Question

我有一张桌子学生：

student_id | name | course
-------------------------------------
  1        | Jack | Comp_Sci
  2        | John | Maths
  3        | Matt | Comp_Sci
  4        | Pete | Biology

表部门：

course   | department
-------------------------
Comp_Sci | Comp_and_Math
Maths    | Comp_and_Math
Biology  | Bio_and_Chem

表using_computers

computer_id | student_id 
-------------------------
1           |  2
2           |  2
2           |  3
2           |  4
3           |  1
4           |  2
4           |  4

和台式电脑

computer_id | name
---------------------
1           | Apple
2           | Dell
3           | Asus
4           | Acer

我想列出像这样的部门中所有使用过的电脑

Comp_and_Maths: Apple 1, Dell 2, Asus 1, Acer 1, sum: 5
Bio_and_Chem: Dell 1, Acer 1 , sum:2

我已经写了2个查询，但我不知道如何连接它们：

SELECT Departments.department, obj_in_class.list_ids
  FROM Departments LEFT JOIN
  (SELECT Departments.course, array_agg(students.student_id) AS list_ids
  FROM Departments
  LEFT JOIN students
      ON Departments.course = students.course GROUP BY Departments.course) AS obj_in_class
   ON Departments.course = obj_in_class.course GROUP BY Departments.department, obj_in_class.list_ids;

(SELECT students.student_id AS id, array_agg(m.name) AS computers
  FROM students LEFT JOIN
    (SELECT computers.name, using_computers.student_id FROM using_computers LEFT JOIN computers ON using_computers.computer_id = computers.computer_id) AS m
    ON students.student_id = m.student_id
      GROUP BY students.student_id) AS students_with_computers;

Answer 1

它不会重复使用您之前的查询：

SELECT 
     -- aggregate computer name and count as a string
    grouped.department || ': ' || 
    array_to_string(array_agg(grouped.name || ' ' || grouped.count), ', ')
     -- sum all the counts
    || ', sum: ' || sum(grouped.count)
  FROM (
    SELECT
        D.department,
        C.name,
        count(C.name) -- count computers' name per department
      FROM Departments D
      JOIN students S USING (course)
      JOIN using_computers UC USING (student_id)
      JOIN computers C USING (computer_id)
      GROUP BY D.department, C.name
      ORDER BY D.department, C.name
    ) grouped
  GROUP BY grouped.department
;

SQLFiddle

我们的想法是按部门和计算机加入每个表和组，以获得每个部门（计算机品牌）的计数。这是在grouped子选择中完成的。我们现在拥有所需的每一项数据和计数。

然后我们按部门分组并汇总所有内容。

从postgresql中的许多表中加入结果

1 个答案: