XSLT有條件的For-Each

Question

如何在XSLT中的條件循環中包含if語句？

我有2或3個不同的場景來定義我想要循環的數據。例如，可能有或沒有產品ID。如果有，我只想遍歷包含該產品ID的記錄，否則我需要所有記錄。我試圖做這樣的事情：

<xsl:variable name="knowledgebaseLoop"> 
     <xsl:choose> 
     <xsl:when test="string-length($productId) &gt; 0"> 
      <xsl:copy-of select="$currentPage/child::* [@parentID = $parentId]"/> 
     </xsl:when> 
     <xsl:otherwise> 
      <xsl:copy-of select="$currentPage/child::* [@parentID = $parentId]"/> 
     </xsl:otherwise> 
     </xsl:choose> 
    </xsl:variable> 

<xsl:for-each select="knowledgebaseLoop"> 
     <xsl:if test="position() &gt; $recordsPerPage * number($pageNumber - 1) and position() &lt;= number($recordsPerPage * number($pageNumber - 1) + $recordsPerPage)"> 
     <div id="main_body_listing"> 
      <div class="listing_title2"><a href="{umbraco.library:NiceUrl(current()/@id)}"><xsl:value-of select="current()/@nodeName"/></a></div> 
      <xsl:value-of select="current()/summary"/> 
     </div> 
     </xsl:if> 
    </xsl:for-each> 

但它無法解析XSLT。我知道這兩個條件是一樣的，一旦我開始工作，我會重寫一個條件。預先感謝任何幫助！

UPDATE：

下面是一些示例XML：

<Knowledgebase id="1073" parentID="-1" level="1" writerID="0" creatorID="0" nodeType="1062" template="1083" sortOrder="5" createDate="2011-04-20T13:43:41" updateDate="2011-05-11T14:34:23" nodeName="Knowledgebase" urlName="knowledgebase" writerName="Chris" creatorName="Chris" path="-1,1073" isDoc=""><pageTitle>Knowledgebase</pageTitle><metaDescription>knowledgebase</metaDescription><metaKeywords>knowledgebase</metaKeywords><umbracoUrlName /><Article id="1074" parentID="1073" level="2" writerID="0" creatorID="0" nodeType="1063" template="1084" sortOrder="1" createDate="2011-04-20T13:44:13" updateDate="2011-05-16T09:39:12" nodeName="What is the answer to this question?" urlName="what-is-the-answer-to-this-question" writerName="Chris" creatorName="Chris" path="-1,1073,1074" isDoc=""><title>Change 2 Test 2</title><summary>Change 2 Test 2</summary><description> 

&lt;p&gt;&lt;strong&gt;Change 2 Test 2&lt;br /&gt; 
&lt;/strong&gt;&lt;/p&gt; 
</description><relatedProducts>1071,1076</relatedProducts><pageTitle>asd</pageTitle><metaDescription>asd</metaDescription><metaKeywords>asd</metaKeywords><umbracoUrlName /></Article><Article id="1082" parentID="1073" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""><title>test</title><summary>test 22</summary><description> 
&lt;p&gt;tst&lt;/p&gt; 

</description><relatedProducts>1075</relatedProducts><pageTitle /><metaDescription></metaDescription><metaKeywords></metaKeywords><umbracoUrlName /></Article><Article id="1146" parentID="1073" level="2" writerID="4" creatorID="3" nodeType="1063" template="1084" sortOrder="3" createDate="2011-05-13T15:36:14" updateDate="2011-05-13T15:37:21" nodeName="Question 3" urlName="question-3" writerName="Content Approver" creatorName="Content Writer" path="-1,1073,1146" isDoc=""><title>Test</title><summary></summary><description> 
&lt;p&gt;Test&lt;/p&gt; 
</description><relatedProducts /><pageTitle>Test</pageTitle><metaDescription>test</metaDescription><metaKeywords>test</metaKeywords><umbracoUrlName /></Article></Knowledgebase> 

我總是需要在父ID是1073.如果物品且僅當一個產品ID存在，我想的文章，其中產品ID位於relatedProducts列表中。再次感謝！

Answer 1

例如，有可能會或可能不會是 產品ID。如果有，我想要 只循環包含 那個產品ID的記錄，否則我想要所有的 記錄。

我總是需要當且僅當一個 產品ID存在，我想文章 該產品ID是在 relatedProducts名單，其中 父ID爲1073的文章。

使用：

<xsl:variable name="vDesiredArticles" select= 
    "/*/Article[@parentID='1073' 
       or 
       contains(concat(',',relatedProducts,','), 
         concat(',', $pProductId, ',') 
         ) 
       ]"/> 

下面是一個完整而簡單的例子：

<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:param name="pProductId" select="1077"/> 

<xsl:template match="/"> 
    <xsl:variable name="vDesiredArticles" select= 
    "/*/Article[@parentID='1073' 
       or 
       contains(concat(',',relatedProducts,','), 
         concat(',', $pProductId, ',') 
         ) 
       ]"/> 
    <xsl:copy-of select="$vDesiredArticles"/> 
</xsl:template> 
</xsl:stylesheet> 

當這種轉變是在下面的XML文檔應用（基於在更新中提供了一個，但變得更有趣）：

<Knowledgebase id="1073" parentID="-1" level="1" writerID="0" 
creatorID="0" nodeType="1062" template="1083" sortOrder="5" 
createDate="2011-04-20T13:43:41" updateDate="2011-05-11T14:34:23" 
nodeName="Knowledgebase" urlName="knowledgebase" writerName="Chris" 
creatorName="Chris" path="-1,1073" isDoc=""> 
    <pageTitle>Knowledgebase</pageTitle> 
    <metaDescription>knowledgebase</metaDescription> 
    <metaKeywords>knowledgebase</metaKeywords> 
    <umbracoUrlName /> 
    <Article id="1074" parentID="1073" level="2" 
    writerID="0" creatorID="0" nodeType="1063" template="1084" 
    sortOrder="1" createDate="2011-04-20T13:44:13" 
    updateDate="2011-05-16T09:39:12" 
    nodeName="What is the answer to this question?" 
    urlName="what-is-the-answer-to-this-question" writerName="Chris" 
    creatorName="Chris" path="-1,1073,1074" isDoc=""> 
     <title>Change 2 Test 2</title> 
     <summary>Change 2 Test 2</summary> 
     <description> &lt;p&gt;&lt;strong&gt;Change 2 Test 2&lt;br /&gt; &lt;/strong&gt;&lt;/p&gt; </description> 
     <relatedProducts>1071,1076</relatedProducts> 
     <pageTitle>asd</pageTitle> 
     <metaDescription>asd</metaDescription> 
     <metaKeywords>asd</metaKeywords> 
     <umbracoUrlName /> 
    </Article> 
    <Article id="1082" parentID="1073" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""> 
     <title>test</title> 
     <summary>test 22</summary> 
     <description> &lt;p&gt;tst&lt;/p&gt; </description> 
     <relatedProducts>1075</relatedProducts> 
     <pageTitle /> 
     <metaDescription></metaDescription> 
     <metaKeywords></metaKeywords> 
     <umbracoUrlName /> 
    </Article> 
    <Article id="1088" parentID="1077" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""> 
     <title>test</title> 
     <summary>test 22</summary> 
     <description> &lt;p&gt;tst&lt;/p&gt; </description> 
     <relatedProducts>1075,1039,1077</relatedProducts> 
     <pageTitle /> 
     <metaDescription></metaDescription> 
     <metaKeywords></metaKeywords> 
     <umbracoUrlName /> 
    </Article> 
    <Article id="1089" parentID="1077" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""> 
     <title>test</title> 
     <summary>test 22</summary> 
     <description> &lt;p&gt;tst&lt;/p&gt; </description> 
     <relatedProducts>1075,1039,1078</relatedProducts> 
     <pageTitle /> 
     <metaDescription></metaDescription> 
     <metaKeywords></metaKeywords> 
     <umbracoUrlName /> 
    </Article> 
    <Article id="1146" parentID="1073" level="2" writerID="4" creatorID="3" nodeType="1063" template="1084" sortOrder="3" createDate="2011-05-13T15:36:14" updateDate="2011-05-13T15:37:21" nodeName="Question 3" urlName="question-3" writerName="Content Approver" creatorName="Content Writer" path="-1,1073,1146" isDoc=""> 
     <title>Test</title> 
     <summary></summary> 
     <description> &lt;p&gt;Test&lt;/p&gt; </description> 
     <relatedProducts /> 
     <pageTitle>Test</pageTitle> 
     <metaDescription>test</metaDescription> 
     <metaKeywords>test</metaKeywords> 
     <umbracoUrlName /> 
    </Article> 
</Knowledgebase> 

想要的，正確的結果產生（僅本刊parentID="1073"或其relatedProducts包含$pProductId（1077））：

<Article id="1074" parentID="1073" level="2" writerID="0" creatorID="0" nodeType="1063" template="1084" sortOrder="1" createDate="2011-04-20T13:44:13" updateDate="2011-05-16T09:39:12" nodeName="What is the answer to this question?" urlName="what-is-the-answer-to-this-question" writerName="Chris" creatorName="Chris" path="-1,1073,1074" isDoc=""> 
    <title>Change 2 Test 2</title> 
    <summary>Change 2 Test 2</summary> 
    <description> &lt;p&gt;&lt;strong&gt;Change 2 Test 2&lt;br /&gt; &lt;/strong&gt;&lt;/p&gt; </description> 
    <relatedProducts>1071,1076</relatedProducts> 
    <pageTitle>asd</pageTitle> 
    <metaDescription>asd</metaDescription> 
    <metaKeywords>asd</metaKeywords> 
    <umbracoUrlName/> 
</Article> 
<Article id="1082" parentID="1073" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""> 
    <title>test</title> 
    <summary>test 22</summary> 
    <description> &lt;p&gt;tst&lt;/p&gt; </description> 
    <relatedProducts>1075</relatedProducts> 
    <pageTitle/> 
    <metaDescription/> 
    <metaKeywords/> 
    <umbracoUrlName/> 
</Article> 
<Article id="1088" parentID="1077" level="2" writerID="0" creatorID="3" nodeType="1063" template="1084" sortOrder="2" createDate="2011-05-04T09:59:31" updateDate="2011-05-11T14:34:23" nodeName="Question 2" urlName="question-2" writerName="Chris" creatorName="Content Writer" path="-1,1073,1082" isDoc=""> 
    <title>test</title> 
    <summary>test 22</summary> 
    <description> &lt;p&gt;tst&lt;/p&gt; </description> 
    <relatedProducts>1075,1039,1077</relatedProducts> 
    <pageTitle/> 
    <metaDescription/> 
    <metaKeywords/> 
    <umbracoUrlName/> 
</Article> 
<Article id="1146" parentID="1073" level="2" writerID="4" creatorID="3" nodeType="1063" template="1084" sortOrder="3" createDate="2011-05-13T15:36:14" updateDate="2011-05-13T15:37:21" nodeName="Question 3" urlName="question-3" writerName="Content Approver" creatorName="Content Writer" path="-1,1073,1146" isDoc=""> 
    <title>Test</title> 
    <summary/> 
    <description> &lt;p&gt;Test&lt;/p&gt; </description> 
    <relatedProducts/> 
    <pageTitle>Test</pageTitle> 
    <metaDescription>test</metaDescription> 
    <metaKeywords>test</metaKeywords> 
    <umbracoUrlName/> 
</Article> 

Answer 2

你可以檢查在foreach循環結構，像這樣：

<xsl:for-each select="item[@productId != '']"> 

中的condtion的[]在選擇項目後，表示您僅希望與屬性匹配條件的項目。因此，像這樣將匹配： <項目的productId =「BR9ES」>

但這並不： <項目>

Answer 3

這是我用它來測試，如果在我的XML文檔中存在的元素：

<xsl:if test="@productID[.!='']">  //if it is not empty 
<xsl:value-of select="@productID"/>  
</xsl:if> 

Answer 4

我會用一個xsl:choose（獲取其他部分與xsl:otherwise）加模板：

<xsl:choose> 
<xsl:when test='@productId'> 
    <xsl:apply-templates select='filter by product id' /> 
</xsl:when> 
<xsl:otherwise> 
    <xsl:apply-templates select='all of them' /> 
</xsl:otherwise> 

然後定義一個模板處理節點列表。