Question

这是我的HTML：

<table dir = "rtl .......">
<tbody>
<script src = "get.aspx?type=js&file=ajax&rev=3"......>
<script language = "JavaScript" src = "get.aspx?type=js&file=mc&rev=6"></script>
<script>..</script>
<tr>
  <td class = "d2"...>..</td>
</tr>
<tr>..</tr> <--
<tr>..</tr> <--
<tr>..</tr> <-- these elements
<tr>..</tr> <--
<tr>..</tr> <--
<tr>..</tr> <--
<tr>..</tr> <--
<tr>
  <td class = "d2"...>..</td>
</tr>
<script>..</script>
<tr>..</tr>
<tr>..</tr>
<tr>..</tr>

如何计算或选择ID为<tr>的两个<td>元素之间的所有d2元素？

Answer 1

xpath将是一个很长的路径，所以支撑自己：

count(//tr[preceding-sibling::tr/td[@class = 'd2']][count(.|//tr[following-sibling::tr/td[@class = 'd2']])=count(//tr[following-sibling::tr/td[@class = 'd2']])])

要选择实际节点而不仅仅是计数，只需删除第一个计数：

//tr[preceding-sibling::tr/td[@class = 'd2']][count(.|//tr[following-sibling::tr/td[@class = 'd2']])=count(//tr[following-sibling::tr/td[@class = 'd2']])]

这里发生了各种各样的事情：

通过选择前一个兄弟的子节点tr / td并选择id ='2'来选择开始节点
通过选择以下兄弟的子节点tr / td并使用id ='2'选择结束节点
使用kaycian方法：http://www.dpawson.co.uk/xsl/sect2/muench.html#d9940e108获取两个节点之间的交集。

xpath选择两个节点之间的元素

1 个答案: