如何选择具有特定节点作为子节点的第一个前兄弟? [英] How to select the first preceding-sibling which has a particular node as child?

问题描述

我一直在尝试编写一个 XPath 语句，该语句将选择节点的第一个前同级.那个兄弟节点应该有一个特定的节点作为它的子节点.

I have been trying to write a XPath statement which will select the first preceding-sibling of a node. That sibling should have a particular node as its child.

例如:

<a>
    <c></c>
    ....
    <b></b>
</a>

<a></a>
....
....
<a></a>

<a>
   <b> start </b>
</a>

在上述 XML 中，如果我的上下文位于节点 a 处，该节点在其子节点 b 中有 start 文本.如何选择子节点为 b 的前一个节点 a?

In the above XML, if my context is at the node a which has start text inside its child b. How can I select the preceding node a whose child is b?

我尝试了 preceding-sibling::*[b][1] 但它只选择节点 a 如果 b 是它的第一个节点孩子.我也试过 preceding-sibling::*[1][b] 但它只检查前面的第一个 a 节点，如果它不包含 b，测试失败.

I tried preceding-sibling::*[b][1] but it only selects the node a if b is its first child. I also tried preceding-sibling::*[1][b] but it only checks the first preceding a node and if it doesn't contains b, the test fails.

注意:我使用的是 XSLT 1.0

Note: I am using XSLT 1.0

谁能给我建议一个解决方案?

Can anyone suggest me a solution?

提前谢谢！！

推荐答案

我相信:

preceding-sibling::*[b][1]

或者最好:

preceding-sibling::a[b][1]

应该可以正常工作.

我尝试过previous-sibling::*[b][1] 但它只选择节点a if b是它的第一个孩子.

I tried preceding-sibling::*[b][1] but it only selects the node a if b is its first child.

我不这么认为.

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