>bd8c65a3ad80</Identifier> 
    <Person Sex="Female" FirstHomeBuyer="Yes" CustomerOfLender="Yes"> 
     <PersonName> 
      <NameTitle Value="Mrs"/> 
      <FirstName>Cheryl</FirstName> 
      <Surname>Bonkers</Surname> 
     </PersonName> 
     <MaritalStatus Status="Married"> 
      <RelatedEntityRef>ee84dc9e38ec</RelatedEntityRef> 
     </MaritalStatus> 
     <NumberOfDependents>2</NumberOfDependents> 
    </Person> 
</Party> 
<Party Type="Guarantor" PrimaryApplicant="No"> 
    <Identifier>ea384b0bf3f5</Identifier> 
    <Person Sex="Male" FirstHomeBuyer="No" CustomerOfLender="No"> 
     <PersonName> 
      <NameTitle Value="Mr"/> 
      <FirstName>Greg</FirstName> 
      <OtherName>Morty</OtherName> 
      <Surname>Sanchez</Surname> 
     </PersonName> 
     <MaritalStatus Status="DeFacto"> 
      <RelatedEntityRef>b8b0f908b08e</RelatedEntityRef> 
     </MaritalStatus> 
     <NumberOfDependents>0</NumberOfDependents> 
    </Person> 
</Party> 
<Party Type="Applicant" PrimaryApplicant="No"> 
    <Identifier>ee84dc9e38ec</Identifier> 
    <Person Sex="Male" FirstHomeBuyer="No" CustomerOfLender="No"> 
     <PersonName> 
      <NameTitle Value="Mr"/> 
      <FirstName>Mark</FirstName> 
      <Surname>Bonkers</Surname> 
     </PersonName> 
     <MaritalStatus Status="Married"> 
      <RelatedEntityRef>bd8c65a3ad80</RelatedEntityRef> 
     </MaritalStatus> 
     <NumberOfDependents>0</NumberOfDependents> 
    </Person> 
</Party> 
</PartySegment>

的願望輸出：

<Household UniqueID="b8b0f908b08e-Household" 
       Name="Sanchez Household" 
       NumberOfAdults="2" 
       NumberOfDependants="1"/> 

<Household UniqueID="bd8c65a3ad80-Household" 
       Name="Bonkers Household" 
       NumberOfAdults="2" 
       NumberOfDependants="2"/>

我到目前爲止（簡化）：這是一對夫婦工作，因爲沒有考慮到MaritalStatus.RelatedEntityRef信息

<xsl:template match="PartySegment" mode="Household_Couple"> 

    <xsl:for-each-group select="Party[Person/MaritalStatus/@Status = ('DeFacto', 'Married')]" group-by="Person/MaritalStatus/@Status = ('DeFacto', 'Married')"> 
     <xsl:variable name="owner_id" select="Identifier"/> 

     <Household UniqueID="{concat(Identifier, '-Household')}" 
      Name="{normalize-space(concat(Person/PersonName/Surname, ' Household'))}" 
      NumberOfAdults="{'2'}" 
      NumberOfDependants="{if(Person/Dependent) then count(current-group()/Person/Dependent) else if(Person/NumberOfDependents) then sum(current-group()/Person/NumberOfDependents) else '0'}"> 

      <xsl:apply-templates select="current-group()/Person/Dependent"/> 

     </Household> 
    </xsl:for-each-group> 
</xsl:template>

來源

2017-10-16 DanielCSD

我不知道它是一個任務組，似乎很要遵循交叉參照的關鍵。是否有某種方法可以區分'Party'元素，處理所有的Party [@Type ='擔保人'和Person/MaritalStatus/@ Status =（'DeFacto'，'Married'）]'就足夠了，然後簡單地搜索引用其他具有'和'key'（'ref'，MaritalStatus/RelatedEntityRef）''的Party' –

黨元素只能是申請人或擔保人。有些合夥人處於一種關係中（DeFacto，已婚）和分擔費用（並且屬於一個家庭），有些是單身，離婚等費用，這創造了他們自己的家庭。我已經實現了爲一個人創建一個家庭的人員的所有信息，但不適用於多個夫婦（正在爲一對夫婦工作） – DanielCSD

如果你可以移動到XSLT 3.0那麼我認爲一個合成分組排序的兩個相關標識的關鍵sort((Identifier, key('ref', Person/MaritalStatus/RelatedEntityRef)/Identifier))可以解決這個問題：

<xsl:key name="ref" match="Party" use="Identifier"/> 

<xsl:template match="PartySegment"> 
    <xsl:for-each-group select="Party[Person/MaritalStatus/@Status = ('DeFacto', 'Married')]" 
     composite="yes" 
     group-by="sort((Identifier, key('ref', Person/MaritalStatus/RelatedEntityRef)/Identifier))"> 
     <Household UniqueID="{concat(Identifier, '-Household')}" 
      Name="{normalize-space(concat(Person/PersonName/Surname, ' Household'))}" 
      NumberOfAdults="{'2'}" 
      NumberOfDependants="{if(Person/Dependent) then count(current-group()/Person/Dependent) else if(Person/NumberOfDependents) then sum(current-group()/Person/NumberOfDependents) else '0'}"> 



     </Household> 
    </xsl:for-each-group> 
</xsl:template>

我得到

<Household UniqueID="b8b0f908b08e-Household" 
      Name="Sanchez Household" 
      NumberOfAdults="2" 
      NumberOfDependants="1"/> 
<Household UniqueID="bd8c65a3ad80-Household" 
      Name="Bonkers Household" 
      NumberOfAdults="2" 
      NumberOfDependants="2"/>

這SaXon 9.8 EE在oXygen內部。

對於撒克遜9.8 HE應該可以重寫，作爲

<xsl:key name="ref" match="Party" use="Identifier"/> 

<xsl:template match="PartySegment"> 
    <xsl:for-each-group select="Party[Person/MaritalStatus/@Status = ('DeFacto', 'Married')]" 
     composite="yes" 
     group-by="Identifier | key('ref', Person/MaritalStatus/RelatedEntityRef)/Identifier"> 
     <Household UniqueID="{concat(Identifier, '-Household')}" 
      Name="{normalize-space(concat(Person/PersonName/Surname, ' Household'))}" 
      NumberOfAdults="{'2'}" 
      NumberOfDependants="{if(Person/Dependent) then count(current-group()/Person/Dependent) else if(Person/NumberOfDependents) then sum(current-group()/Person/NumberOfDependents) else '0'}"> 



     </Household> 
    </xsl:for-each-group> 
</xsl:template>

對於XSLT 2.0，我們沒有，我們需要確保構建一個鍵與這兩個組件組合鍵：

<xsl:template match="PartySegment"> 
    <xsl:for-each-group select="Party[Person/MaritalStatus/@Status = ('DeFacto', 'Married')]" 
     group-by="string-join((Identifier | key('ref', Person/MaritalStatus/RelatedEntityRef)/Identifier), '|')"> 
     <Household UniqueID="{concat(Identifier, '-Household')}" 
      Name="{normalize-space(concat(Person/PersonName/Surname, ' Household'))}" 
      NumberOfAdults="{'2'}" 
      NumberOfDependants="{if(Person/Dependent) then count(current-group()/Person/Dependent) else if(Person/NumberOfDependents) then sum(current-group()/Person/NumberOfDependents) else '0'}"> 



     </Household> 
    </xsl:for-each-group> 
</xsl:template>

來源

2017-10-16 10:23:15

謝謝。這是工作。我一直在想，解決方案需要成爲分組的唯一鍵，現在我知道這有一個名稱（組合鍵）。 – DanielCSD

XSLT 2，按屬性和元素分組多個

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