我的表格有类似下面的信息
Emp Date START_TIME END_TIME Code Minutes
--- -------- ------------------- ------------------- ---- -------
E1 11/1/2012 11/1/2012 6:55:00 AM 11/1/2012 7:01:00 AM C1 6
E1 11/1/2012 11/1/2012 6:57:00 AM 11/1/2012 8:01:00 AM C2 64
E2 11/1/2012 11/1/2012 6:57:00 AM 11/1/2012 8:00:00 AM C2 63
E1 11/2/2012 11/2/2012 7:35:00 AM 11/2/2012 8:01:00 AM C1 26
预期输出
Date Code Range Minutes
--------- ---- ----------------------- -------
11/1/2012 C1 6:30:00 AM-7:00:00 AM 5
11/1/2012 C1 7:00:00 AM-7:30:00 AM 1
11/1/2012 C2 6:30:00 AM-7:00:00 AM 6
11/1/2012 C2 7:00:00 AM-7:30:00 AM 60
11/1/2012 C2 7:30:00 AM-8:00:00 AM 60
11/1/2012 C2 8:00:00 AM-8:30:00 AM 1
11/2/2012 C1 7:30:00 AM-8:00:00 AM 25
11/2/2012 C1 8:00:00 AM-8:30:00 AM 1
离开Emp字段，我希望按日期分组，并且代码的总时间分别为30分钟。我所拥有的限制是使用select语句实现这一点，即仅通过SQL查询而不允许使用PL / SQL 。提前谢谢！
答案 0 :(得分：2)
涉及模型条款的解决方案。
首先计算每个条目我们需要的30分钟块数。
SQL> select emp, start_time, end_time, code,
2 trunc(start_time, 'mi')
3 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440) start_block,
4 ceil(2*24*(end_time-(trunc(start_time, 'mi')
5 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440)))) blocks
6 from tab f
7 /
EM START_TIME END_TIME CO START_BLOCK BLOCKS
-- ---------------------- ---------------------- -- ---------------------- ----------
E1 11/01/2012 06:55:00 am 11/01/2012 07:01:00 am C1 11/01/2012 06:30:00 am 2
E1 11/01/2012 06:57:00 am 11/01/2012 08:01:00 am C2 11/01/2012 06:30:00 am 4
E2 11/01/2012 06:57:00 am 11/01/2012 08:00:00 am C2 11/01/2012 06:30:00 am 3
E1 11/02/2012 07:35:00 am 11/02/2012 08:01:00 am C1 11/02/2012 07:30:00 am 2
现在，我们使用model子句生成行以将其分解为30分钟。
SQL> with foo as (select rownum id, emp, start_time, end_time, code,
2 trunc(start_time, 'mi')
3 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440) start_block,
4 ceil(2*24*(end_time-(trunc(start_time, 'mi')
5 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440)))) blocks
6 from tab f)
7 select trunc(start_time) thedate, code, emp, range, minutes
8 from foo
9 model partition by(id)
10 dimension by(0 as f)
11 measures(code, emp, start_time, end_time, start_block, blocks,
12 sysdate as start_range,
13 sysdate as end_range,
14 cast(0 as number) minutes,
15 cast('' as varchar2(50)) range)
16 rules (start_range [for f from 0 to blocks[0]-1 increment 1] = start_block[0] + (30*cv(f)/1440),
17 end_range[any] = start_range[cv()] + (30/1440),
18 code[any] = code[0],
19 emp[any] = emp[0],
20 start_time[any] = start_time[0],
21 end_time[any] = end_time[0],
22 range [any] = to_char(start_range[cv()], 'dd/mm/yyyy hh:mi:ss am') || ' - ' || to_char(end_range[cv()], 'dd/mm/yyyy hh24:mi:ss am'),
23 minutes [any] = case
24 when start_time[0] between start_range[cv()] and end_range[cv()]
25 then 1440 *(end_range[cv()] - start_time[0])
26 when end_time[0] between start_range[cv()] and end_range[cv()]
27 then 1440 *(end_time[0] - start_range[cv()])
28 else 1440 * (end_range[cv()] - start_range[cv()])
29 end );
CO EM RANGE MINUTES
-- -- -------------------------------------------------- ----------
C2 E2 11/01/2012 06:30:00 am - 11/01/2012 07:00:00 am 3
C2 E2 11/01/2012 07:00:00 am - 11/01/2012 07:30:00 am 30
C2 E2 11/01/2012 07:30:00 am - 11/01/2012 08:00:00 am 30
C1 E1 11/01/2012 06:30:00 am - 11/01/2012 07:00:00 am 5
C1 E1 11/01/2012 07:00:00 am - 11/01/2012 07:30:00 am 1
C1 E1 11/02/2012 07:30:00 am - 11/02/2012 08:00:00 am 25
C1 E1 11/02/2012 08:00:00 am - 11/02/2012 08:30:00 am 1
C2 E1 11/01/2012 06:30:00 am - 11/01/2012 07:00:00 am 3
C2 E1 11/01/2012 07:00:00 am - 11/01/2012 07:30:00 am 30
C2 E1 11/01/2012 07:30:00 am - 11/01/2012 08:00:00 am 30
C2 E1 11/01/2012 08:00:00 am - 11/01/2012 08:30:00 am 1
11 rows selected.
所以我们分区：
partition by(id)
即通过唯一的参考。那么我们将用维度
生成行dimension by(0 as f)
与部分规则相结合：
for f from 0 to blocks[0]-1 increment 1
因此生成了start_range列 start_range [对于f从0到块[0] -1增量1] = start_block [0] +（30 * cv（f）/ 1440），
start_block [0]位于第一个查询中，例如：
EM START_TIME END_TIME CO START_BLOCK BLOCKS
-- ---------------------- ---------------------- -- ---------------------- ----------
E1 11/01/2012 06:55:00 am 11/01/2012 07:01:00 am C1 11/01/2012 06:30:00 am 2
因此对于此行，它的计算结果为
start_range[0 to 1] = 11/01/2012 06:30:00 am + (30minutes * the value of f)
即
start_range[0] = 11/01/2012 06:30:00 am + (30min*0) = 11/01/2012 06:30:00 am
start_range[1] = 11/01/2012 06:30:00 am + (30min*1) = 11/01/2012 07:00:00 am
其余的很直接：
end_range[any] = start_range[cv()] + (30/1440),
表示对于当前行的end-range，我们会start_range并添加30分钟。
range列是start_range和end_range的串联：
range [any] = to_char(start_range[cv()], 'dd/mm/yyyy hh:mi:ss am') || ' - ' || to_char(end_range[cv()], 'dd/mm/yyyy hh24:mi:ss am'),
最后为了计算该范围内的分钟数：
minutes [any] = case
when start_time[0] between start_range[cv()] and end_range[cv()]
then 1440 *(end_range[cv()] - start_time[0])
when end_time[0] between start_range[cv()] and end_range[cv()]
then 1440 *(end_time[0] - start_range[cv()])
else 1440 * (end_range[cv()] - start_range[cv()])
end );
1440只是得到答案的分钟。
现在我们可以将所有内容分组：
SQL> with foo as (select rownum id, emp, start_time, end_time, code,
2 trunc(start_time, 'mi')
3 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440) start_block,
4 ceil(2*24*(end_time-(trunc(start_time, 'mi')
5 - (mod(to_char(trunc(start_time, 'mi'), 'mi'), 30) / 1440)))) blocks
6 from tab f)
7 select thedate, code, range, sum(minutes) minutes
8 from (select trunc(start_time) thedate, code, emp, range, minutes
9 from foo
10 model partition by(id)
11 dimension by(0 as f)
12 measures(code, emp, start_time, end_time, start_block, blocks,
13 sysdate as start_range,
14 sysdate as end_range,
15 cast(0 as number) minutes,
16 cast('' as varchar2(50)) range)
17 rules (start_range [for f from 0 to blocks[0]-1 increment 1] = start_block[0] + (30*cv(f)/1440),
18 code[any] = code[0],
19 emp[any] = emp[0],
20 end_range[any] = start_range[cv()] + (30/1440),
21 start_time[any] = start_time[0],
22 end_time[any] = end_time[0],
23 range [any] = to_char(start_range[cv()], 'dd/mm/yyyy hh:mi:ss am') || ' - ' || to_char(end_range[cv()], 'dd/mm/yyyy hh24:mi:ss am'),
24 minutes [any] = case
25 when start_time[0] between start_range[cv()] and end_range[cv()]
26 then 1440 *(end_range[cv()] - start_time[0])
27 when end_time[0] between start_range[cv()] and end_range[cv()]
28 then 1440 *(end_time[0] - start_range[cv()])
29 else 1440 * (end_range[cv()] - start_range[cv()])
30 end ))
31 group by thedate, code, range
32 order by thedate, code, range;
THEDATE CO RANGE MINUTES
---------- -- -------------------------------------------------- ----------
11/01/2012 C1 11/01/2012 06:30:00 am - 11/01/2012 07:00:00 am 5
11/01/2012 C1 11/01/2012 07:00:00 am - 11/01/2012 07:30:00 am 1
11/01/2012 C2 11/01/2012 06:30:00 am - 11/01/2012 07:00:00 am 6
11/01/2012 C2 11/01/2012 07:00:00 am - 11/01/2012 07:30:00 am 60
11/01/2012 C2 11/01/2012 07:30:00 am - 11/01/2012 08:00:00 am 60
11/01/2012 C2 11/01/2012 08:00:00 am - 11/01/2012 08:30:00 am 1
11/02/2012 C1 11/02/2012 07:30:00 am - 11/02/2012 08:00:00 am 25
11/02/2012 C1 11/02/2012 08:00:00 am - 11/02/2012 08:30:00 am 1
答案 1 :(得分：1)
我很确定这可以清理，并且更清晰，更有效，因为Oracle不是我的强项之一，但是它可以工作并且应该知道如何完成任务。
这里的关键是加入一个数字列表，将你的记录分成半小时。
SELECT "Date",
"Code",
"RangeStart" + ((r - 1) / 48.0) AS "RangeStart",
"RangeStart" + (r / 48.0) AS "RangeEnd",
SUM(CASE WHEN r = 1 THEN "StartMinutes"
WHEN "END_TIME" >= "RangeStart" + ((r - 1) / 48.0) AND "END_TIME" < "RangeStart" + (r / 48.0) THEN "EndMinutes"
ELSE 30
END) AS "TotalMinutes"
FROM ( SELECT "Emp",
"Date",
"START_TIME",
"END_TIME",
"Code",
CASE WHEN EXTRACT(MINUTE from "START_TIME") > 30 THEN 60 ELSE 30 END - EXTRACT(MINUTE from "START_TIME") AS "StartMinutes",
EXTRACT(MINUTE from END_TIME) - CASE WHEN EXTRACT(MINUTE from "END_TIME") > 30 THEN 30 ELSE 0 END AS "EndMinutes",
"START_TIME" - (EXTRACT(MINUTE from "START_TIME") - CASE WHEN EXTRACT(MINUTE from "START_TIME") > 30 THEN 30 ELSE 0 END) / (60 * 24.0) AS "RangeStart"
FROM T
) T
INNER JOIN
( SELECT Rownum r
FROM dual
CONNECT BY Rownum <= 100
) r
ON "END_TIME" > ("RangeStart" + ((r - 1) / 48.0))
GROUP BY "Date", "Code", "RangeStart" + ((r - 1) / 48.0), "RangeStart" + (r / 48.0)
ORDER BY "Code", "Date", "RangeStart";
答案 2 :(得分：1)
这是另一种解决方案（它不是很优雅，并且使用硬编码的日期文字来获取存储桶的边界 - 应该用子查询替换它们来获取它们）：
with v_data as (
select 1 pk, 'E1' emp, to_date('2012-11-01', 'YYYY-MM-DD') as date1, to_date('2012-11-01 06:55:00', 'YYYY-MM-DD hh24:mi:ss') as start_time, to_date('2012-11-01 07:01:00', 'YYYY-MM-DD hh24:mi:ss') as end_time, 'C1' as code, 6 as minutes from dual union all
select 2 pk, 'E1' emp, to_date('2012-11-01', 'YYYY-MM-DD') as date1, to_date('2012-11-01 06:57:00', 'YYYY-MM-DD hh24:mi:ss') as start_time, to_date('2012-11-01 08:01:00', 'YYYY-MM-DD hh24:mi:ss') as end_time, 'C2' as code, 64 as minutes from dual union all
select 3 pk, 'E2' emp, to_date('2012-11-01', 'YYYY-MM-DD') as date1, to_date('2012-11-01 06:57:00', 'YYYY-MM-DD hh24:mi:ss') as start_time, to_date('2012-11-01 08:00:00', 'YYYY-MM-DD hh24:mi:ss') as end_time, 'C2' as code, 63 as minutes from dual union all
select 4 pk, 'E1' emp, to_date('2012-11-02', 'YYYY-MM-DD') as date1, to_date('2012-11-02 07:35:00', 'YYYY-MM-DD hh24:mi:ss') as start_time, to_date('2012-11-02 08:01:00', 'YYYY-MM-DD hh24:mi:ss') as end_time, 'C1' as code, 26 as minutes from dual),
v_buckets as (
select
to_date('2012-11-01', 'YYYY-MM-DD') + (rownum-1)/48 as bucket_start,
to_date('2012-11-01', 'YYYY-MM-DD') + rownum/48 as bucket_end
from dual
connect by rownum <96
)
select v3.date1, v3.bucket_start, v3.bucket_end, v3.code, sum(v3.time_spent_in_bucket) as minutes
from (
select v2.*, (least(end_time, bucket_end) - greatest(start_time, bucket_start))*1440 as time_spent_in_bucket from
(
select buck.*,
v1.*
from v_buckets buck
join v_data v1
on (
-- time slot completely contained in one bucket
(v1.start_time >= buck.bucket_start and v1.start_time < buck.bucket_end and
v1.end_time >= buck.bucket_start and v1.end_time < buck.bucket_end)
-- time slot starts in bucket, expands to next bucket
or (v1.start_time >= buck.bucket_start and v1.start_time < buck.bucket_end and
v1.end_time >= buck.bucket_end)
-- time slot started in previous bucket, ends in this bucket)
or (v1.start_time < buck.bucket_start and v1.end_time > buck.bucket_start and
v1.end_time <= buck.bucket_end)
-- time slot began in previous bucket, expands to next bucket
or (v1.start_time < buck.bucket_start and v1.end_time >= buck.bucket_end)
)
) v2
) v3
where start_time is not null
group by date1, bucket_start, bucket_end, code
order by bucket_start, code
答案 3 :(得分：1)
这是我的尝试：
select trunc(trunc_start) as datetime, code, range , sum(duration) minutes
from (
select code, end_time, start_time, TRUNC_START ,
to_char(trunc_start,'hh:mi:ss AM')||'-'||to_char(trunc_start+1/48,'hh:mi:ss AM') as range,
case
when end_time-trunc_start between 0 and 1/48 then (end_time-trunc_start)*1440
when start_time-trunc_start between 0 and 1/48 then (trunc_start-start_time)*1440+30
else 30
end as duration
from(
select s.*, n ,
trunc(start_time) + trunc((start_time-trunc(start_time))*48)/48 + (n-1)/48 as trunc_start
from s
join (select level n from dual connect by level <=48) a
on n-2 <= (end_time-start_time)*100
)b
)
where trunc_start < end_time --eliminating fake intervals
group by code, trunc(trunc_start), range
order by 1, 3
;
抱歉where：）
答案 4 :(得分：-1)
以下是您的范围的一些一般示例：
SELECT job
, sum(decode(greatest(sal,2999), least(sal,6000), 1, 0)) "Range 3000-6000"
, sum(decode(greatest(sal,1000), least(sal,2999), 1, 0)) "Range 1000-3000"
, sum(decode(greatest(sal,0), least(sal,999), 1, 0)) "Range 0-1000"
FROM scott.emp
GROUP BY job
/