XQuery连接结果

时间:2016-03-14 12:38:42

标签: xml xquery basex

我有一个看起来像这样的XML:

<?xml version="1.0"?>
<root>
  <flight>
    <number>10001</number>
    <airport>LAX</airport>
    <dest>
      <airport>SFO</airport>
    </dest>
  </flight>
  <flight>
    <number>10002</number>
    <airport>LAX</airport>
    <dest>
      <airport>JFK</airport>
    </dest>
  </flight>
  <flight>
    <number>10003</number>
    <airport>JFK</airport>
    <dest>
      <airport>LAX</airport>
    </dest>
  </flight>
</root>

使用XQuery我需要得到这样的东西:

<res>
    <airport code="LAX">
        <deps>2</deps>
        <dests>1</deps>
    </airport>
    <airport code="JFK">
        <deps>1</deps>
        <dests>1</deps>
    </airport>
    <airport code="SFO">
        <deps>0</deps>
        <dests>1</deps>
    </airport>
</res>

我做到了,并且可以得到正确的结果,但是,我的查询只能找到depsdests,但不能同时找到。{/ p>

以下是我解决问题的方法。

let $all := doc("flights.xml")/root
for $airports in distinct-values($all/flight//*/airport) (:here I get all airport codes:)
order by $airports 

for $nr-dep in $all/flight/airport

where $nr-dep = $airports 
group by $airports 

return <res>
          <airport name="{$airports}"><deps>{count($nr-dep)}</deps></airport>
       </res>

我在这里得到了离境计数。我可以通过将for $nr-dep in $all/flight/airport替换为for $nr-dep in $all/flight/dest/airport来轻松获取destionation但是我找不到像预期的XML一样显示两者的方法。

2 个答案:

答案 0 :(得分:4)

以下是使用group by的版本:

<res>{
    for $airport in //airport
    group by $code := $airport/text()
    order by $code
    return <airport code="{$code}">
        <deps>{ count($airport/parent::flight) }</deps>
        <dests>{ count($airport/parent::dest) }</dests>
    </airport>
}</res>

答案 1 :(得分:3)

为什么不简单:

for $airport in distinct-values($all//airport)
order by $airport
return <airport code="{$airport}">
  <deps>{count($all//flight/airport[. = $airport])}</deps>
  <dests>{count($all//dest/airport[. = $airport])}</dests>
</airport>