在XQUERY中排序和合並

Question

我很難根據需要應用的某些合併規則對一組XML進行排序。我在XQuery 1.0中這樣做，並且一直在玩獨特值函數，但卻很難將我需要的所有記錄聯繫起來。

我嘗試應用的邏輯是

• 需要有存在的頻率和付款人和地方貢獻類型是一樣的每個組合中的一種發生。
• 出現一次時是具有最近的日期是組合的實例
• 如果細節出現在使用相同的，最近的日期，日期，付款人和頻率則需要對它們進行合併，使得同一記錄顯示該頻率和該特定日期的付款人的金額是該日期支付的總金額。
• 如果列表中存在超過記錄，則需要合併詳細信息以得出具有最近日期的組合的實例。

任何幫助將非常感激。

 <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

<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> 

的例子會導致：

<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

您可以在其他以後做兩個嵌套集團之一：

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> 

由於您的日期格式與lexicog正確排序raphic排序順序，也可以只直接比較字符串：

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> 

由於您的處理器似乎只支持node() S作爲參數傳遞給fn:min(...)，您可以使用此快速更換：

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'))