Question

我有两个问题：

第一：

select year, sum(count(id)*3);

我得到的错误是“字段列表中的未知列x.year”。我希望查询充当select instructor.id, name, count(teaches.id) as 'Number of Sections' from instructor natural left outer join teaches group by teaches.id, name, instructor.id order by instructor.id asc;

第二

我需要编写一个与没有外连接的情况相同的查询：

(select instructor.id, name, count(teaches.id) as 'Number of Sections'
from instructor, teaches
where instructor.id = teaches.id
group by teaches.id, name, instructor.id)
union
(select i.id, i.name, count(t.id)
from instructor i, teaches t
where i.id not in (select id from teaches)
group by t.id, i.name, i.id);

我尝试了以下但是没有用：

{{1}}

谢谢！

Answer 1

尝试下面的第一个

select x.year, sum(x.cnt)
from (select count(id)*3 as cnt, year from takes where year = 2009 group by year) x
group by x.year;

Answer 2

我假设你有以下表::

mysql> select * from takes;
+------+------------+
| id   | year       |
+------+------------+
|  101 | 2017-01-01 |
|  102 | 2015-01-01 |
|  103 | 2014-01-01 |
|  103 | 2015-01-01 |
|  101 | 2015-01-01 |
|  101 | 2009-01-01 |
|  102 | 2009-01-01 |
|  102 | 2009-01-01 |
|  101 | 2009-01-01 |
|  101 | 2009-01-01 |
|  103 | 2009-01-01 |
+------+------------+

select x.year, sum(x.cnt) from (select count(id)*3 as cnt, year from takes where year(year) = "2009" group by year)x group by x.year;

+------------+------------+
| year       | sum(x.cnt) |
+------------+------------+
| 2009-01-01 |         18 |
+------------+------------+

那应该是你期待的！

Answer 3

尝试以下

-- 1
select x.year, sum(x.cnt)
from
  (
    select year,count(id)*3 as cnt
    from takes
    where year = 2009
    group by year
  ) x
group by x.year;

-- 2.1
select i.id, i.name, count(t.id) as 'Number of Sections'
from instructor i
join teaches t on i.id = t.id
group by i.id, i.name

union all

select id, name, 0 -- I think here is 0 sections
from instructor
where id not in (select id from teaches);

-- 2.2 - I think it also should work
select i.id, i.name, count(t.id) as 'Number of Sections'
from instructor i
left join teaches t on i.id = t.id
group by i.id, i.name;

MySQL多个聚合函数和外连接

3 个答案: