XSL将带有属性的xml行转换为无序列表树视图

时间:2016-03-10 15:57:11

标签: xml xslt xslt-1.0 xslt-grouping muenchian-grouping

我有来自SharePoint控件的以下XML。我想使用XSLT进行转换以生成嵌套的ul> li列表。但我遇到问题,因为我迭代每一行,主文件夹重复每一行,而不是制作一个主文件夹节点,并在该节点下添加Menu_Display_name的值来模仿树视图... 我的XML就像:

<dsQueryResponse>
    <NewDataSet>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ CORP HR TIME SELF SERVICE" SECONDARY_FOLDER="Time" MENU_DISPLAY_NAME="Create Timecard"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ CORP HR TIME SELF SERVICE" SECONDARY_FOLDER="Time" MENU_DISPLAY_NAME="Recent Timecards"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ CORP HR TIME SELF SERVICE" SECONDARY_FOLDER="Time" MENU_DISPLAY_NAME="Templates"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ CORP HR TIME SELF SERVICE" SECONDARY_FOLDER="Time" MENU_DISPLAY_NAME="Timecard Search"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ CORP EXP ENTRY" SECONDARY_FOLDER="" MENU_DISPLAY_NAME="Expenses Home"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ HR EMP SELF SERVICE" SECONDARY_FOLDER="" MENU_DISPLAY_NAME="Accommodation Request"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ HR EMP SELF SERVICE" SECONDARY_FOLDER="" MENU_DISPLAY_NAME="Additional Personal Information" ></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ HR EMP SELF SERVICE" SECONDARY_FOLDER="" MENU_DISPLAY_NAME="All Actions Awaiting Your Attention"></Row>
        <Row DOMAIN_USERNAME="someUser" PRIMARY_FOLDER="XYZ HR EMP SELF SERVICE" SECONDARY_FOLDER="" MENU_DISPLAY_NAME="Appraisals"></Row>
    </NewDataSet>
</dsQueryResponse>

我在XSL上的“穷人的尝试”是这样的:

 <xsl:template name="dvt_1.body">
    <xsl:param name="Rows" />
    <xsl:param name="FirstRow" />
    <xsl:param name="LastRow" />
    <ul>
      <xsl:for-each select="$Rows">
        <xsl:if test="position() &gt;= $FirstRow and position() &lt;= $LastRow">
          <xsl:call-template name="dvt_1.rowview" />
        </xsl:if>
      </xsl:for-each>
    </ul>
  </xsl:template>


  <xsl:template name="dvt_1.rowview">

    <li class="isFolder isExpanded">
       <xsl:value-of select="@PRIMARY_FOLDER" />
       <xsl:choose>
         <xsl:when test="@SECONDARY_FOLDER != ''">
           <ul>
             <li class="isFolder isExpanded">
               <xsl:value-of select="@SECONDARY_FOLDER" />
               <ul>
                  <li><xsl:value-of select="@MENU_DISPLAY_NAME" /></li>
               </ul>
             </li>
           </ul>
         </xsl:when>
         <xsl:otherwise>
           <ul>
              <li><xsl:value-of select="@MENU_DISPLAY_NAME" /></li>
           </ul>
          </xsl:otherwise>
        </xsl:choose>
    </li>

  </xsl:template>

我想要的xsl输出是这样的:

<div id="navigator">
        <ul>
            <li class="isFolder isExpanded">
               XYZ CORP HR TIME SELF SERVICE    
                <ul>
                    <li class="isFolder isExpanded">
                        Time
                            <ul>
                                <li><a href="#" target="_tab">Create Timecard</a></li>
                                <li><a href="#" target="_tab">Recent Timecards</a></li>
                                <li><a href="#" target="_tab">Templates</a></li>
                                <li><a href="#" target="_tab">Timecard Search</a></li>                                   
                            </ul>
                    </li>
                </ul>        
            </li>
            <li class="isFolder isExpanded">
                XYZ CORP EXP ENTRY
                <ul>
                    <li><a href="#" target="_tab">Expense Home</a></li>      
                </ul>
            </li> 
            <li class="isFolder isExpanded">
                XYZ HR EMP SELF SERVICE
                <ul>
                    <li><a href="#" target="_tab">Accommodation Request</a></li>
                    <li><a href="#" target="_tab">Additional Personal Information</a></li>
                    <li><a href="#" target="_tab">All Actions Awaiting Your Attention</a></li>
                    <li><a href="#" target="_tab">Appraisals</a></li>
                </ul>
            </li>
        </ul>
    </div>

有人可以帮我用xslt帮助我实现这个目标吗?

1 个答案:

答案 0 :(得分:4)

Muenchian grouping确实是你需要在XSLT 1.0中看到的(这是Sharepoint使用的,我相信)。首先按PRIMARY_VALUE属性进行分组,因此您有一个这样的键:

<xsl:key name="primary" match="Row" use="@PRIMARY_FOLDER" />

但我会假设您可能会为给定的SECONDARY_FOLDER提供多个PRIMARY_VALUE,因此您需要第二个密钥:

<xsl:key name="secondary" match="Row" use="concat(@PRIMARY_FOLDER, '|', @SECONDARY_FOLDER)" />

首先选择第一次出现PRIMARY_FOLDER

的行
<xsl:for-each select="$rows[generate-id() = generate-id(key('primary', @PRIMARY_FOLDER)[1])]">

然后在其中,您选择具有不同SECONDARY_VALUE的行以构成嵌套列表的基础

<xsl:apply-templates select="key('primary', @PRIMARY_FOLDER)[generate-id() = generate-id(key('secondary', concat(@PRIMARY_FOLDER, '|', @SECONDARY_FOLDER))[1])]" mode="secondary" />

唯一的额外工作是根据是否填充SECONDARY_FOLDER,您的行为略有不同。您可以使用两个单独的模板执行此操作。

试试这个XSLT:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="html" indent="yes" />

    <xsl:key name="primary" match="Row" use="@PRIMARY_FOLDER" />
    <xsl:key name="secondary" match="Row" use="concat(@PRIMARY_FOLDER, '|', @SECONDARY_FOLDER)" />

    <xsl:variable name="rows" select="//Row" />

    <xsl:template match="/">
      <ul>
          <xsl:for-each select="$rows[generate-id() = generate-id(key('primary', @PRIMARY_FOLDER)[1])]">
              <li class="isFolder isExpanded">
                  <xsl:value-of select="@PRIMARY_FOLDER" />
                  <xsl:apply-templates select="key('primary', @PRIMARY_FOLDER)[generate-id() = generate-id(key('secondary', concat(@PRIMARY_FOLDER, '|', @SECONDARY_FOLDER))[1])]" mode="secondary" />
              </li>
          </xsl:for-each>
      </ul>
    </xsl:template>

    <xsl:template match="Row[@SECONDARY_FOLDER != '']" mode="secondary">
        <li class="isFolder isExpanded">
            <xsl:value-of select="@SECONDARY_FOLDER" />
            <ul>
                <xsl:apply-templates select="key('secondary', concat(@PRIMARY_FOLDER, '|', @SECONDARY_FOLDER))" />
            </ul>
        </li>
    </xsl:template>

    <xsl:template match="Row" mode="secondary">
        <xsl:apply-templates select="key('primary', @PRIMARY_FOLDER)" />
    </xsl:template>

    <xsl:template match="Row">
        <li>
            <a href="#" target="_tab">
                <xsl:value-of select="@MENU_DISPLAY_NAME" />
            </a>
        </li>
    </xsl:template>
</xsl:stylesheet>

http://xsltransform.net/pPzifq9处查看此操作。

请注意,如果您的实际XML具有名称空间,则需要修改XSLT以将其考虑在内。