SQL用于获取两个日期范围之间的计数。

huangapple go评论59阅读模式
英文:

SQL to fetch the count between tow date range

问题

我需要获取在00:15:00和01:15:00之间以及随后的任何日期范围内的PRODUCT_ID计数。

示例脚本:-

我的数据库结构和数据如下。

CREATE TABLE time1 (cr_date date , product_id number );

insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD-MON-YYYY HH:MI:SS') , 12345);
insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD-MON-YYYY HH:MI:SS') , 12346);
insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD-MON-YYYY HH:MI:SS') , 12347);
insert into time1 values (to_date ('01-JAN-2022 03:30:00', 'DD-MON-YYYY HH:MI:SS') , 42345);
insert into time1 values (to_date ('01-JAN-2022 03:30:00', 'DD-MON-YYYY HH:MI:SS') , 42346);
insert into time1 values (to_date ('01-JAN-2022 03:35:00', 'DD-MON-YYYY HH:MI:SS') , 42347);
insert into time1 values (to_date ('01-JAN-2022 03:40:00', 'DD-MON-YYYY HH:MI:SS') , 42348);
insert into time1 values (to_date ('01-JAN-2022 10:40:00', 'DD-MON-YYYY HH:MI:SS') , 10348);
insert into time1 values (to_date ('01-JAN-2022 10:42:00', 'DD-MON-YYYY HH:MI:SS') , 10349);
insert into time1 values (to_date ('01-JAN-2022 10:43:00', 'DD-MON-YYYY HH:MI:SS') , 11348);

COMMIT;

输出如下所示:-

| 小时      | 计数    |
|:--------  |:-------|
|00:15:00  |3     |
|01:15:00  |0     |
|02:15:00  |0     |
|03:15:00  |0     |
|04:15:00  |4     |
|05:15:00  |0     |
|06:15:00  |0     |
|07:15:00  |0     |
|08:15:00  |0     |
|09:15:00  |0     |
|10:15:00  |0     |
|11:15:00  |3     |
|..
|...
|23:15:00  |0     |
英文:

I require to fetch the count of PRODUCT_ID between 00:15:00 and 01:15:00 and subsequentially for any date range.

Example Scripts:-

My DB structure and data is as follows.

CREATE TABLE time1 (cr_date date , product_id number );

insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD_MON-YYYY HH:MI:SS') , 12345);
insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD_MON-YYYY HH:MI:SS') , 12346);
insert into time1 values (to_date ('01-JAN-2022 01:00:00', 'DD_MON-YYYY HH:MI:SS') , 12347);
insert into time1 values (to_date ('01-JAN-2022 03:30:00', 'DD_MON-YYYY HH:MI:SS') , 42345);
insert into time1 values (to_date ('01-JAN-2022 03:30:00', 'DD_MON-YYYY HH:MI:SS') , 42346);
insert into time1 values (to_date ('01-JAN-2022 03:35:00', 'DD_MON-YYYY HH:MI:SS') , 42347);
insert into time1 values (to_date ('01-JAN-2022 03:40:00', 'DD_MON-YYYY HH:MI:SS') , 42348);
insert into time1 values (to_date ('01-JAN-2022 10:40:00', 'DD_MON-YYYY HH:MI:SS') , 10348);
insert into time1 values (to_date ('01-JAN-2022 10:42:00', 'DD_MON-YYYY HH:MI:SS') , 10349);
insert into time1 values (to_date ('01-JAN-2022 10:43:00', 'DD_MON-YYYY HH:MI:SS') , 11348);

COMMIT;

Output is required to be as below:-

| hours   | count |
|:------  |:------|
|00:15:00 |3|
|01:15:00 |0|
|02:15:00 |0|
|03:15:00 |0|
|04:15:00 |4|
|05:15:00 |0|
|06:15:00 |0|
|07:15:00 |0|
|08:15:00 |0|
|09:15:00 |0|
|10:15:00 |0|
|11:15:00 |3|
|..
|...
|23:15:00 |0|

答案1

得分: 1

你最初发布的示例数据基本无用,没有涉及时间组件。

在我的一个表中,有一个 datum 列,值看起来像这样:

这是查询:

  • fmin CTE 中创建 24 行(00:15, 01:15, ... 23:15)
  • datum 上进行联接到前一个15分钟的“舍入”值(无论是在同一小时内还是在前一个小时内 - 取决于分钟)

所以:

结果:

 C_TIM        CNT
----- ----------
00:15          0
01:15          0
02:15          0
03:15          0
04:15          0
05:15          0
06:15          0
07:15          2
08:15         10
09:15          1
10:15         14
11:15          6
12:15         10
13:15         38
14:15          5
15:15          0
16:15          0
17:15          0
18:15          0
19:15          0
20:15          0
21:15          0
22:15          0
23:15          0

24 rows selected.
英文:

Sample data you (initially) posted is pretty much useless, there's no time component involved.

In one of my tables, there's a datum column and values look like this:

SQL> SELECT id, TO_CHAR (datum, 'hh24:mi') hrs FROM obr WHERE rownum <= 10;

        ID HRS
---------- -----
     21547 08:41
     21541 08:17
     21563 09:03
     21614 10:46
     21618 11:01
     21620 11:04
     21622 11:05
     21626 11:10
     21629 11:14
     21642 13:35

10 rows selected.

SQL>

This is query which

  • in fmin CTE creates 24 rows (00:15, 01:15, ... 23:15)
  • join is done on datum "rounded" to previous 15-minute value (whether in the same hour, or in previous hour - depends on minutes)

So:

SQL> WITH
  2     fmin
  3     AS
  4        (    SELECT TRUNC (SYSDATE) + (LEVEL - 1) / 24 + INTERVAL '15' MINUTE c_time
  5               FROM DUAL
  6         CONNECT BY LEVEL <= 24)
  7    SELECT TO_CHAR (f.c_time, 'hh24:mi') c_time, COUNT (z.id) cnt
  8      FROM fmin f
  9           LEFT JOIN obr z
 10              ON TO_CHAR (f.c_time, 'hh24:mi') =
 11                 TO_CHAR (
 12                      TRUNC (datum, 'hh24')
 13                    + CASE
 14                         WHEN TO_NUMBER (TO_CHAR (datum, 'mi')) >= 15
 15                         THEN
 16                            INTERVAL '15' MINUTE
 17                         WHEN TO_NUMBER (TO_CHAR (datum, 'mi')) < 15
 18                         THEN
 19                            INTERVAL '-45' MINUTE
 20                      END,
 21                    'hh24:mi')
 22  GROUP BY TO_CHAR (f.c_time, 'hh24:mi')
 23  ORDER BY 1;

Result:

C_TIM        CNT
----- ----------
00:15          0
01:15          0
02:15          0
03:15          0
04:15          0
05:15          0
06:15          0
07:15          2
08:15         10
09:15          1
10:15         14
11:15          6
12:15         10
13:15         38
14:15          5
15:15          0
16:15          0
17:15          0
18:15          0
19:15          0
20:15          0
21:15          0
22:15          0
23:15          0

24 rows selected.

SQL>

答案2

得分: 0

根据时间来计算数量是不可能的,因为你只在CR_DATE列中存储了日期。

即使你将date转换为timestamp,时间始终会被设置为默认值,即00:00:00.00000012:00:00.000000

要实现这个目标,你需要像下面这样重新创建你的表:

CREATE TABLE time1 (cr_date timestamp, product_id number );

然后,你需要像下面的示例一样将实际的timestamp插入表中:

INSERT INTO TIME1 (CR_DATE,PRODUCT_ID) VALUES (to_timestamp('01-01-22 14:32:62.258152','DD-MM-RR HH24:MI:SS.FF'),12345);

这里是to_timestamp的文档

在此之后,你可以根据时间查询,就像你的问题中所描述的那样。

SELECT TRUNC(CR_DATE,'HH'), count(product_id) 
FROM TIME1 
GROUP BY TRUNC(CR_DATE,'HH');

这里是TRUNC()函数的文档

从上面的查询中可能会得到以下结果。

TRUNC(CR_DATE,'HH')   COUNT(product_id)
------------------- ----------
2022-01-01 09:00:00        748
2022-01-01 16:00:00         24
2022-01-01 17:00:00         12
2022-01-01 22:00:00        737
2022-01-01 23:00:00        182
2022-01-01 20:00:00         16
2022-01-01 21:00:00        293
2022-01-01 22:00:00        610

如果你发现任何问题,请在下面评论以进一步讨论。

英文:

You wont get count based on time as you are storing only date in CR_DATE column.

Even if you convert date to timestamp, the time will always be set to default value i.e., 00:00:00.000000 or 12:00:00.000000.

To achieve this, you have to re-create your table like below :

CREATE TABLE time1 (cr_date timestamp, product_id number );

And then, you have to insert actual timestamp into the table like the one example below.

INSERT INTO TIME1 (CR_DATE,PRODUCT_ID) VALUES (to_timestamp('01-01-22 14:32:62.258152','DD-MM-RR HH24:MI:SS.FF'),12345);

Here is the documentations for to_timestamp.

After this, you can query based on time as per your question.

SELECT TRUNC(CR_DATE,'HH'), count(product_id) 
FROM TIME1 
GROUP BY TRUNC(CR_DATE,'HH');

Here is the documentation for TRUNC() function.

The following result can be possibly obtained from above query.

TRUNC(CR_DATE,'HH')   COUNT(product_id)
------------------- ----------
2022-01-01 09:00:00        748
2022-01-01 16:00:00         24
2022-01-01 17:00:00         12
2022-01-01 22:00:00        737
2022-01-01 23:00:00        182
2022-01-01 20:00:00         16
2022-01-01 21:00:00        293
2022-01-01 22:00:00        610

If you find any issue, please comment below for further discussion.

huangapple
  • 本文由 发表于 2023年6月2日 13:59:45
  • 转载请务必保留本文链接:https://go.coder-hub.com/76387483.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定