I have the following date field format: "HH:MM", which represents the duration of an event, and I'd like to extract the equivalent value in HOUR (example: 2h:30min ==> result : 2,5 hour).
Ideally, change your table to store the data as an INTERVAL DAY TO SECOND data type and then you can just store INTERVAL '2:30' HOUR TO MINUTE and use date arithmetic to get your answer.
SELECT ( DATE '1970-01-01' + your_interval_value - DATE '1970-01-01' ) * 24
FROM DUAL;
Since you are storing strings instead of intervals then you can use NUMTODSINTERVAL and string functions to convert the hours and minutes to intervals and then use the same date arithmetic:
db<>fiddle here
Oracle Setup:
CREATE TABLE table_name ( value ) AS
SELECT '2h:30min' FROM DUAL UNION ALL
SELECT '15min' FROM DUAL UNION ALL
SELECT '3h' FROM DUAL UNION ALL
SELECT '26 h : 20 min' FROM DUAL UNION ALL
SELECT '-2h:30min' FROM DUAL UNION ALL
SELECT '-4h' FROM DUAL UNION ALL
SELECT '-45min' FROM DUAL UNION ALL
SELECT '0h:0min' FROM DUAL UNION ALL
SELECT '0h' FROM DUAL UNION ALL
SELECT '-0min' FROM DUAL;
Query:
SELECT value,
( DATE '1970-01-01'
+ NUMTODSINTERVAL(
CASE WHEN INSTR( value, '-' ) > 0 THEN -1 ELSE 1 END
*
TO_NUMBER( COALESCE( REGEXP_SUBSTR( value, '(\d*)\s*h', 1, 1, 'i', 1 , '0' ) ),
'HOUR'
)
+ NUMTODSINTERVAL(
CASE WHEN INSTR( value, '-' ) > 0 THEN -1 ELSE 1 END
*
TO_NUMBER( COALESCE( REGEXP_SUBSTR( value, '(\d*)\s*min', 1, 1, 'i', 1 ), '0' ) ),
'MINUTE'
)
- DATE '1970-01-01'
) * 24 AS hours_difference
FROM table_name;
Output:
VALUE | HOURS_DIFFERENCE
:------------ | ----------------------------------------:
2h:30min | 2.5
15min | .2500000000000000000000000000000000000008
3h | 3
26 h : 20 min | 26.33333333333333333333333333333333333328
-2h:30min | -2.5
-4h | -4
-45min | -.75
0h:0min | 0
0h | 0
-0min | 0
An oracle way:
SELECT (TO_DATE('2h:30min', 'HH24"h":MI"min"') - TO_DATE('0:00', 'HH24:MI')) * 24.0 from dual
How it works:
Parses 2h:30min into a date, and generates a date of 01-JAN-19 (all date parsing without a day, month, year becomes the 1st of the current month/year)
Parses 0:00 to a date -> same day, same month, same year as above, but a time of midnight
Subtracts them, resulting in a floating point number that represents the fractions of a day (a day is 1.0, 12 hours is 0.5 etc) between the two datetimes. This works out in this case as approx 0.1041667
We multiply by 24 to turn our 0.1041667 days into hours -> 2.5
If the value is stored as a string, then you can extract the components and use arithmetic.
One method to extract the components would be to take the first two character and then the 4th and 5th. regexp_substr() provides a bit more flexibility:
select cast(regexp_substr(yyyymm, '[^:]+', 1, 1) as number) + cast(regexp_substr(yyyymm, '[^:]+', 1, 2) as number) / 60
from (select '24:30' as yyyymm from dual) x
Related
I need to return the time between two dates in Oracle except the time during the weekends, I could return the minute. But when I set a weekend date, I receive a null result instead of the remaining time in workweek.
First, we need to create a function:
CREATE OR REPLACE FUNCTION get_bus_minutes_between (start_dt DATE, end_dt DATE)
RETURN NUMBER
IS
v_return NUMBER;
BEGIN
select sum(greatest(end_dt - start_dt,0)) * 24 * 60 work_minutes
into v_return
from dual
where trunc(start_dt) - trunc(start_dt,'iw') < 5; -- exclude weekends
RETURN v_return;
END;
Case 1 - Return the minutes in the workweek - Ok
Starting and ending in the workweek.
SELECT
"GET_BUS_MINUTES_BETWEEN"(TO_DATE('14-09-2020 06:00:00', 'dd-mm-yyyy hh24:mi:ss'),
TO_DATE('14-09-2020 10:00:00', 'dd-mm-yyyy hh24:mi:ss')) "WORK_MINUTES"
FROM
"SYS"."DUAL";
Case 2 - Return the remaining minutes in the workweek - Fail
Starting at the weekend and ending in workweek.
SELECT
"GET_BUS_MINUTES_BETWEEN"(TO_DATE('13-09-2020 06:00:00', 'dd-mm-yyyy hh24:mi:ss'),
TO_DATE('14-09-2020 10:00:00', 'dd-mm-yyyy hh24:mi:ss')) "WORK_MINUTES"
FROM
"SYS"."DUAL";
13-09-2020 is Sunday, therefore I was expected the return as 600 minutes related the Monday.
In these possibilities, we can start at the workweek and end at weekend.
You don't need to use SQL or a row generator and can do it with a simple calculation using only PL/SQL. Adapted from my answers here and here.
CREATE OR REPLACE FUNCTION get_bus_minutes_between (start_dt DATE, end_dt DATE)
RETURN NUMBER
IS
p_start_date DATE;
p_end_date DATE;
p_working_days NUMBER;
BEGIN
IF start_dt IS NULL OR end_dt IS NULL THEN
RETURN NUll;
END IF;
-- Enforce that the values are earliest start date to latest end date.
p_start_date := LEAST( start_dt, end_dt );
p_end_date := GREATEST( start_dt, end_dt );
-- Calculate the number of days from the beginning of the ISO week containing
-- the start date and the beginning of the ISO week containing the end date
-- and then multiply this by 5/7 to get the number of full business days.
--
-- Then add on the extra days from the beginining of the ISO week containing
-- the end date and the end date and subtract the extra days from the
-- beginning of the ISO week containing the start date to the start date.
p_working_days := ( TRUNC( p_end_date, 'IW' ) - TRUNC( p_start_date, 'IW' ) ) * 5 / 7
+ LEAST( p_end_date - TRUNC( p_end_date, 'IW' ), 5 )
- LEAST( p_start_date - TRUNC( p_start_date, 'IW' ), 5 );
-- If the start date and end date are reversed then return a negative value.
IF start_dt > end_dt THEN
RETURN -ROUND( p_working_days * 24 * 60, 3 );
ELSE
RETURN +ROUND( p_working_days * 24 * 60, 3 );
END IF;
END;
/
Then:
SELECT GET_BUS_MINUTES_BETWEEN(
DATE '2020-09-14' + INTERVAL '6' HOUR,
DATE '2020-09-14' + INTERVAL '10' HOUR
) AS minutes_between
FROM DUAL;
Outputs:
| MINUTES_BETWEEN |
| --------------: |
| 240 |
and:
SELECT GET_BUS_MINUTES_BETWEEN(
DATE '2020-09-13' + INTERVAL '6' HOUR,
DATE '2020-09-14' + INTERVAL '10' HOUR
) AS minutes_between
FROM DUAL;
outputs:
| MINUTES_BETWEEN |
| --------------: |
| 600 |
db<>fiddle here
If the intervals are not too big, one method uses a brute force approach to generate all minutes in the range, then exclude week-ends:
with cte(dt, end_dt) as (
select start_dt, end_dt from dual
union all
select dt + 1 / 24 / 60, end_dt from cte where dt < end_dt
)
select count(*) work_minutes
from cte
where trunc(dt) - trunc(dt,'iw') < 5
If the intervals are not too big, one method uses a brute force approach to generate all minutes in the range, then exclude week-ends:
with cte(dt, end_dt) as (
select start_dt, end_dt from dual
union all
select dt + 1 / 24 / 60, end_dt from cte where dt < end_dt
)
select count(*) work_minutes
from cte
where to_char(dt, 'IW') <= 5
If you have large intervals, we can reduce the number of iterations by pre-generating minutes / hours series:
with
params (start_dt, end_dt) as (
select start_dt, end_dt from dual
)
minutes (mi) as (
select 0 from dual
union all select mi + 1 from minutes where mi < 59
),
hours (hr) as (
select 0 from dual
union all select hr + 1 from hours where hr < 23
)
select count(*) work_minutes
from params p
cross join minutes m
cross join hours h
where
p.start_dt + h.hr / 24 + m.mi / 24 / 60 <= end_dt
and trunc(p.start_dt + h.hr / 24 + m.mi / 24 / 60) - trunc(p.start_dt + h.hr / 24 + m.mi / 24 / 60,'iw') < 5
how to convert varchar(hh:mm) to minutes in oracle sql.
For example:
HH:MM Minutes
08:00 480
08:45 525
07:57 477
This will work even if the duration is 24 hours or greater:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE durations ( duration ) AS
SELECT '00:30' FROM DUAL UNION ALL
SELECT '07:57' FROM DUAL UNION ALL
SELECT '08:00' FROM DUAL UNION ALL
SELECT '12:00' FROM DUAL UNION ALL
SELECT '20:01' FROM DUAL UNION ALL
SELECT '23:59' FROM DUAL UNION ALL
SELECT '24:00' FROM DUAL UNION ALL
SELECT '24:59' FROM DUAL;
Query 1:
SELECT duration,
( (
DATE '1970-01-01'
+ NUMTODSINTERVAL( SUBSTR( duration, 1, INSTR( duration, ':' ) - 1 ), 'HOUR' )
+ NUMTODSINTERVAL( SUBSTR( duration, INSTR( duration, ':' ) + 1 ), 'MINUTE' )
)
- DATE '1970-01-01'
) * 24 * 60 AS Minutes
FROM durations
Results:
| DURATION | MINUTES |
|----------|---------|
| 00:30 | 30 |
| 07:57 | 477 |
| 08:00 | 480 |
| 12:00 | 720 |
| 20:01 | 1201 |
| 23:59 | 1439 |
| 24:00 | 1440 |
| 24:59 | 1499 |
However, there is an INTERVAL DAY TO SECOND data type that would be better suited to your data:
CREATE TABLE your_table (
duration INTERVAL DAY TO SECOND
);
Then you can just do:
INSERT INTO your_table ( duration ) VALUES ( INTERVAL '08:00' HOUR TO MINUTE );
To get the number of minutes you can then simply do:
SELECT ( ( DATE '1970-01-01' + duration ) - DATE '1970-01-01' ) *24*60 AS minutes
FROM your_table
Try this
TO_NUMBER(SUBSTR('(08:00)',2,INSTR('(08:00)',':')-2))*60+TO_NUMBER(SUBSTR('(08:00)',INSTR('(08:00)',':')+1,2))
If you can convert your input to a real date first, the task becomes much easier. Here, I have shamelessly appended the time to a fake date to create a date such as 2017-01-01 00:30. To find out the number of minutes since midnight, you simply subtract the date for "midnight". It will return the difference in days, so you need to multiply by number of minutes per day to get what you want.
select time
,(to_date('2017-01-01 ' || time, 'yyyy-mm-dd hh24:mi') - date '2017-01-01') * 24 * 60 as minutes
from (select '00:30' as time from dual union all
select '08:00' as time from dual union all
select '08:30' as time from dual union all
select '12:00' as time from dual union all
select '23:59' as time from dual
);
Here is some sample input and output
time minutes
==== =======
00:30 30
08:00 480
08:30 510
12:00 720
23:59 1 439
If you require to Print 08:00 hours as 480 minutes,
Extract the Digit before : and multply with 60 and add the digit after :. So you can convert the HH:MM representation in to minutes.
SELECT REGEXP_SUBSTR(ATT.workdur,'[^:]+',1,1)*60 + REGEXP_SUBSTR(ATT.workdur,'[^:]+',1,2) MINUTES FROM DUAL;
I have a table that looks like this:
+--------------------+---------+
| Month (date) | amount |
+--------------------+---------+
| 2016-10-01 | 20 |
| 2016-08-01 | 10 |
| 2016-07-01 | 17 |
+--------------------+---------+
I'm looking for a query (sql statement) which satisfies the following conditions:
Give me the value of the previous month.
If there is no value for the previous month lock back in time until one can be found.
If there is just a value for the current month give me this value.
In the example table the row I'm looking for would be this:
+--------------------+---------+
| 2016-08-01 | 10 |
+--------------------+---------+
Has anyone a idea for a non complex select query?
Thanks in advance,
Peter
You may need the following:
SELECT *
FROM ( SELECT *
FROM test
WHERE TRUNC(SYSDATE, 'month') >= month
ORDER BY CASE
WHEN TRUNC(SYSDATE, 'month') = month
THEN 0 /* if current month, ordered last */
ELSE 1 /* previous months are ordered first */
END DESC,
month DESC /* among previous months, the greatest first */
)
WHERE ROWNUM = 1
Another way using MAX
WITH tbl AS (
SELECT TO_DATE('2016-10-01', 'YYYY-MM-DD') AS "month", 20 AS amount FROM dual
UNION
SELECT TO_DATE('2016-08-01', 'YYYY-MM-DD') AS "month", 10 AS amount FROM dual
UNION
SELECT TO_DATE('2016-07-01', 'YYYY-MM-DD') AS "month", 5 AS amount FROM dual
)
SELECT *
FROM tbl
WHERE TRUNC("month", 'MONTH') = NVL((SELECT MAX(t."month")
FROM tbl t
WHERE t."month" < TRUNC(SYSDATE, 'MONTH')),
TRUNC(SYSDATE, 'MONTH'));
I would use row_number():
select t.*
from (select t.*,
row_number() over (order by (case when to_char(dte, 'YYYY-MM') = to_char(sysdate, 'YYYY-MM') then 1 else 2 end) desc,
dte desc
) as seqnum
from t
) t
where seqnum = 1;
Actually, you don't need row_number() for this:
select t.*
from (select t.*
from t
order by (case when to_char(dte, 'YYYY-MM') = to_char(sysdate, 'YYYY-MM') then 1 else 2 end) desc,
dte desc
) t
where rownum = 1;
It's not the nicest query but it should work.
select amount, date from (
select amount, date, row_number over(partition by HERE_PUT_ID order by
case trunc(date, 'month') when trunc(sysdate, 'month') then to_date('00010101', 'yyyymmdd') else trunc(date, 'month') end
desc) r)
where r = 1;
I guess you have some id in table so put id column instead of HERE_PUT_ID if you want query for whole table just delete: partition by HERE_PUT_ID
I added more data for testing, and an "id" column (a more realistic scenario) to show how this would work. If there is no "id" in your data, simply delete any reference to it from the solution.
Notes - month is a reserved Oracle word, don't use it as a column name. The solution assumes the date column contains dates that are already truncated to the beginning of the month. The trick in "order by" in the dense_rank last is to assign a value (ANY value!) when the month is the current month; by default, the value assigned to all other months is NULL, which by default come after any non-null value in an ascending order.
You may want to test the various solutions for efficiency if execution time is important.
with
inputs ( id, mth, amount ) as (
select 1, date '2016-10-01', 20 from dual union all
select 1, date '2016-08-01', 10 from dual union all
select 1, date '2016-07-01', 17 from dual union all
select 2, date '2016-10-01', 30 from dual union all
select 2, date '2016-09-01', 25 from dual union all
select 3, date '2016-10-01', 20 from dual union all
select 4, date '2016-08-01', 45 from dual union all
select 4, date '2016-06-01', 30 from dual
)
-- end of TEST DATA - the solution (SQL query) is below this line
select id,
max(mth) keep(dense_rank last order by
case when mth = trunc(sysdate, 'mm') then 0 end, mth) as mth,
max(amount) keep(dense_rank last order by
case when mth = trunc(sysdate, 'mm') then 0 end, mth) as amount
from inputs
group by id
order by id -- ORDER BY is optional
;
ID MTH AMOUNT
--- ---------- -------
1 2016-08-01 10
2 2016-09-01 25
3 2016-10-01 20
4 2016-08-01 45
You could sort the data in the direction you want to:
with MyData as
(
SELECT to_date('2016-10-01','YYYY-MM-DD') MY_DATE, 20 AMOUNT FROM DUAL UNION
SELECT to_date('2016-08-01','YYYY-MM-DD') MY_DATE, 10 AMOUNT FROM DUAL UNION
SELECT to_date('2016-07-01','YYYY-MM-DD') MY_DATE, 17 AMOUNT FROM DUAL
),
MyResult AS (
SELECT
D.*
FROM MyData D
ORDER BY
DECODE(
12*TO_CHAR(MY_DATE,'YYYY') + TO_CHAR(MY_DATE,'MM'),
12*TO_CHAR(SYSDATE,'YYYY') + TO_CHAR(SYSDATE,'MM'),
-1,
12*TO_CHAR(MY_DATE,'YYYY') + TO_CHAR(MY_DATE,'MM'))
DESC
)
SELECT * FROM MyResult WHERE RowNum = 1
I have a monthly amount that I need to spread equally over the number of days in the month. The data looks like this:
Month Value
----------- ---------------
01-Jan-2012 100000
01-Feb-2012 121002
01-Mar-2012 123123
01-Apr-2012 118239
I have to spread the Jan amount over 31 days, the Feb amount over 29 days and the March amount over 31 days.
How can I use PL/SQL to find out how many days there are in the month given in the month column?
SELECT CAST(to_char(LAST_DAY(date_column),'dd') AS INT)
FROM table1
Don't use to_char() and stuff when doing arithmetics with dates.
Strings are strings and dates are dates. Please respect the data types and use this instead:
1+trunc(last_day(date_column))-trunc(date_column,'MM')
Indeed, this is correct. It computes the difference between the value of the last day of the month and the value of the first (which is obviously always 1 and therefore we need to add this 1 again).
You must not forget to use the trunc() function if your date columns contains time, because last_day() preserves the time component.
SELECT EXTRACT(DAY FROM LAST_DAY(SYSDATE)) num_of_days FROM dual;
/
SELECT SYSDATE, TO_CHAR(LAST_DAY(SYSDATE), 'DD') num_of_days FROM dual
/
-- Days left in a month --
SELECT SYSDATE, LAST_DAY(SYSDATE) "Last", LAST_DAY(SYSDATE) - SYSDATE "Days left"
FROM DUAL
/
You can add a month and subtract the two dates
SQL> ed
Wrote file afiedt.buf
1 with x as (
2 select date '2012-01-01' dt from dual union all
3 select date '2012-02-01' from dual union all
4 select date '2012-03-01' from dual union all
5 select date '2012-01-31' from dual
6 )
7 select dt, add_months(trunc(dt,'MM'),1) - trunc(dt,'MM')
8* from x
SQL> /
DT ADD_MONTHS(TRUNC(DT,'MM'),1)-TRUNC(DT,'MM')
--------- -------------------------------------------
01-JAN-12 31
01-FEB-12 29
01-MAR-12 31
31-JAN-12 31
select add_months(my_date, 1)-my_date from dual;
SELECT TO_CHAR(LAST_DAY(SYSDATE), 'fmday-mon-rr dd') as LastDayOfMonth
FROM dual;
Use the following Oracle query:
select to_number(to_char(last_day(sysdate),'dd')) TotalDays from dual
Date_Parameter='01-Oct-2017'
select to_number(to_char(last_day('Date_Parameter'),'dd')) TotalDays from dual
I have a SQL query that takes a date parameter (if I were to throw it into a function) and I need to run it on every day of the last year.
How to generate a list of the last 365 days, so I can use straight-up SQL to do this?
Obviously generating a list 0..364 would work, too, since I could always:
SELECT SYSDATE - val FROM (...);
There's no need to use extra large tables or ALL_OBJECTS table:
SELECT TRUNC (SYSDATE - ROWNUM) dt
FROM DUAL CONNECT BY ROWNUM < 366
will do the trick.
Recently I had a similar problem and solved it with this easy query:
SELECT
(to_date(:p_to_date,'DD-MM-YYYY') - level + 1) AS day
FROM
dual
CONNECT BY LEVEL <= (to_date(:p_to_date,'DD-MM-YYYY') - to_date(:p_from_date,'DD-MM-YYYY') + 1);
Example
SELECT
(to_date('01-05-2015','DD-MM-YYYY') - level + 1) AS day
FROM
dual
CONNECT BY LEVEL <= (to_date('01-05-2015','DD-MM-YYYY') - to_date('01-04-2015','DD-MM-YYYY') + 1);
Result
01-05-2015 00:00:00
30-04-2015 00:00:00
29-04-2015 00:00:00
28-04-2015 00:00:00
27-04-2015 00:00:00
26-04-2015 00:00:00
25-04-2015 00:00:00
24-04-2015 00:00:00
23-04-2015 00:00:00
22-04-2015 00:00:00
21-04-2015 00:00:00
20-04-2015 00:00:00
19-04-2015 00:00:00
18-04-2015 00:00:00
17-04-2015 00:00:00
16-04-2015 00:00:00
15-04-2015 00:00:00
14-04-2015 00:00:00
13-04-2015 00:00:00
12-04-2015 00:00:00
11-04-2015 00:00:00
10-04-2015 00:00:00
09-04-2015 00:00:00
08-04-2015 00:00:00
07-04-2015 00:00:00
06-04-2015 00:00:00
05-04-2015 00:00:00
04-04-2015 00:00:00
03-04-2015 00:00:00
02-04-2015 00:00:00
01-04-2015 00:00:00
SELECT (sysdate-365 + (LEVEL -1)) AS DATES
FROM DUAL connect by level <=( sysdate-(sysdate-365))
if a 'from' and a 'to' date is replaced in place of sysdate and sysdate-365, the output will be a range of dates between the from and to date.
A method quite frequently used in Oracle is something like this:
select trunc(sysdate)-rn
from
( select rownum rn
from dual
connect by level <= 365)
/
Personally, if an application has a need for a list of dates then I'd just create a table with them, or create a table with a series of integers up to something ridiculous like one million that can be used for this sort of thing.
Oracle specific, and doesn't rely on pre-existing large tables or complicated system views over data dictionary objects.
SELECT c1 from dual
MODEL DIMENSION BY (1 as rn) MEASURES (sysdate as c1)
RULES ITERATE (365)
(c1[ITERATION_NUMBER]=SYSDATE-ITERATION_NUMBER)
order by 1
About a year and a half too late, but for posterity here is a version for Teradata:
SELECT calendar_date
FROM SYS_CALENDAR.Calendar
WHERE SYS_CALENDAR.Calendar.calendar_date between '2010-01-01' (date) and '2010-01-03' (date)
Date range between 12/31/1996 and 12/31/2020
SELECT dt, to_char(dt, 'MM/DD/YYYY') as date_name,
EXTRACT(year from dt) as year,
EXTRACT(year from fiscal_dt) as fiscal_year,
initcap(to_char(dt, 'MON')) as month,
to_char(dt, 'YYYY') || ' ' || initcap(to_char(dt, 'MON')) as year_month,
to_char(fiscal_dt, 'YYYY') || ' ' || initcap(to_char(dt, 'MON')) as fiscal_year_month,
EXTRACT(year from dt)*100 + EXTRACT(month from dt) as year_month_id,
EXTRACT(year from fiscal_dt)*100 + EXTRACT(month from fiscal_dt) as fiscal_year_month_id,
to_char(dt, 'YYYY') || ' Q' || to_char(dt, 'Q') as quarter,
to_char(fiscal_dt, 'YYYY') || ' Q' || to_char(fiscal_dt, 'Q') as fiscal_quarter
--, EXTRACT(day from dt) as day_of_month, to_char(dt, 'YYYY-WW') as week_of_year, to_char(dt, 'D') as day_of_week
FROM (
SELECT dt, add_months(dt, 6) as fiscal_dt --starts July 1st
FROM (
SELECT TO_DATE('12/31/1996', 'mm/dd/yyyy') + ROWNUM as dt
FROM DUAL CONNECT BY ROWNUM < 366 * 30 --30 years
)
WHERE dt <= TO_DATE('12/31/2020', 'mm/dd/yyyy')
)
Ahahaha, here's a funny way I just came up with to do this:
select SYSDATE - ROWNUM
from shipment_weights sw
where ROWNUM < 365;
where shipment_weights is any large table;
I had the same requirement - I just use this. User enters the number of days by which he/she wants to limit the calendar range to.
SELECT DAY, offset
FROM (SELECT to_char(SYSDATE, 'DD-MON-YYYY') AS DAY, 0 AS offset
FROM DUAL
UNION ALL
SELECT to_char(SYSDATE - rownum, 'DD-MON-YYYY'), rownum
FROM all_objects d)
where offset <= &No_of_days
I use the above result set as driving view in LEFT OUTER JOIN with other views involving tables which have dates.
A week from 6 months back
SELECT (date'2015-08-03' + (LEVEL-1)) AS DATES
FROM DUAL
where ROWNUM < 8
connect by level <= (sysdate-date'2015-08-03');
if you omit ROWNUM you get 50 rows only, independent of the value.
Better late than never. Here's a method that I devised (after reading this post) for returning a list of dates that includes: (a) day 1 of of the current month through today, PLUS (b) all dates for the past two months:
select (sysdate +1 - rownum) dt
from dual
connect by rownum <= (sysdate - add_months(sysdate - extract(day from sysdate),-2));
The "-2" is the number of prior full months of dates to include. For example, on July 10th, this SQL returns a list of all dates from May 1 through July 10 - i.e. two full prior months plus the current partial month.
Another simple way to get 365 days from today would be:
SELECT (TRUNC(sysdate) + (LEVEL-366)) AS DATE_ID
FROM DUAL connect by level <=( (sysdate)-(sysdate-366));
For the fun of it, here's some code that should work in SQL Server, Oracle, or MySQL:
SELECT current_timestamp - CAST(d1.digit + d2.digit + d3.digit as int)
FROM
(
SELECT digit
FROM
(
select '1' as digit
union select '2'
union select '3'
union select '4'
union select '5'
union select '6'
union select '7'
union select '8'
union select '9'
union select '0'
) digits
) d1
CROSS JOIN
(
SELECT digit
FROM
(
select '1' as digit
union select '2'
union select '3'
union select '4'
union select '5'
union select '6'
union select '7'
union select '8'
union select '9'
union select '0'
) digits
) d2
CROSS JOIN
(
SELECT digit
FROM
(
select '1' as digit
union select '2'
union select '3'
union select '4'
union select '5'
union select '6'
union select '7'
union select '8'
union select '9'
union select '0'
) digits
) d3
WHERE CAST(d1.digit + d2.digit + d3.digit as int) < 365
ORDER BY d1.digit, d2.digit, d3.digit -- order not really needed here
Bonus points if you can give me a cross-platform syntax to re-use the digits table.
I do this so often for a scheduling app I work on that I created a pipelined table function. Sometimes I need days, hours or 15 minutes between times. This is not exactly the same function I use, because my code is in a package. However, here, I'm getting days between Jan 1 2020 and Jan 10 2020:
SELECT
days.date_time
FROM
table(between_times(TO_DATE('2020-01-01'),TO_DATE('2020-01-10'),(60*24), 'Y')) days
The pipelined function:
function between_times(i_start_time TIMESTAMP, i_end_time TIMESTAMP, i_interval_in_minutes NUMBER, include_end_time VARCHAR2 := 'N')
RETURN DateTableType PIPELINED
AS
time_counter TIMESTAMP := i_start_time;
BEGIN
IF i_start_time IS NULL OR i_end_time IS NULL or i_start_time > i_end_time OR i_interval_in_minutes IS NULL OR
i_interval_in_minutes <= 0 THEN
RETURN;
END IF;
LOOP
-- by default does not include end time
if (include_end_time = 'Y') THEN
exit when time_counter > i_end_time;
ELSE
exit when time_counter >= i_end_time;
END IF;
pipe row(DateType( time_counter ));
time_counter := time_counter + i_interval_in_minutes/(60*24);
END LOOP;
EXCEPTION WHEN NO_DATA_NEEDED THEN NULL;
END;
WITH Date_Table (Dates, Heading) AS -- Using Oracle SQL
(SELECT TRUNC(SYSDATE) Dates, ' Start' as Heading FROM dual
UNION ALL
SELECT TRUNC(DATES-1) , ' Inside recursion' as Heading FROM Date_Table
WHERE Dates > sysdate-365 ) -- Go back one year
SELECT TO_CHAR(Dates,'MM/DD/YYYY')
FROM Date_Table
ORDER BY Dates DESC;
I don't have the answer to re-use the digits table but here is a code sample that will work at least in SQL server and is a bit faster.
print("code sample");
select top 366 current_timestamp - row_number() over( order by l.A * r.A) as DateValue
from (
select 1 as A union
select 2 union
select 3 union
select 4 union
select 5 union
select 6 union
select 7 union
select 8 union
select 9 union
select 10 union
select 11 union
select 12 union
select 13 union
select 14 union
select 15 union
select 16 union
select 17 union
select 18 union
select 19 union
select 20 union
select 21
) l
cross join (
select 1 as A union
select 2 union
select 3 union
select 4 union
select 5 union
select 6 union
select 7 union
select 8 union
select 9 union
select 10 union
select 11 union
select 12 union
select 13 union
select 14 union
select 15 union
select 16 union
select 17 union
select 18
) r
print("code sample");
This query generates a list of dates 4000 days in the future and 5000 in the past as of today (inspired on http://blogs.x2line.com/al/articles/207.aspx):
SELECT * FROM (SELECT
(CONVERT(SMALLDATETIME, CONVERT(CHAR,GETDATE() ,103)) + 4000 -
n4.num * 1000 -
n3.num * 100 -
n2.num * 10 -
n1.num) AS Date,
year(CONVERT(SMALLDATETIME, CONVERT(CHAR,GETDATE() ,103)) + 4000 -
n4.num * 1000 -
n3.num * 100 -
n2.num * 10 -
n1.num) as Year,
month(CONVERT(SMALLDATETIME, CONVERT(CHAR,GETDATE() ,103)) + 4000 -
n4.num * 1000 -
n3.num * 100 -
n2.num * 10 -
n1.num) as Month,
day(CONVERT(SMALLDATETIME, CONVERT(CHAR,GETDATE() ,103)) + 4000 -
n4.num * 1000 -
n3.num * 100 -
n2.num * 10 -
n1.num) as Day
FROM (SELECT 0 AS num union ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9) n1
,(SELECT 0 AS num UNION ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9) n2
,(SELECT 0 AS num union ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9) n3
,(SELECT 0 AS num UNION ALL
SELECT 1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8) n4
) GenCalendar ORDER BY 1