Question

我希望有人可以帮助我，因为我一直试图弄清楚如何将数据集转换为在多个级别上分组的表结构。第一级是事件日期，第二级是公司名称，最后一级是用户。

以下是XML。

<xmlData>
    <records>
      <record>
        <userid>1</userid>
        <usersname>Jane Doe</usersname>
        <companyname>Company A</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
      <record>
        <userid>3</userid>
        <usersname>Jane Doe</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 DEC 2016</eventdate>
        <jeventdate>2457724</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 JAN 2017</eventdate>
        <jeventdate>2457755</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
    </records>
</xmlData>

我试图获得的结果显示在以下简单的HTML输出中。

<h1>01 DEC 2016</h1>
  <h2>Company B</h2>
  <table>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>
<h1>01 JAN 2017</h1>
  <h2>Company B</h2>
  <table>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>
<h1>01 FEB 2017</h1>
  <h2>Company A</h2>
  <table>
    <tr><td>1</td><td>Jane Doe</td></tr>
  </table>
  <h2>Company B</h2>
  <table>
    <tr><td>3</td><td>Dave Dodd</td></tr>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>

我遇到的问题实际上是两个部分。首先是获得三个层次的深度。第二个是获取所有记录，而不仅仅是前几个记录。

这是我一直在使用的XSL。

<xsl:key name="monthof" match="record" use="eventdate"/>
<xsl:key name="companyof" match="record" use="concat(eventdate,'|',companyname)"/>
<xsl:key name="userof" match="record" use="concat(eventdate,'|',companyname,'|',usersname)"/>

<xsl:template match="xmlData/records">
    <xsl:for-each select="record[key('monthof',eventdate)]">
        <xsl:sort select="jeventdate"/>
        <xsl:variable name="lstEventDate" select="key('monthof',eventdate)" />
        <h2><xsl:value-of select="eventdate"/></h2>
        <xsl:for-each select="key('companyof',concat(eventdate,'|',companyname))">
            <xsl:sort select="companyname"/>
            <h3>
                <xsl:value-of select="companyname"/>
            </h3>
            <table>
                <xsl:for-each select="key('userof',concat(eventdate,'|',companyname,'|',usersname))">
                    <tr>
                        <td><xsl:value-of select="userid"/></td>
                        <td><xsl:value-of select="usersname"/></td>
                    </tr>                                   
                </xsl:for-each>
            </table>
        </xsl:for-each>
    </xsl:for-each>
</xsl:template>

这是我得到的输出。

01 DEC 2016

Company B

2 Jane Doe 

01 JAN 2017

Company B

2 Jane Doe 

01 FEB 2017

Company A

1 Joe Smith 

01 FEB 2017

Company B

1 Joe Smith 

Company B

3 Dave Dodd 

01 FEB 2017

Company B

1 Joe Smith 

Company B

2 Jane Doe

Answer 1

您定义的键很好，但您需要使用Muenchian grouping方法使用它们，如评论中已经指出的那样：

Customer c1 = new Customer(1,"Raja",10);
Customer c2 = new Customer(1,"Raja",15);
Customer c3 = new Customer(2,"Raja",10);

您可能想要检查是否无法移动到XSLT 2.0，这样可以更容易地进行

<head>
<link rel="stylesheet" type="text/css" href="style.css">
<script type="text/javascript" src="/js/jquery-3.1.1.js">    </script>
<script type="text/javascript" src="/js/menu.js"></script>
</head>

xsl转换用于分组多层结果

1 个答案: