1

我正在尝试使用 sum 和 count 变量对每个地方进行“平均目标得分”排序：

我的 XML 是：

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="rugby.xsl"?>

<rugby>
    <games>
        <game>
            <place>USA</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>20-1</score>
        </game>
        <game>
            <place>USA</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>2-20</score>
        </game>
        <game>
            <place>USA</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>24-11</score>
        </game>
        <game>
            <place>USA</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>12-41</score>
        </game>
        <game>
            <place>USA</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>20-100</score>
        </game>
        <game>
            <place>Mexico</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>1-32</score>
        </game>
        <game>
            <place>Mexico</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>2-100</score>
        </game>
        <game>
            <place>Peru</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>2-10</score>
        </game>
        <game>
            <place>Peru</place>
            <team1>Team1</team1>
            <team1>Team2</team1>
            <score>100-2</score>
        </game>
    </games>
</rugby>

和 XSLT：

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">

    <xsl:template match="/rugby">
        <html>
            <head>
                <link rel="stylesheet" href="my_style.css" type="text/css"/>
            </head>
            <body>
                <table>
                    <tr>
                        <td>Place</td>
                        <td>Average Goals Score</td>
                        <td>Average Goals Conceded</td>
                    </tr>
                    <xsl:variable name="numOfGames" select="count(//games/game)"/>
                    <xsl:for-each select="//games/game/place">
                        <xsl:variable name="nameOfPlace" select="text()"/>
                        <xsl:variable name="score" select="sum(//games/game/place[contains(., $nameOfPlace)]/following-sibling::score/number(substring-before(.,'-')))"/>
                        <xsl:variable name="conceded" select="sum(//games/game/place[contains(., $nameOfPlace)]/following-sibling::score/number(substring-after(.,'-')))"/>

                        <tr>
                            <td><xsl:value-of select="place"/></td>
                            <td><xsl:value-of select="$score div $numOfGames"/></td>
                            <td><xsl:value-of select="$conceded div $numOfGames"/></td>
                        </tr>
                    </xsl:for-each>
                </table>
            </body>
        </html>
    </xsl:template>
</xsl:stylesheet>

我试图对变量进行排序：

<xsl:sort select="$score div $numOfGames" order="ascending" data-type="number"/>

但它不起作用，因为排序必须在 for each 下（我必须先创建变量）。我试图删除变量 sum 和 count 但我无法删除变量 nameOfPlace，所以排序根本不起作用。

<xsl:for-each select="//games/game/place">
                        <xsl:variable name="nameOfPlace" select=""/>
                        <xsl:sort select="sum(//games/game/place[contains(., $nameOfPlace)]/following-sibling::score/number(substring-before(.,'-'))) div count(//games/game)" order="ascending" data-type="number"/>
                        <tr>
                            <td><xsl:value-of select="place"/></td>
                            <td><xsl:value-of select="sum(//games/game/place[contains(., $nameOfPlace)]/following-sibling::score/number(substring-before(.,'-'))) div count(//games/game)"/></td>
                            <td><xsl:value-of select="sum(//games/game/place[contains(., $nameOfPlace)]/following-sibling::score/number(substring-after(.,'-'))) div count(//games/game)"/></td>
                        </tr>
                    </xsl:for-each>

我怎样才能避免这种情况？

4

1 回答 1

0

使用变量的一种方法是使用以下命令在 XPath 表达式中声明变量for：

<xsl:sort select="for $nameOfPlace in text(),
      $score in sum(//games/game/place[contains(., $nameOfPlace)]/
                     following-sibling::score/number(substring-before(.,'-')))
        return $score div $numOfGames"
   order="ascending" data-type="number"/>

类似的东西。（未经测试。）

是的，它很笨重。不幸的是，这些变量在 XPath 表达式之外将不可用。因此，如果您想在排序后使用它们，则必须再次声明它们。以上是否是对没有变量的改进取决于您。

更新： 关于在没有变量的情况下进行操作。

您无法弄清楚如何删除nameOfPlace变量声明。这个怎么样：

<xsl:for-each select="//games/game/place">
  <xsl:sort select="sum(//games/game/place[contains(., current()/text())]/
        following-sibling::score/number(substring-before(.,'-')))
      div count(//games/game)"
    order="ascending" data-type="number"/>

current()允许您引用当前 XPath 表达式 ( )之外sort select的任何上下文项。在这种情况下，这将是迭代的<place>元素。<xsl:-for-each>

于 2013-10-30T11:24:06.870 回答