Question

我在mysql中有一个表，其值为给定值

+--------+---+---+---+---+----+
+ emp_no + A + B + C + D +  E +
+--------+---+---+---+---+----+
+ 100    + 1 + 0 + 2 + 1 +  1 +
+ 101    + 2 + 1 + 1 + 3 +  1 +
+ 102    + 3 + 4 + 4 + 2 +  5 +
+ 103    + 5 + 3 + 4 + 1 +  2 +
+--------+---+---+---+---+----+

我是否可以编写一个sql查询，它将从上述表中获取总计数为1，总计数为2，总计数为3 ....对此非常感谢。

谢谢

Answer 1

试试这个：

SELECT num, SUM(counts) FROM (
SELECT A num, COUNT(A) counts FROM table1 GROUP BY A
union all
SELECT B num, COUNT(B) counts FROM table1 GROUP BY B
union all
SELECT C num, COUNT(C) counts FROM table1 GROUP BY C
union all
SELECT D num, COUNT(D) counts FROM table1 GROUP BY D
union all
SELECT E num, COUNT(E) counts FROM table1 GROUP BY E
) temp 
GROUP BY num

Answer 2

是的，但有点复杂。这是一种方法，通过取消隐藏值然后聚合：

select val, count(*)
from (select emp_no,
             (case when n.n = 1 then A
                   when n.n = 2 then B
                   when n.n = 3 then C
                   when n.n = 4 then D
                   when n.n = 5 then E
              end) as val
      from table t cross join
           (select 1 as n union all select 2 union all select 3 union all select 4 union all select 5
           ) n
     ) e;

Answer 3

可能SUM()具有布尔条件的总和。看看这个：

SELECT
    SUM(
        (`a` = 1)
        +
        (`b` = 1)
        +
        (`c` = 1)
        +
        (`d` = 1)
        +
        (`e` = 1)
    ) as `1`,
    SUM(
        (`a` = 2)
        +
        (`b` = 2)
        +
        (`c` = 2)
        +
        (`d` = 2)
        +
        (`e` = 2)
    ) as `2`,
    SUM(
        (`a` = 3)
        +
        (`b` = 3)
        +
        (`c` = 3)
        +
        (`d` = 3)
        +
        (`e` = 3)
    ) as `3`,
    SUM(
        (`a` = 4)
        +
        (`b` = 4)
        +
        (`c` = 4)
        +
        (`d` = 4)
        +
        (`e` = 4)
    ) as `4`,
    SUM(
        (`a` = 5)
        +
        (`b` = 5)
        +
        (`c` = 5)
        +
        (`d` = 5)
        +
        (`e` = 5)
    ) as `5`
FROM
    `table`;

MySQL Query从整个表中获取值的计数

3 个答案: