如果元素屬性有注意，我想省略addressLine1

Question

如果Address元素的type屬性=「Standard With Attention」，我不想顯示AddressLine1。相反，我只想顯示AddressLine2，AddressLine3和AddressLine4。 我的輸出

經辦人：邁克爾，134 w ^百老匯ST NW，APT 3B，華盛頓特區，12345

預期輸出： 134W¯¯百老匯ST NW，APT 3B，華盛頓特區12345

我的XML

<ProtectedAddresses> 
    <Address InternalAddressID="1618613567" Type="Standard With Attention"> 
     <AddressLine1>Attn: Michael</AddressLine1> 
     <AddressLine2>134 W Broadway ST NW</AddressLine2> 
     <AddressLine3>APT 3B</AddressLine3> 
     <AddressLine4>Washington, DC, 12345</AddressLine4> 
     <Attention>James</Attention> 
     </Address> 
</ProtectedAddresses> 

XSLT

<xsl:for-each select="Addresses/Address"> 
    <xsl:for-each select="ancestor::ProtectionOrder/ProtectionOrderParties/ProtectionOrderParty/DCProtectionOrderPartyAdditional/ProtectedAddresses/Address[@InternalAddressID=current()/@InternalAddressID]"> 
    <xsl:for-each select="AddressLine1 | AddressLine2 | AddressLine3 | AddressLine4"> 
     <xsl:value-of select="."/> 
     <xsl:if test="position()!=last()">, </xsl:if> 
    </xsl:for-each> 
    <xsl:text>; </xsl:text> 
    </xsl:for-each> 
</xsl:for-each> 

Answer 1

如果您要更換<xsl:for-each select="AddressLine1 | AddressLine2 | AddressLine3 | AddressLine4">與<xsl:for-each select="AddressLine1[not(../@Type = 'Standard With Attention')] | AddressLine2 | AddressLine3 | AddressLine4">在構造節點設定如果條件../@Type = 'Standard With Attention'是真實的，只有三個要素。

Answer 2

你的例子相當混亂，因爲你只向我們展示了一部分XML。因此，很難給出準確的答案，因爲背景不夠清楚，而且不可能根據您的輸入測試給定的答案。

恕我直言，你應該使用關鍵，而不是這樣的：

<xsl:for-each select="ancestor::ProtectionOrder/ProtectionOrderParties/ProtectionOrderParty/DCProtectionOrderPartyAdditional/ProtectedAddresses/Address[@InternalAddressID=current()/@InternalAddressID]"> 

現在，爲了解決您的問題（儘可能地）：排除AddressLine1在某些情況下，你應該追加它一個謂語 - 即是這樣的：

<xsl:for-each select="AddressLine1[not(../@Type='Standard With Attention')] | AddressLine2 | AddressLine3 | AddressLine4"> 

這將工作，如果 - 因爲它似乎 - 條件在實際012的父元素存在。