XML: Return 个满足给定条件的序列元素

Question

我有一个包含电影信息的 XML 文件。我想表达我对 return XML 导演超过 1 部电影的导演的表达。

我现在拥有的是：

for $director in //director
where count($director) > 1
return $director

这不会产生任何结果。我想做的是：

对于 XML 中的每一位董事： return导演的名字如果导演的名字在//director

中被多次提及

你会如何解决这个问题？

编辑：XML 的样本：

<result>
    <videos>
        <video id="id1235AA0">
            <title>The Fugitive</title>
            <genre>action</genre>
            <rating>PG-13</rating>
            <summary>Tommy Lee Jones and Harrison Ford are the hunter and the hunted in this fast-paced story of a falsely convicted man who escapes to find his wife's true killer.</summary>
            <details>Harrison Ford and Tommy Lee Jones race through the breathless manhunt movie based on the classic TV series. Ford is prison escapee Dr. Richard Kimble, a Chicago surgeon falsely convicted of killing his wife and determined to prove his innocence by leading his pursuers to the one-armed man who actually commited the crime.</details>
            <year>1997</year>
            <director>Andrew Davis</director>
            <studio>Warner</studio>
            <user_rating>4</user_rating>
            <runtime>110</runtime>
            <actorRef>00000003</actorRef>
            <actorRef>00000006</actorRef>
            <vhs>13.99</vhs>
            <vhs_stock>206</vhs_stock>
            <dvd>14.99</dvd>
            <dvd_stock>125</dvd_stock>
            <beta>1.03</beta>
            <beta_stock>12</beta_stock>
            <LaserDisk>12.00</LaserDisk>
            <LaserDisk_stock>10</LaserDisk_stock>
        </video>
    </videos>
</result>

Answer 1

您根本没有执行聚合。如果您的 XQuery 引擎支持 XQuery 3.0，请使用 group by:

for $director in //director
where count($director) > 1
group by $director
return $director

否则，遍历所有 distinct-values(//director)，找到每个名称的所有匹配导演标签并计算它们。

XML: Return 个满足给定条件的序列元素

XML: Return elements of a sequence that satisfy given condition

xml

xquery