我的SAMPLE表格包含以下五列：
sample_id (PK) (NUMBER)
sampled_on (DATE)
received_on (DATE)
completed_on (DATE)
authorized_on (DATE)
我想查询每小时一行（受给定日期范围约束）和五列：
YYYY-MM-DD HH24 请提供正确的查询或至少一个点。
以赏金重新开启：
+300声誉，第一个将 Rob van Wijk 的答案（单次访问样本）合并到一个视图中，我可以按日期范围有效查询{{1 }或start_date/end_date）。
答案 0 :(得分：2)
这可能不是最漂亮或最优的解决方案，但似乎有效。说明：首先将所有日期转换为YYYY-MM-DD HH24格式，下一个收集号码采样/接收/完成/授权日期+ HH24，最后加入。
with sample_hour as
(select sample_id,
to_char(sampled_on, 'YYYY-MM-DD HH24') sampled_on,
to_char(received_on, 'YYYY-MM-DD HH24') received_on,
to_char(completed_on, 'YYYY-MM-DD HH24') completed_on,
to_char(authorized_on, 'YYYY-MM-DD HH24') authorized_on
from sample),
s as
(select sampled_on thedate, count(*) num_sampled
from sample_hour
group by sampled_on),
r as
(select received_on thedate, count(*) num_received
from sample_hour
group by received_on),
c as
(select completed_on thedate, count(*) num_completed
from sample_hour
group by completed_on),
a as
(select authorized_on thedate, count(*) num_authorized
from sample_hour
group by authorized_on)
select s.thedate, num_sampled, num_received, num_completed, num_authorized
from s
left join r on s.thedate = r.thedate
left join c on s.thedate = c.thedate
left join a on s.thedate = a.thedate
;
这假设表格“样本”创建如下：
create table sample
(sample_id number not null primary key,
sampled_on date,
received_on date,
completed_on date,
authorized_on date);
答案 1 :(得分：2)
这是一个例子。首先创建表并插入一些随机数据。
SQL> create table sample
2 ( sample_id number primary key
3 , sampled_on date
4 , received_on date
5 , completed_on date
6 , authorized_on date
7 )
8 /
Tabel is aangemaakt.
SQL> insert into sample
2 select level
3 , trunc(sysdate) + dbms_random.value(0,2)
4 , trunc(sysdate) + dbms_random.value(0,2)
5 , trunc(sysdate) + dbms_random.value(0,2)
6 , trunc(sysdate) + dbms_random.value(0,2)
7 from dual
8 connect by level <= 1000
9 /
1000 rijen zijn aangemaakt.
然后介绍给定日期范围的变量并填充它们。
SQL> var DATE_RANGE_START varchar2(10)
SQL> var DATE_RANGE_END varchar2(10)
SQL> exec :DATE_RANGE_START := '2009-10-23'
PL/SQL-procedure is geslaagd.
SQL> exec :DATE_RANGE_END := '2009-10-24'
PL/SQL-procedure is geslaagd.
首先，您必须在给定的日期范围内生成所有小时数。这样可以确保如果您有一个没有日期的小时，您仍然会有一个包含4个零的记录。实现在all_hours查询中。查询的其余部分（只有一个表访问您的样本表！）可以非常简单。
SQL> with all_hours as
2 ( select to_date(:DATE_RANGE_START,'yyyy-mm-dd') + numtodsinterval(level-1,'hour') hour
3 from dual
4 connect by level <=
5 ( to_date(:DATE_RANGE_END,'yyyy-mm-dd')
6 - to_date(:DATE_RANGE_START,'yyyy-mm-dd')
7 + 1
8 ) * 24
9 )
10 select h.hour
11 , count(case when h.hour = trunc(s.sampled_on,'hh24') then 1 end) sampled#
12 , count(case when h.hour = trunc(s.received_on,'hh24') then 1 end) received#
13 , count(case when h.hour = trunc(s.completed_on,'hh24') then 1 end) completed#
14 , count(case when h.hour = trunc(s.authorized_on,'hh24') then 1 end) authorized#
15 from all_hours h
16 cross join sample s
17 group by h.hour
18 /
HOUR SAMPLED# RECEIVED# COMPLETED# AUTHORIZED#
------------------- ---------- ---------- ---------- -----------
23-10-2009 00:00:00 18 25 20 20
23-10-2009 01:00:00 26 24 16 13
23-10-2009 02:00:00 16 26 17 15
23-10-2009 03:00:00 19 18 27 13
23-10-2009 04:00:00 28 20 18 23
23-10-2009 05:00:00 17 13 19 21
23-10-2009 06:00:00 18 23 16 15
23-10-2009 07:00:00 19 24 14 22
23-10-2009 08:00:00 21 19 23 22
23-10-2009 09:00:00 25 20 23 24
23-10-2009 10:00:00 16 21 25 18
23-10-2009 11:00:00 21 29 21 18
23-10-2009 12:00:00 33 28 24 20
23-10-2009 13:00:00 24 19 15 15
23-10-2009 14:00:00 20 27 16 25
23-10-2009 15:00:00 15 25 27 13
23-10-2009 16:00:00 19 14 27 18
23-10-2009 17:00:00 22 22 15 27
23-10-2009 18:00:00 20 19 29 23
23-10-2009 19:00:00 20 18 17 23
23-10-2009 20:00:00 11 18 20 27
23-10-2009 21:00:00 13 25 24 19
23-10-2009 22:00:00 22 13 22 29
23-10-2009 23:00:00 20 20 19 24
24-10-2009 00:00:00 18 17 18 29
24-10-2009 01:00:00 23 30 26 21
24-10-2009 02:00:00 28 19 28 25
24-10-2009 03:00:00 21 21 11 23
24-10-2009 04:00:00 23 20 21 17
24-10-2009 05:00:00 24 16 23 23
24-10-2009 06:00:00 23 26 22 30
24-10-2009 07:00:00 25 26 18 12
24-10-2009 08:00:00 24 20 23 17
24-10-2009 09:00:00 18 26 15 19
24-10-2009 10:00:00 20 19 25 18
24-10-2009 11:00:00 19 27 17 20
24-10-2009 12:00:00 23 16 18 20
24-10-2009 13:00:00 15 15 22 19
24-10-2009 14:00:00 23 23 16 29
24-10-2009 15:00:00 18 31 32 28
24-10-2009 16:00:00 22 15 18 13
24-10-2009 17:00:00 25 17 20 26
24-10-2009 18:00:00 19 20 21 16
24-10-2009 19:00:00 22 13 28 29
24-10-2009 20:00:00 23 17 23 14
24-10-2009 21:00:00 18 18 21 22
24-10-2009 22:00:00 22 20 18 21
24-10-2009 23:00:00 21 18 22 22
48 rijen zijn geselecteerd.
希望这有帮助。
此致 罗布。
答案 2 :(得分：2)
尝试：
CREATE OR REPLACE VIEW my_view AS
WITH date_bookends AS (
SELECT LEAST(MIN(t.sampled_on), MIN(t.received_on), MIN(t.completed_on), MIN(t.authorized_on)) 'min_date'
GREATEST(MAX(t.sampled_on), MAX(t.received_on), MAX(t.completed_on), MAX(t.authorized_on)) 'max_date'
FROM SAMPLE t),
all_hours AS (
SELECT t.min_date + numtodsinterval(LEVEL - 1,'hour') date_by_hour
FROM date_bookends t
CONNECT BY LEVEL <= ( t.max_date - t.min_date + 1) * 24)
SELECT h.date_by_hour,
COUNT(CASE WHEN h.hour = TRUNC(s.sampled_on,'hh24') THEN 1 END) sampled#
COUNT(CASE WHEN h.hour = TRUNC(s.received_on,'hh24') THEN 1 END) received#
COUNT(CASE WHEN h.hour = TRUNC(s.completed_on,'hh24') THEN 1 END) completed#
COUNT(CASE WHEN h.hour = TRUNC(s.authorized_on,'hh24') THEN 1 END) authorized#
FROM all_hours h
CROSS JOIN sample s
GROUP BY h.hour
不使用子查询因子：
CREATE OR REPLACE VIEW my_view AS
SELECT h.date_by_hour,
COUNT(CASE WHEN h.hour = TRUNC(s.sampled_on,'hh24') THEN 1 END) sampled#
COUNT(CASE WHEN h.hour = TRUNC(s.received_on,'hh24') THEN 1 END) received#
COUNT(CASE WHEN h.hour = TRUNC(s.completed_on,'hh24') THEN 1 END) completed#
COUNT(CASE WHEN h.hour = TRUNC(s.authorized_on,'hh24') THEN 1 END) authorized#
FROM (SELECT t.min_date + numtodsinterval(LEVEL - 1,'hour') date_by_hour
FROM (SELECT LEAST(MIN(t.sampled_on), MIN(t.received_on), MIN(t.completed_on), MIN(t.authorized_on)) 'min_date'
GREATEST(MAX(t.sampled_on), MAX(t.received_on), MAX(t.completed_on), MAX(t.authorized_on)) 'max_date'
FROM SAMPLE t) t
CONNECT BY LEVEL <= ( t.max_date - t.min_date + 1) * 24) h
CROSS JOIN sample s
GROUP BY h.hour
查询两次访问SAMPLES表 - 第一次获得最早的＆amp;构建date_by_hour值的最新日期。
答案 3 :(得分：0)
我会这样做4个查询（每个日期一个）：
SELECT <date to hour>, count(*) FROM sample GROUP BY <date to hour>
然后将数据放在应用程序中。如果您确实需要单个查询，则可以在hour上加入各个查询。
答案 4 :(得分：0)
试试这个......
WITH src_data AS
( SELECT sample_id
, TRUNC( sampled_on, 'HH24' ) sampled_on
, TRUNC( received_on, 'HH24' ) received_on
, TRUNC( completed_on, 'HH24' ) completed_on
, TRUNC( authorized_on, 'HH24' ) authorized_on
FROM sample
)
, src_hours AS
( SELECT sampled_on the_date
FROM src_data
WHERE sampled_on IS NOT NULL
UNION
SELECT received_on the_date
FROM src_data
WHERE received_on IS NOT NULL
UNION
SELECT completed_on the_date
FROM src_data
WHERE completed_on IS NOT NULL
UNION
SELECT authorized_on the_date
FROM src_data
WHERE authorized_on IS NOT NULL
)
SELECT h.the_date
, ( SELECT COUNT(*)
FROM src_data s
WHERE s.sampled_on = h.the_date ) num_sampled_on
, ( SELECT COUNT(*)
FROM src_data r
WHERE r.received_on = h.the_date ) num_received_on
, ( SELECT COUNT(*)
FROM src_data c
WHERE c.completed_on = h.the_date ) num_completed_on
, ( SELECT COUNT(*)
FROM src_data a
WHERE a.authorized_on = h.the_date ) num_authorized_on
FROM src_hours h
答案 5 :(得分：0)
也许就像创建这个视图一样：
{{2p>然后从视图中选择：create view hours as
select hour, max(cnt_sample) cnt_sample, max(cnt_received) cnt_received, max(cnt_completed) cnt_completed, max(cnt_authorized) cnt_authorized
from (
select to_char(sampled_on , 'yyyymmddhh24') hour,
count(sample_id) over (partition by to_char(sampled_on ,'yyyymmddhh24')) cnt_sample,
0 cnt_received,
0 cnt_completed,
0 cnt_authorized from sample union all
select to_char(received_on , 'yyyymmddhh24') hour,
0 cnt_sample,
count(sample_id) over (partition by to_char(received_on ,'yyyymmddhh24')) cnt_received,
0 cnt_completed,
0 cnt_authorized from sample union all
select to_char(completed_on , 'yyyymmddhh24') hour,
0 cnt_sample,
0 cnt_received,
count(sample_id) over (partition by to_char(completed_on ,'yyyymmddhh24')) cnt_completed,
0 cnt_authorized from sample union all
select to_char(authorized_on, 'yyyymmddhh24') hour,
0 cnt_sample,
0 cnt_received,
0 cnt_completed,
count(sample_id) over (partition by to_char(authorized_on ,'yyyymmddhh24')) cnt_authorized from sample
)
group by hour
;
答案 6 :(得分：0)
这就是我的想法，但我不确定它是否足够适合观看。
select
the_date,
sum(decode(the_type,'S',the_count,0)) samples,
sum(decode(the_type,'R',the_count,0)) receipts,
sum(decode(the_type,'C',the_count,0)) completions,
sum(decode(the_type,'A',the_count,0)) authorizations
from(
select
trunc(sampled_on,'HH24') the_date,
'S' the_type,
count(1) the_count
FROM sample
group by trunc(sampled_on,'HH24')
union all
select
trunc(received_on,'HH24'),
'R',
count(1)
FROM sample
group by trunc(received_on,'HH24')
union all
select
trunc(completed_on,'HH24'),
'C',
count(1)
FROM sample
group by trunc(completed_on,'HH24')
union all
select
trunc(authorized_on,'HH24'),
'A',
count(1)
FROM sample
group by trunc(authorized_on,'HH24')
)
group by the_date
然后，要查询，您可以使用正常的日期结构进行查询：
select * from magic_view where the_date > sysdate-1;
好的，所以我创建了一个示例表并做了一些指标：
create table sample (
sample_id number primary key,
sampled_on date,
received_on date,
completed_on date,
authorized_on date
);
insert into sample (
select
level,
trunc(sysdate) + dbms_random.value(0,2),
trunc(sysdate) + dbms_random.value(0,2),
trunc(sysdate) + dbms_random.value(0,2),
trunc(sysdate) + dbms_random.value(0,2),
from dual
connect by level <= 1000
);
解释计划是：
---------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)|
---------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 4000 | 97K| 25 (20)|
| 1 | HASH GROUP BY | | 4000 | 97K| 25 (20)|
| 2 | VIEW | | 4000 | 97K| 24 (17)|
| 3 | UNION-ALL | | | | |
| 4 | HASH GROUP BY | | 1000 | 9000 | 6 (17)|
| 5 | TABLE ACCESS FULL| SAMPLE | 1000 | 9000 | 5 (0)|
| 6 | HASH GROUP BY | | 1000 | 9000 | 6 (17)|
| 7 | TABLE ACCESS FULL| SAMPLE | 1000 | 9000 | 5 (0)|
| 8 | HASH GROUP BY | | 1000 | 9000 | 6 (17)|
| 9 | TABLE ACCESS FULL| SAMPLE | 1000 | 9000 | 5 (0)|
| 10 | HASH GROUP BY | | 1000 | 9000 | 6 (17)|
| 11 | TABLE ACCESS FULL| SAMPLE | 1000 | 9000 | 5 (0)|
---------------------------------------------------------------------
在我的计算机上，过去24小时内针对此视图的查询在23毫秒内完成。不错，但它只有1000行。在对4个单独的查询进行折扣之前，您需要对各个解决方案进行性能分析。
答案 7 :(得分：0)
我现在提议：
create view hours_ as
with four as (
select 1 as n from dual union all
select 2 as n from dual union all
select 3 as n from dual union all
select 4 as n from dual )
select
case when four.n = 1 then trunc(sampled_on , 'hh24')
when four.n = 2 then trunc(received_on , 'hh24')
when four.n = 3 then trunc(completed_on , 'hh24')
when four.n = 4 then trunc(authorized_on, 'hh24')
end hour_,
sum ( case when four.n = 1 then 1
else 0
end ) sample_,
sum ( case when four.n = 2 then 1
else 0
end ) receive_,
sum ( case when four.n = 3 then 1
else 0
end ) complete_,
sum ( case when four.n = 4 then 1
else 0
end ) authorize_
from
four cross join sample
group by
case when four.n = 1 then trunc(sampled_on , 'hh24')
when four.n = 2 then trunc(received_on , 'hh24')
when four.n = 3 then trunc(completed_on , 'hh24')
when four.n = 4 then trunc(authorized_on, 'hh24')
end ;
为了查看视图是否确实只访问过一次：
explain plan for select * from hours_
where hour_ between sysdate -1 and sysdate;
select * from table (dbms_xplan.display);
结果是：
--------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 61 | 16 (7)| 00:00:01 |
| 1 | VIEW | HOURS_ | 1 | 61 | 16 (7)| 00:00:01 |
| 2 | HASH GROUP BY | | 1 | 39 | 16 (7)| 00:00:01 |
|* 3 | FILTER | | | | | |
| 4 | NESTED LOOPS | | 1 | 39 | 15 (0)| 00:00:01 |
| 5 | VIEW | | 4 | 12 | 8 (0)| 00:00:01 |
| 6 | UNION-ALL | | | | | |
| 7 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
| 8 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
| 9 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
| 10 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
|* 11 | TABLE ACCESS FULL| SAMPLE | 1 | 36 | 2 (0)| 00:00:01 |
--------------------------------------------------------------------------------
答案 8 :(得分：0)
与RenéNyffenegger的想法相似。按每种类型的日期字段过滤，然后合并计数。
请注意，在一个Select中无法执行此查询，因为您需要同时按每个日期字段进行分组和排序，如果不拆分为单独的子查询，则无法进行此操作。
对于此示例，我已将“2009-11-04”的日期范围编码为“2009-11-04 23:59:59”：
SELECT
DateHour,
SUM(sampled) total_sampled,
SUM(received) total_received,
SUM(completed) total_completed,
SUM(authorized) total_authorized
FROM
(SELECT
TO_CHAR(CREATED_DATE, 'YYYY-MM-DD HH24') DateHour,
1 sampled,
0 received,
0 completed,
0 authorized
FROM
SAMPLE
WHERE
sampled_on >= TO_DATE('2009-11-04', 'YYYY-MM-DD')
AND sampled_on <= TO_DATE('2009-11-04 23:59:59', 'YYYY-MM-DD HH24:MI:SS')
UNION ALL
SELECT
TO_CHAR(CREATED_DATE, 'YYYY-MM-DD HH24') DateHour,
0 sampled,
1 received,
0 completed,
0 authorized
FROM
SAMPLE
WHERE
received_on >= TO_DATE('2009-11-04', 'YYYY-MM-DD')
AND received_on <= TO_DATE('2009-11-04 23:59:59', 'YYYY-MM-DD HH24:MI:SS')
UNION ALL
SELECT
TO_CHAR(CREATED_DATE, 'YYYY-MM-DD HH24') DateHour,
0 sampled,
0 received,
1 completed,
0 authorized
FROM
SAMPLE
WHERE
completed_on >= TO_DATE('2009-11-04', 'YYYY-MM-DD')
AND completed_on <= TO_DATE('2009-11-04 23:59:59', 'YYYY-MM-DD HH24:MI:SS')
UNION ALL
SELECT
TO_CHAR(CREATED_DATE, 'YYYY-MM-DD HH24') DateHour,
0 sampled,
0 received,
0 completed,
1 authorized
FROM
SAMPLE
WHERE
authorized_on >= TO_DATE('2009-11-04', 'YYYY-MM-DD')
AND authorized_on <= TO_DATE('2009-11-04 23:59:59', 'YYYY-MM-DD HH24:MI:SS'))
GROUP BY
DateHour
ORDER BY
DateHour