XQUERY中的排序与合并

Question

我很难根据需要应用的一些合并规则对一组XML进行排序。我是在XQuery 1.0中这样做的，并且一直在使用不同的值函数，但是我发现很难把我所需要的所有记录联系起来。

我试图应用的逻辑是

• 对于频率和支付者以及缴款类型相同的情况，每个组合都需要出现一次。
• 其中一个事件是具有最近日期的组合的实例。
• 如果在同一记录上出现了与同一日期、最近日期、付款人和频率相同的详细信息，则需要将它们合并，以便为该频率显示的金额和该特定日期的付款人是在该日期支付的总额。
• 如果列表中存在比记录上更多的内容，则需要合并详细信息，以获得具有最近日期的组合的实例。

任何帮助都会很感激的。

     <transaction_history>
        <contribution_transaction contribution_type="Type A" currency="GBP">
           <date>2015-12-31</date>
           <payer>Employer</payer>
           <amount>5042.38</amount>
           <frequency>Quarterly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type A" currency="GBP">
           <date>2015-12-31</date>
           <payer>Employer</payer>
           <amount>4676.94</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type B" currency="GBP">
           <date>2015-09-08</date>
           <payer>Employer</payer>
           <amount>4317.52</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type A" currency="GBP">
           <date>2015-12-31</date>
           <payer>Employer</payer>
           <amount>9393.11</amount>
           <frequency>Quarterly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type A" currency="GBP">
           <date>2015-10-02</date>
           <payer>Employee</payer>
           <amount>3944.49</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type B" currency="GBP">
           <date>2015-12-31</date>
           <payer>Employee</payer>
           <amount>3675.85</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type B" currency="GBP">
           <date>2015-10-02</date>
           <payer>Employee</payer>
           <amount>3385.69</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
        <contribution_transaction contribution_type="Type A" currency="GBP">
           <date>2015-12-31</date>
           <payer>Employee</payer>
           <amount>7369.64</amount>
           <frequency>Monthly</frequency>
        </contribution_transaction>
     </transaction_history>

在…的例子中

<contribution_transaction contribution_type="Type A" currency="GBP">
       <date>2015-12-31</date>
       <payer>Employer</payer>
       <amount>5042.38</amount>
       <frequency>Quarterly</frequency>
    </contribution_transaction>
<contribution_transaction contribution_type="Type A" currency="GBP">
       <date>2015-12-31</date>
       <payer>Employer</payer>
       <amount>5042.38</amount>
       <frequency>Quarterly</frequency>
    </contribution_transaction>

将导致：

 <contribution_transaction contribution_type="Type A" currency="GBP">
       <date>2015-12-31</date>
       <payer>Employer</payer>
       <amount>10084.76</amount>
       <frequency>Quarterly</frequency>
    </contribution_transaction>

Answer 1

您可以一个接一个地进行两个嵌套分组：

for $frequency in distinct-values(//frequency)
let $contrib := //contribution_transaction[frequency = $frequency]
for $payer in distinct-values($contrib/payer)
let $res := $contrib[payer = $payer]
let $min-date := min($res/date/xs:date(.))
return <result frequency="{$frequency}" payer="{$payer}">{
  (: do something clever with the relevant transactions here :)
  $res[xs:date(date) = $min-date]
}</result>

由于您的日期格式与字典排序顺序是正确排序的，所以您也可以直接比较字符串：

for $frequency in distinct-values(//frequency)
let $contrib := //contribution_transaction[frequency = $frequency]
for $payer in distinct-values($contrib/payer)
let $res := $contrib[payer = $payer]
let $min-date := min($res/date/string())
return <result frequency="{$frequency}" payer="{$payer}">{
  (: do something clever with the relevant transactions here :)
  $res[date = $min-date]
}</result>

由于您的处理器似乎只支持node()的参数作为fn:min(...)的参数，所以可以使用以下快速替换：

declare function local:min($seq) {
  local:min((), $seq)
};

declare function local:min($min, $seq) {
  if(empty($seq)) then $min
  else (
    let $first := $seq[1], $rest := $seq[position() > 1]
    return (
      if($min < $first) then local:min($min, $rest)
      else local:min($first, $rest)
    )
  )
};

(: example usage :)
local:min(('B', 'A', 'b'))