Question

我有一个MySQL表：

create table tbl (
  amount int
);

insert into tbl (amount) values (1);
insert into tbl (amount) values (2);
insert into tbl (amount) values (3);
insert into tbl (amount) values (4);

我的目标是通过使用案例陈述来报告以下存储桶中有多少值。

铲斗A：值0-1
铲斗B：值2-5
铲斗C：值6-9

首先让我们尝试一个简单的查询：

select "Bucket A" as Bucket, count(amount) "Count"
from tbl
where amount in (0,1)
union
select "Bucket B" as Bucket, count(amount) "Count"
from tbl
where amount in (2,3,4,5)
union
select "Bucket C" as Bucket, count(amount) "Count"
from tbl
where amount in (6,7,8,9);

结果：

+----------+-------+
| Bucket   | Count |
+----------+-------+
| Bucket A |     1 |
| Bucket B |     3 |
| Bucket C |     0 |
+----------+-------+

结果很完美，但我想要一个案例陈述。
所以我试试这个：

select 
sum(case when amount in (0,1) then 1 else 0 end) as "Bucket A",
sum(case when amount in (2,3,4,5) then 1 else 0 end) as "Bucket B",
sum(case when amount in (6,7,8,9) then 1 else 0 end) as "Bucket C"
from tbl;

结果：

+----------+----------+----------+
| Bucket A | Bucket B | Bucket C |
+----------+----------+----------+
|        1 |        3 |        0 |
+----------+----------+----------+

值是正确的，而且我有一个案例陈述，但问题是价值被转动了。

我怎么能 1.使用案例陈述
2.没有支点？

Answer 1

您可以使用聚合执行此操作：

select (case when amount in (0, 1) then 'Bucket A'
             when amount in (2, 3,4, 5) then 'Bucket B'
             when amount in (6, 7, 8, 9) then 'Bucket C'
        end) as bucket, count(*) as `count`
from tbl
where amount in (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
group by (case when amount in (0, 1) then 'Bucket A'
               when amount in (2,3,4,5) then 'Bucket B'
               when amount in (6,7,8,9) then 'Bucket C'
           end);

编辑：

数字克里斯提出了一个非常好的观点。这可以通过使用left outer join：

来解决

select (case when tbl.amount in (0, 1) then 'Bucket A'
             when tbl.amount in (2, 3,4, 5) then 'Bucket B'
             when tbl.amount in (6, 7, 8, 9) then 'Bucket C'
        end) as bucket, count(tbl.amount) as `count`
from (select 0 as amount union all
      select 2 as amount union all
      select 6 as amount
     ) throwaway left outer join
     tbl
     on throwaway.amount = tbl.amount
where tbl.amount in (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
group by (case when tbl.amount in (0, 1) then 'Bucket A'
               when tbl.amount in (2,3,4,5) then 'Bucket B'
               when tbl.amount in (6,7,8,9) then 'Bucket C'
           end);

或者更清楚地说，将原始查询用作子查询：

select buckets.bucket, coalesce(`count`, 0) as `count`
from (select 'Bucket A' as bucket union all
      select 'Bucket B' union all
      select 'Bucket C'
     ) buckets left outer join
     (select (case when amount in (0, 1) then 'Bucket A'
                   when amount in (2, 3,4, 5) then 'Bucket B'
                   when amount in (6, 7, 8, 9) then 'Bucket C'
              end) as bucket, count(*) as `count`
      from tbl
      where amount in (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
      group by (case when amount in (0, 1) then 'Bucket A'
                     when amount in (2,3,4,5) then 'Bucket B'
                     when amount in (6,7,8,9) then 'Bucket C'
                 end)
     ) g
     on buckets.bucket = g.bucket;

Answer 2

select 
"Bucket A" as "Bucket", sum(case when amount in (0,1) then 1 else 0 end) as "Count" from tbl
UNION
select "Bucket B", sum(case when amount in (2,3,4,5) then 1 else 0 end) from tbl
UNION
select "Bucket C", sum(case when amount in (6,7,8,9) then 1 else 0 end) from tbl;

喜欢这个？ sqlfiddle

Answer 3

SELECT "Bucket A" AS Bucket ,
(SELECT SUM(CASE WHEN amount IN (0,1) THEN 1 ELSE 0 END) FROM tbl) AS "COUNT" 
UNION
SELECT "Bucket B" AS Bucket ,
(SELECT SUM(CASE WHEN amount IN (2,3,4,5) THEN 1 ELSE 0 END) FROM tbl) AS "COUNT" 
UNION
SELECT "Bucket C" AS Bucket ,
(SELECT SUM(CASE WHEN amount IN (6,7,8,9) THEN 1 ELSE 0 END) FROM tbl) AS "COUNT"

sqlfiddle demo

Answer 4

使用制造的存储桶名称列表，然后将其连接到表格：

select concat('Bucket ', b) bucket, count(amount) count
from (select 'A' as b union select 'B' union select 'C') a
left join tbl on b = 
  case when amount in (0, 1) then 'A'
       when amount in (2,3,4,5) then 'B'
       when amount in (6,7,8,9) then 'C' end
group by 1

当没有找到存储桶的行时，这将产生一个零计数的行。

请参阅SQLFiddle

在MySQL中如何使用case语句重写查询？

4 个答案: