如何将一个查询的计数添加到另一个查询中答案

【问题标题】：How to add the count from one query into another query如何将一个查询的计数添加到另一个查询中
【发布时间】：2022-08-17 11:39:22
【问题描述】：

我的第一个查询如下所示：

SELECT location, COUNT(*) as sections
FROM section
GROUP BY location

这给了我：

标签： sql oracle

【解决方案1】：

只需加入查询：

SELECT *
FROM
(
  SELECT location, COUNT(*) as sections
  FROM section
  GROUP BY location
)
FULL OUTER JOIN
(
  SELECT s.location, COUNT(*) as students
  FROM enrollment e 
  INNER JOIN section s ON s.section_id = e.section_id
  GROUP BY s.location
) USING (location)
ORDER BY location;

另一种选择是按部分对注册进行分组，然后按位置加入和分组。

SELECT
  location,
  COUNT(*) as sections,
  SUM(students_in_section) AS students
FROM section s
LEFT JOIN
(
  SELECT section_id, COUNT(*) as students_in_section
  FROM enrollment
  GROUP BY section_id
) e ON e.section_id = s.section_id
GROUP BY s.location
ORDER BY s.location;

另一种选择是加入表格并计算不同的部分和不同的注册。

SELECT
  location,
  COUNT(DISTINCT s.section_id) as sections,
  COUNT(DISTINCT e.enrollment_id) AS students
FROM section s
LEFT JOIN enrollment e ON e.section_id = s.section_id
GROUP BY s.location
ORDER BY s.location;

【讨论】：

计算每个表的不同 rowid 将提供更准确的结果，独立于数据模型和每个表的主键标识
@astentx：这是一个很好的建议。
...但应注意，它仅对普通表有效

【解决方案2】：

您可以使用COUNT(DISTINCT ...) 计算每个位置的唯一部分

  SELECT location, COUNT (DISTINCT s.section_id) AS sections, COUNT (*) AS students
    FROM enrollment e INNER JOIN section s ON s.section_id = e.section_id
GROUP BY location

【讨论】：

这不是相同的结果。它只会计算注册中存在的部分。结合这两个查询将给出一个位置中所有部分的计数，无论它们在注册中的表示如何