Question

有没有办法使用xPath从第一个'row'到第二个和前面'row'节点的每个节点的正确节点？
XML数据：

<extracted>
     <row>
        <item>Hour1</item>
        <item>Hour2</item>
        <item>Hour3A</item>
        <item>Hour3B</item>
        <item>Hour4</item>
        <item>Hour5</item>
        <item>Hour6</item>
        <item>Hour7</item>
        <item>Hour8</item>
        <item>Hour9</item>
        <item>Hour10</item>
        <item>Hour11</item>
        <item>Hour12</item>
        <item>Hour13</item>
        <item>Hour14</item>
        <item>Hour15</item>
        <item>Hour16</item>
        <item>Hour17</item>
        <item>Hour18</item>
        <item>Hour19</item>
        <item>Hour20</item>
        <item>Hour21</item>
        <item>Hour22</item>
        <item>Hour23</item>
        <item>Hour24</item>
        <item>Aver./Sum</item>
      </row>        
      <row>
        <item>12,43</item>
        <item>12,40</item>
        <item>12,40</item>
        <item/>
        <item>12,40</item>
        <item>12,40</item>
        <item>12,48</item>
        <item>14,35</item>
        <item>15,48</item>
        <item>22,79</item>
        <item>24,16</item>
        <item>24,35</item>
        <item>28,25</item>
        <item>24,68</item>
        <item>23,30</item>
        <item>21,93</item>
        <item>16,76</item>
        <item>15,08</item>
        <item>15,06</item>
        <item>14,89</item>
        <item>14,79</item>
        <item>16,06</item>
        <item>15,45</item>
        <item>14,57</item>
        <item>14,57</item>
        <item>17,13</item>
    </row>
    <row>
        <item>12,43</item>
        <item>12,40</item>
        <item>12,40</item>
        <item/>
        <item>12,40</item>
        <item>12,40</item>
        <item>12,48</item>
        <item>14,35</item>
        <item>15,48</item>
        <item>22,79</item>
        <item>24,16</item>
        <item>24,35</item>
        <item>28,25</item>
        <item>24,68</item>
        <item>23,30</item>
        <item>21,93</item>
        <item>16,76</item>
        <item>15,08</item>
        <item>15,06</item>
        <item>14,89</item>
        <item>14,79</item>
        <item>16,06</item>
        <item>15,45</item>
        <item>14,57</item>
        <item>14,57</item>
        <item>17,13</item>
    </row>
</extracted>

例如：12,43匹配“Hour1”。只有这一行有“小时”但有几个有值。

我想做的事情：

以某种方式获取节点的位置并在第一个“行”节点的谓词中使用位置。

Answer 1

答案可能涉及position()函数，但魔鬼在细节中，这一切都取决于您当时正在操作的上下文。以XSLT为例，这有效：

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:template match="extracted">
    <ex>
      <xsl:apply-templates select="row[position() > 1]" />
    </ex>
  </xsl:template>

  <xsl:template match="row">
    <r>
      <xsl:apply-templates select="item" />
    </r>
  </xsl:template>

  <xsl:template match="item">
    <!-- the current node is an item, and the current node list is the set
         of item children of the parent row element.  Therefore position()
         is 1 for the first item, 2 for the second item, etc. -->
    <xsl:variable name="myPosition" select="position()" />
    <!-- so here we select the nth item from the first row of the table -->
    <entry key="{../../row[1]/item[$myPosition]}" value="{.}" />
  </xsl:template>
</xsl:stylesheet>

但这不是

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:template match="extracted">
    <ex>
      <xsl:apply-templates select="row[position() > 1]" />
    </ex>
  </xsl:template>

  <xsl:template match="row">
    <r>
      <xsl:apply-templates select="node()" />
    </r>
  </xsl:template>

  <xsl:template match="item">
    <!-- the current node is an item, and the current node list is the set
         of all child nodes of the parent row element, including text nodes.
         Therefore position() is 1 for the whitespace before the first item,
         2 for the first item itself, 3 for the whitespace between the first and
         second items, 4 for the second item, etc. -->
    <xsl:variable name="myPosition" select="position()" />
    <!-- so here we select the wrong item from the first row of the table -->
    <entry key="{../../row[1]/item[$myPosition]}" value="{.}" />
  </xsl:template>
</xsl:stylesheet>

如果您无法确定是否在一个有效position()的上下文中执行XPath表达式，那么您将不得不使用其他方法，例如

count(preceding-sibling::item) + 1

通过计算此前item个元素的数量来计算正确的位置。

XPath：选择具有非唯一节点名称的另一个节点的位置节点

1 个答案: