Question

我想按流派对所有电影进行分组，然后列出此类型的所有电影。

我的XML电影数据库如下所示：

<movies>
  <movie>
    <title>A History of Violence</title>
    <year>2005</year>
    <country>USA</country>
    <genre>Crime</genre>
    <summary>Tom Stall, a humble family man and owner of a 
    popular neighborhood restaurant, lives a quiet but 
    fulfilling existence in the Midwest. One night Tom 
    foils a crime at his place of business and, to his 
    chagrin, is plastered all over the news for his 
    heroics. Following this, mysterious people follow 
    the Stalls' every move, concerning Tom more than 
    anyone else. As this situation is confronted, more 
    lurks out over where all these occurrences have 
    stemmed from compromising his marriage, family 
    relationship and the main characters' former 
    relations in the process.</summary>
 <director>     
        <last_name>Cronenberg</last_name>
        <first_name>David</first_name>
        <birth_date>1943</birth_date>
</director> 
<actor>
        <first_name>Vigo</first_name>
        <last_name>Mortensen</last_name>
        <birth_date>1958</birth_date>
        <role>Tom Stall</role>
</actor>
<actor>
        <first_name>Maria</first_name>
        <last_name>Bello</last_name>
        <birth_date>1967</birth_date>
        <role>Eddie Stall</role>
</actor>
<actor>
        <first_name>Ed</first_name>
        <last_name>Harris</last_name>
        <birth_date>1950</birth_date>
        <role>Carl Fogarty</role>
</actor>
<actor>
        <first_name>William</first_name>
        <last_name>Hurt</last_name>
        <birth_date>1950</birth_date>
        <role>Richie Cusack</role>
</actor>
 </movie>

这是我的表达：

xquery version "3.0";

let $movie := collection ('/db/Movie/data')/movies/movie

return

<html>
<head>

     </head>
     <body>
        <h1>Movies grouped by genre:</h1>

        <ol>{
              for $m in $movie
              let $g := $m/genre
              let $t := distinct-values($m/title/text())
              group by $g 
  return
                <li>{$g}  <p> <ol>Title: {$t}</ol> </p></li>



        }</ol>
   </body>
</html>

但结果会给我一行中的所有标题但我想要它们也列出了分数点。

这是实际输出：

<li>
<genre>Crime</genre>
<p>
<ol>Title: A History of Violence Heat Match Point</ol>
</p>
</li>
<li>

应该是这样的：

<li>
<genre>Crime</genre>
<p>
<ol>Title: A History of Violence 
           Heat 
           Match Point
</ol>
</p>
</li>
<li>

我如何调整查询？

提前致谢。迎接

Answer 1

只需在里面添加另一个循环。我重新格式化并重命名了一些变量以便更具描述性。一般情况下，如果有充分理由这样做，请不要使用text()，大多数时候最好使用data()来聚合元素中的所有文本节点。

xquery version "3.0";

let $movies := collection ('/db/Movie/data')/movies/movie
return
  <html>
    <head></head>
    <body>
      <h1>Movies grouped by genre:</h1>
      <ol>{
        for $movie in $movies
        let $genre := $movie/genre
        group by $genre 
        let $titles := distinct-values($movie/title/data())
        return
          <li>
            <h2>{$genre} Titles</h2>
            <ol>{
              for $title in $titles
              return <li>{$title}</li>
            }</ol>
          </li>
        }</ol>
     </body>
   </html>

你可以使用一个使用元素构造函数作为轴步骤的隐式循环，但这需要删除distinct-values调用（你真的需要吗？）我只是重复了$movie循环：< / p>

for $movie in $movies
let $genre := $movie/genre
group by $genre 
return
  <li>
    <h2>{$genre} Titles</h2>
    <ol>{ $movie/title/element li { data() } }</ol>
  </li>

顺便说一句，HTML不允许段落内的列表。它仍然是格式良好的XML，但不是有效的HTML。我也解决了这个问题。

Xquery FLWOR与group by

1 个答案: