Question

我有xml文档，如:(这可以嵌套多次）

<?xml version = "1.0" encoding = "UTF -8"?>
<person xmlns:xsi = " http://www.w3.org/2001/XMLSchema-instance"
    name = "Adrian"
    gender = "male">
    <children>
        <person name = "Sabina" gender = "female">
            <children>
            <person name = "Dolores" gender = "female"/>
            <person name = "Ahmed" gender = "male"/>
            <person name = "Nusrat" gender = "female"/>
            </children>
        </person>
        <person name = "Alfonso" gender = "male">
            <children>
            <person gender = "female" name = "Perdita"/>
            <person gender = "male" name = "Ioan "> </person>
            </children>
        </person>
    </children>
</person>

我正在提供原始XML的简短形式。 xml可以是任意深度。我必须找到关系。（1）兄弟关系（2）侄子侄子关系。（巴黎acutally）

我的兄弟巴黎：

<sibling>Sabrina,ALfonso</sibling> <sibling>Dolores,Ahmed</sibling>
<sibling>Dolores,Nusrat</sibling> <sibling>Ahmed,Nusrat</sibling> 
<sibling>Perdita,Ioan</sibling>

这不应该是艰难的。但是对于Aunt-Nephew对，我们也必须检查性别。阿姨 - 侄子对输出

<AuntNeph>Sabrina,Ioan</AuntNeph>

感谢。

Answer 1

对于每个person，请检查following-sibling轴以查找其他人（兄弟姐妹）。只检查一种方式（不是后面和前面的）以省略重复的兄弟对（例如“Sabina Alfonso”和“Alfonso Sabina”）。

for $person in //person
for $sibling in $person/following-sibling::person
return element sibling { string-join(($person, $sibling)/@name, ',') }

对于侄子，查询非常相似，我认为不需要进一步解释：

for $aunt in //person[@gender="female"]
for $nephew in $aunt/(preceding-sibling::person, following-sibling::person)/children/person[@gender="male"]
return element AuntNeph { string-join(($aunt, $nephew)/@name, ',') }

Xquery - 找到如何找到任意深度节点

1 个答案: