XQuery 3合并两个节点序列并去重

Question

在 XQuery 3.1 中（在 eXistDB 4.4 下）我有两个函数，其中 return 两组节点描述了一个人的家庭关系。两个函数的结果可以重叠。

第一个函数person:person-relationship-as-object("#Guilhem_Faure_SML-AU")returns:

<person relation="Peire_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Arnald_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Esteve_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Martin_de_Verazilh_SML-AU">
  <span class="en">Aunt(s)/Uncle(s)</span>
  <span class="fr">Tante(s)/Oncle(s)</span>
</person>

<person relation="Raimund_de_Verazilh_SML-AU">
  <span class="en">Aunt(s)/Uncle(s)</span>
  <span class="fr">Tante(s)/Oncle(s)</span>
</person>

第二个函数person:person-relationship-as-subject("#Guilhem_Faure_SML-AU")returns

<person relation="Arnald_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Peire_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Esteve_Faure_SML-AU">
  <span class="en">Sibling(s)</span>
  <span class="fr">Frère(s)/sœur(s)</span>
</person>

<person relation="Bernarda_Guilhem_Faure_SML-AU">
  <span class="en">Spouse(s)</span>
  <span class="fr">Époux/épouse(s)</span>
</person>

现在我想将两组节点合并为一组，并且return每个不同的节点只合并一次。

我尝试了一个粗略的类似 SQL 的结构，但失败了：

let $x := person:person-relationship-as-object("#Guilhem_Faure_SML-AU") 
          | person:person-relationship-as-subject("#Guilhem_Faure_SML-AU")

for $y in $x

order by $y/@relation ascending

group by $y/@relation, $y/span[@class="en"], $y/span[@class="fr"]

return $y

我想我真的不明白为什么 group by 在这里不起作用。此外，如果 group by 没有完成这项工作，我还能如何处理（我认为？）应该是 grouping/distinct 结果中的简单练习？

Answer 1

使用 XQuery 和 group by，return 子句中的变量 $y 绑定到您创建的整个组，因此要消除重复项，您需要使用 return $y[1].见 https://www.w3.org/TR/xquery-31/#id-group-by 说

In the post-grouping tuple generated for a given group, each non-grouping variable is bound to a sequence containing the concatenated values of that variable in all the pre-grouping tuples that were assigned to that group.

然后

This behavior may be surprising to SQL programmers, since SQL reduces the equivalent of a non-grouping variable to one representative value.

XQuery 3合并两个节点序列并去重

XQuery 3 merging two node sequences and de-duplicating

xquery

exist-db