oracle sql中有没有办法通过交叉连接获得计数？答案

【问题标题】：Is there a way in oracle sql to get a count with cross join?oracle sql中有没有办法通过交叉连接获得计数？
【发布时间】：2020-03-23 04:53:22
【问题描述】：

我需要获取专业列表和字母等级列表，并且我需要获取与字母等级相关的专业内完成的课程数量。所以如果我有：

Major
Mechanical Engineer
Electrical Engineer
Civil Engineer
Chemical Engineer 
...

Grade
A
B
C
...

Student
100 - Alice - Mechanical Engineer
101 - Tom - Mechanical Engineer
102 - Rex - Mechanical Engineer
103 - Bob - Mechanical Engineer
104 - John - Civil Engineer 
105 - Alex - Electrical Engineer

Course
001 - Solid Mechanics - 100 - A
002 - Thermodynamics - 100 - A
003 - Heat Transfer - 100 - A 
004 - Heat Transfer - 101 - A
005 - Gadgetry - 100 - A
006 - Gadgetry - 101 - A
007 - Gadgetry - 102 - A
008 - Dynamics - 102 - A
009 - Gadgetry - 101 - C
010 - Heat Transfer - 102 - C
011 - Fluid Mechanics - 100 - B 
012 - Materials - 102 -B 
013 - Intro to EE - 105 - B
014 - Embedded Systems - 105 - B
015 - Analog Circuits - 105 - B
...

我需要一个输出：

Mechanical Engineer - A - 8
Mechanical Engineer - B - 2
Mechanical Engineer - C - 2
Electrical Engineer - A - 0
Electrical Engineer - B - 3
...

还有一张学生表，上面写着每个学生的专业，还有一张课程表，学生用什么课程（有些学生可以不参加任何课程）以及他们从该课程获得的成绩。

到目前为止我有：

SELECT major_name, grade, COUNT(grade_code) OVER (PARTITION BY major_name)
FROM Major CROSS JOIN ((grade LEFT JOIN course USING (grade_code)) RIGHT JOIN student USING (st_id))
ORDER BY major_name, grade;

但是计数不起作用（每个专业得到相同的错误计数），基本上是课程表中字母等级的总计数。

Mechanical Engineer - A - 10
Mechanical Engineer - B - 10
Mechanical Engineer - C - 10
Electrical Engineer - A - 10
...

【问题讨论】：

有匹配的样本表数据和预期结果。（我们还能如何验证计数？）
我临时编造了一些样本数据，希望这很有意义

标签： sql count cross-join

【解决方案1】：

不需要CROSS JOIN 或聚合COUNT...OVER() 窗口函数。只需加入主表（student 和 course），然后加入查找表（major 和 grade）。然后，聚合计数。因为major链接到不在FROM子句中的student，所以用相应的别名替换传统的ON的USING：

SELECT m.major_name, 
       g.grade, 
       count(*) as major_grade_count
FROM course c
LEFT JOIN student s USING (st_id)
LEFT JOIN grade g USING (grade_code)
LEFT JOIN major m ON g.major_code = m.major_code      -- ADJUST FIELD HERE
GROUP BY  m.major_name, 
          g.grade;

【讨论】：