Question

如何在循环条件中在XSLT中包含if语句？

我有2或3个不同的场景定义了我想要循环的数据。例如，可能有也可能没有产品ID。如果有，我想只循环包含该产品ID的记录，否则我想要所有记录。我试图做这样的事情：

<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>

但它无法解析XSLT。我知道两个条件是相同的，一旦我开始工作，我就会重写一个。在此先感谢您的帮助！

更新：

以下是一些示例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>

我总是需要父ID为1073的文章。当且仅当存在产品ID时，我想要产品ID在relatedProducts列表中的文章。再次感谢！

Answer 1

例如，可能有也可能没有   产品ID。如果有，我想   只循环包含的记录   那个产品ID，否则我想要所有   记录。

我总是需要那些文章   父ID是1073.当且仅当a   产品ID存在，我想要文章   产品ID在哪里   relatedProducts列表。

使用：

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

以下是一个完整且简短的示例：

<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>

将此转换应用于以下XML文档（基于更新中提供的转换，但更有趣）：

<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> <p><strong>Change 2 Test 2<br /> </strong></p> </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> <p>tst</p> </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> <p>tst</p> </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> <p>tst</p> </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> <p>Test</p> </description> <relatedProducts /> <pageTitle>Test</pageTitle> <metaDescription>test</metaDescription> <metaKeywords>test</metaKeywords> <umbracoUrlName /> </Article> </Knowledgebase>

生成了需要的，正确的结果（只有包含parentID="1073"或relatedProducts包含$pProductId（1077）的文章）：

<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> <p><strong>Change 2 Test 2<br /> </strong></p> </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> <p>tst</p> </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> <p>tst</p> </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> <p>Test</p> </description> <relatedProducts/> <pageTitle>Test</pageTitle> <metaDescription>test</metaDescription> <metaKeywords>test</metaKeywords> <umbracoUrlName/> </Article>

Answer 2

我会使用xsl:choose（使用xsl:otherwise获取其他部分）以及模板：

<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>

然后定义一个模板来处理节点列表。

Answer 3

您可以在foreach循环结构中进行检查，如下所示：

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

选择后面的[]后面的条件表示您只需要具有与该条件匹配的属性的项目。所以像这样的东西会匹配：＆LT; item productId =“BR9ES”＆gt;

但这不会：＆LT; item＆gt;

Answer 4

这是我用来测试XML文档中是否存在元素的内容：

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

在For-Each中的XSLT条件

4 个答案: