xml - Xquery中的分组依据和计数

我正在尝试计算类似的“itemid_ref”，以获取其中每个items ref的编号：

<?xml version="1.0" encoding="utf-8"?>
<electrystore>
  <itembill>
   <itembill_id>1</itembill_id>
   <itemid_ref>2</itemid_ref>
   <billid_ref>2</billid_ref>
   <price>20000</price>
  </itembill>
   <itembill>
   <itembill_id>2</itembill_id>
   <itemid_ref>3</itemid_ref>
   <billid_ref>3</billid_ref>
   <price>250000</price>
  </itembill>
  <itembill>
   <itembill_id>3</itembill_id>
   <itemid_ref>3</itemid_ref>
   <billid_ref>3</billid_ref>
   <price>30000</price>
  </itembill>
  <itembill>
   <itembill_id>4</itembill_id>
   <itemid_ref>1</itemid_ref>
   <billid_ref>2</billid_ref>
   <price>140000</price>
  </itembill>
  <itembill>
   <itembill_id>5</itembill_id>
   <itemid_ref>2</itemid_ref>
   <billid_ref>1</billid_ref>
   <price>10000</price>
  </itembill>
</electrystore>

我试图将xquery的输出如下：

<?xml version="1.0" encoding="UTF-8"?>
<results>
<result>
    <itemid_ref>
       2
    </itemid_ref>
    <numberOfitemes>2</numberOfitemes>
</result>
<result>
    <itemid_ref>
       3
    </itemid_ref>
    <numberOfitemes>2</numberOfitemes>
</result>
<result>
   <itemid_ref>
       1
    </itemid_ref>
    <numberOfitemes>1</numberOfitemes>
</result>

我该怎么做？

最佳答案

XQuery1.0中没有group by，但通常可以通过使用distinct-values，只要数据集不太大。

for $itemid_ref in distinct-values(/electrystore/itembill/itemid_ref)
let $count := count(/electrystore/itembill[itemid_ref = $itemid_ref])
return element result {
  element itemid_ref { $itemid_ref },
  element numberOfitemes { $count }
}

要在大型数据集上执行，您可能需要查看xquery处理器中允许访问索引的扩展。
在XQuery3.0中有点简单：

for $itembill in /electrystore/itembill
let $itemid_ref := $itembill/itemid_ref
group by $itemid_ref
return element result {
  element itemid_ref { $itemid_ref },
  element numberOfitemes { count($itembill) }
}

关于xml - Xquery中的分组依据和计数，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/20483524/