Question

我需要选择属于这个范围的项目数

create table numbers (val int);

insert into numbers(val) values (2), (3), (11), (12), (13), (31);

select count(1) as qty , val / 10 as range
from numbers
group by val / 10;

显然，如果范围内没有项目，则不会包含在输出中。我可以想到一些不太优雅的方法来包含输出中的所有范围，但是有一个优雅而快速的方法（在PostgreSQL或MS SQL Server方言中）

Answer 1

您似乎想要创建结果的直方图。

的PostgreSQL：

select x, count(val)
from generate_series(1,6) x 
left outer join numbers on (x = width_bucket(val, 0, 60, 6))
group by x;

我使用了width_bucket而不是简单的除法和模数，因为它更通用，更容易适应更复杂的范围。而且，它太棒了。

Mark Bannister用于序列生成的递归CTE可以集成并作为x加入，而不是generate_series，以便在需要时增加可移植性，并且可以自动确定限制：

with recursive ranges(rangev) as (
    select 0 rangev union all select rangev+1 as rangev from ranges where rangev < 4
), bounds(lower_bucket, upper_bucket) as (
    select (min(val))/10, (max(val)/10)+1 from numbers
)
select
    rangev as bucket,
    rangev*10 AS lower_bound,
    (rangev+1)*10-1 AS upper_bound,
    count(val) AS num_in_bucket
from ranges cross join bounds
left outer join numbers on (rangev = width_bucket(val, lower_bucket, upper_bucket*10, upper_bucket))
group by rangev
order by rangev asc;

如果您希望/10超过width_bucket（例如，如果MS SQL中没有width_bucket），那么很容易更改回来。

输出：

 bucket | lower_bound | upper_bound | num_in_bucket 
--------+-------------+-------------+---------------
      0 |           0 |           9 |             0
      1 |          10 |          19 |             2
      2 |          20 |          29 |             3
      3 |          30 |          39 |             0
      4 |          40 |          49 |             1
(5 rows)

Answer 2

Postgresql 9.2还实现了范围类型：

  SELECT 
    range.r, 
    count(val.n) 
  FROM 
    (VALUES (int4range '[0, 10)'), (int4range '[10, 20)'), (int4range '[20, 30)'), (int4range '[30, 40)')) range (r) 
     LEFT JOIN (VALUES (2), (3), (9), (11), (17), (31), (33)) AS val (n) ON  val.n <@   range.r
  GROUP BY
    range.r
  ORDER BY r ASC

┌─────────┬───────┐
│    r    │ count │
├─────────┼───────┤
│ [0,10)  │     3 │
│ [10,20) │     2 │
│ [20,30) │     0 │
│ [30,40) │     2 │
└─────────┴───────┘

Answer 3

尝试：

with ranges as 
(select 0 rangev union all 
 select rangev+1 as rangev 
 from ranges 
 where rangev < 4)
select count(val) as qty , rangev
from ranges 
left join numbers on rangev = val / 10
group by rangev;

（对于PostgresQL，SQLServer版本 - 将with ranges更改为with recursive ranges）

Answer 4

对于MS SQL Server

declare @min int, @max int
    select @min = MIN(val/10), @max = MAX(val/10) from numbers

    select ranges.number as range, COUNT(val) as qty from master.dbo.spt_values ranges
        left join numbers on ranges.number = numbers.val/10
    where type='p' and number between @min and @max
    group by number

同样的原则可以适用于Postgres，尽管数字列表的生成会有所不同

按范围分组记录

4 个答案: