Sql 时间戳变量的不规则分组

我有一张如下安排的桌子：

id       lateAt
1231235  2019/09/14
1242123  2019/09/13
3465345  NULL
5676548  2019/09/28
8986475  2019/09/23

其中lateAt是某笔贷款的付款延迟的时间戳。因此，对于当前的每个日期-我需要每天查看这些数字-有一定数量的条目延迟0-15、15-30、30-45、45-60、60-90和90天以上

这是我想要的输出：

lateGroup   Count
0-15        20
15-30       22
30-45       25
45-60       32
60-90       47
90+         57

---

这是我可以在R中轻松计算的东西，但要将结果返回到BI仪表板，我必须在数据库中创建一个新表，我认为这不是一个好的做法。解决此问题的SQL本机方法是什么？

您没有提到正在使用的DBMS，但几乎所有DBMS都有一个称为值构造函数的构造，如下所示：

select bins.lateGroup, bins.minVal, bins.maxVal FROM
    (VALUES 
        ('0-15',0,15),
        ('15-30',15.0001,30),  -- increase by a small fraction so bins don't overlap
        ('30-45',30.0001,45),
        ('45-60',45.0001,60),
        ('60-90',60.0001,90),
        ('90-99999',90.0001,99999)
    ) AS bins(lateGroup,minVal,maxVal)

--- example from SQL Server 2012 SP1
--- first let's set up some sample data
create table #temp (id int, lateAt datetime);
INSERT #temp (id, lateAt) values
   (1231235,'2019-09-14'),
   (1242123,'2019-09-13'),
   (3465345,NULL),
   (5676548,'2019-09-28'),
   (8986475,'2019-09-23');

--- here's the actual query
select lateGroup, count(*) as Count
from #temp as T,
    (VALUES
        ('0-15',0,15),
        ('15-30',15.0001,30),  -- increase by a small fraction so bins don't overlap
        ('30-45',30.0001,45),
        ('45-60',45.0001,60),
        ('60-90',60.0001,90),
        ('90-99999',90.0001,99999)
    ) AS bins(lateGroup,minVal,maxVal)
    ) AS bins(lateGroup,minVal,maxVal)
where datediff(day,lateAt,getdate()) between minVal and maxVal
group by lateGroup
order by lateGroup

--- remove our sample data
drop table #temp;

如果您的DBMS没有，那么您可能可以使用UNION ALL：

然后，您的完整查询以及您提供的示例数据如下所示：

select bins.lateGroup, bins.minVal, bins.maxVal FROM
    (VALUES 
        ('0-15',0,15),
        ('15-30',15.0001,30),  -- increase by a small fraction so bins don't overlap
        ('30-45',30.0001,45),
        ('45-60',45.0001,60),
        ('60-90',60.0001,90),
        ('90-99999',90.0001,99999)
    ) AS bins(lateGroup,minVal,maxVal)

--- example from SQL Server 2012 SP1
--- first let's set up some sample data
create table #temp (id int, lateAt datetime);
INSERT #temp (id, lateAt) values
   (1231235,'2019-09-14'),
   (1242123,'2019-09-13'),
   (3465345,NULL),
   (5676548,'2019-09-28'),
   (8986475,'2019-09-23');

--- here's the actual query
select lateGroup, count(*) as Count
from #temp as T,
    (VALUES
        ('0-15',0,15),
        ('15-30',15.0001,30),  -- increase by a small fraction so bins don't overlap
        ('30-45',30.0001,45),
        ('45-60',45.0001,60),
        ('60-90',60.0001,90),
        ('90-99999',90.0001,99999)
    ) AS bins(lateGroup,minVal,maxVal)
    ) AS bins(lateGroup,minVal,maxVal)
where datediff(day,lateAt,getdate()) between minVal and maxVal
group by lateGroup
order by lateGroup

--- remove our sample data
drop table #temp;

以下是输出： 晚群计数 15-30 2 30-45 2

---

注意：不计算延迟为null的行。

在SQL中执行此操作的快捷方式是：

SELECT '0-15'   AS lateGroup, 
       COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 0
  AND (CURRENT_DATE - t.lateAt) <  15

UNION

SELECT '15-30'  AS lateGroup, 
       COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 15
  AND (CURRENT_DATE - t.lateAt) <  30

UNION

SELECT '30-45'  AS lateGroup, 
       COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 30
  AND (CURRENT_DATE - t.lateAt) <  45

-- Etc...

---

对于生产代码，您可能希望执行更像Ross答案的操作。

我认为您可以在一个清晰的查询中完成所有操作：

with cte_lategroup as
(
    select *
    from (values(0,15,'0-15'),(15,30,'15-30'),(30,45,'30-45')) as t (mini, maxi, designation)
)
select 
    t2.designation
    , count(*)
from test t
    left outer join cte_lategroup t2
    on current_date - t.lateat >= t2.mini
    and current_date - lateat < t2.maxi
group by t2.designation;

---

我将使用a定义延迟组，根据天数加入：

with groups (grp) as (
  values 
    (int4range(0,15, '[)')),
    (int4range(15,30, '[)')),
    (int4range(30,45, '[)')),
    (int4range(45,60, '[)')),
    (int4range(60,90, '[)')),
    (int4range(90,null, '[)'))
)
select grp, count(t.user_id)
from groups g
  left join the_table t on g.grp @> current_date - t.late_at
group by grp
order by grp;

int4range0,15，“[”创建一个从0（包含）到15（独占）的范围

---

在线示例：

那么迟到组是根据今天和迟到列之间的差异计算出来的？因此id=1231235属于30-45组，因为今天晚了32天2019-10-16我在使用postgresql，您的第二条评论是对的。您的组重叠。迟到15天的id是属于0-15组还是属于15-30组oup或两者都有疑问？很好。它应该在下确界包含，在上确界独占。如果不使用between，则不需要此小部分的增加part@a_horse_with_no_name-你当然是对的。这归结为在不需要的详细内容之间的选择：在每个上限中添加分数，或者复制datediff是where子句中表达式的一部分。我发现，当每一行上的更改相同时，重复它实际上会增加重新阅读时的清晰度。我真希望SQL Server具有范围。