I'm using Oracle SQL Developer and I would like to make a SELECT that brings all the products sold from the last day of the previous month (but only products sold from 4 pm on the last day of the previous month) to the current day (only until 8 am of the current day).
For example, today is 7/21/2022. If I run this query today, it should bring data from:
06/30/2022 above 16:00hrs -> 07/21/2022 until 08:00hrs
You can use TRUNC(value, 'MM') to find midnight of the 1st day of the current month and then subtract 8 hours to find the start of the range and then use TRUNC(value) to find midnight of today and add 8 hours to find the end of the range:
SELECT *
FROM table_name
WHERE date_column >= TRUNC(SYSDATE, 'MM') - INTERVAL '8' HOUR
AND date_column <= TRUNC(SYSDATE) + INTERVAL '8' HOUR;
You can use TRUNC to get to the first day of a date's month. Then subtract one day and add sixteen hours. And it's again TRUNC that you use to get back to the beginning of a day (midnight) to which you can add eight hours.
select *
from mytable
where dt >= trunc(sysdate, 'mm') - interval '1' day + interval '16' hour
and dt < trunc(sysdate, 'dd') + interval '8' hour
order by dt;
Related
So I've been trying to fetch some daily data with SYSDATE on a date type YYYYMMDD as following:
SELECT dates, trunc(calendar_date, 'DD') calendar_dates, weekday_nbr
FROM db.date
WHERE dates BETWEEN to_char(TRUNC(SYSDATE)-2, 'YYYYMMDD') AND to_char(TRUNC(SYSDATE)-1, 'YYYYMMDD')
But now I'm trying to use the same but on a YYYY+MM+Week date type with not much success
I tried using:
SELECT T time, period, fiscal_week
FROM db.time
WHERE time BETWEEN to_char(TRUNC(SYSDATE)-2, 'W') AND to_char(TRUNC(SYSDATE)-1,'W')
With time as a 7 digit number, and period and fiscal week as a 2 digit number
Knowing that there's no way I can truncate such date type, how can TRUNC SYSDATE YYYY+MM+Week to get the data on the last 2 weeks?
Also I was thinking about maybe getting the totals from a set day and then dropping all but the last 2 weeks, but on the long run maybe that would be time consuming.
Knowing that there's no way I can truncate such date type, how can TRUNC SYSDATE YYYY+MM+Week to get the data on the last 2 weeks?
Assuming that your fiscal weeks are from Monday-Sunday then you can truncate to the start of the ISO week (which is always Midnight on Monday) and use that for the basis of the comparison:
SELECT *
FROM db.time
WHERE dates >= TRUNC(SYSDATE, 'IW') - INTERVAL '14' DAY
AND dates < TRUNC(SYSDATE, 'IW')
If you have a column that is for weeks then you should still use a DATE data type and add a CHECK constraint (and can use virtual columns to generate the week and the year):
CREATE TABLE time (
dt DATE
CHECK (dt = TRUNC(dt, 'IW')),
year NUMBER(4,0)
GENERATED ALWAYS AS (EXTRACT(YEAR FROM dt)),
month NUMBER(2,0)
GENERATED ALWAYS AS (EXTRACT(MONTH FROM dt)),
week NUMBER(1,0)
GENERATED ALWAYS AS (FLOOR((dt - TRUNC(dt, 'MM'))/7) + 1),
time VARCHAR2(7)
GENERATED ALWAYS AS (
CAST(
TO_CHAR(dt, 'YYYYMM')
|| (FLOOR((dt - TRUNC(dt, 'MM'))/7) + 1)
AS VARCHAR2(7)
)
)
-- ...
);
fiddle
Then you can use the logic above on the date column.
If you do not have a DATE column then you will need to convert your YYYYMMW number into a DATE and then use the logic above.
For example, if the logic for your fiscal weeks (which you have not described) is that the first week of each month starts on the first Monday of the month then you can convert the YYYYMMW number to a DATE using:
SELECT NEXT_DAY(
TO_DATE(SUBSTR(time, 1, 6), 'YYYYMM') - INTERVAL '1' DAY,
'MONDAY'
) + INTERVAL '7' DAY * (SUBSTR(time, 7, 1) - 1) AS week_start
FROM db.time
and then could use it to filter the table using:
SELECT *
FROM (
SELECT t.*,
NEXT_DAY(
TO_DATE(SUBSTR(time, 1, 6), 'YYYYMM') - INTERVAL '1' DAY,
'MONDAY'
) + INTERVAL '7' DAY * (SUBSTR(time, 7, 1) - 1) AS week_start
FROM db.time t
)
WHERE week_start >= TRUNC(SYSDATE, 'IW') - INTERVAL '14' DAY
AND week_start < TRUNC(SYSDATE, 'IW')
If you have different logic for calculating when fiscal weeks start then you will need to apply that logic to the conversion.
I want to get data from last month day by day, I can get the last 30 days but I just want the month as it may be less or more than 30 days,
this is the query for getting the last 30 days
SELECT Trunc(timestamp),
Count(*)
FROM table1
WHERE Trunc(timestamp) > Trunc(sysdate - 30)
GROUP BY Trunc(timestamp)
ORDER BY 1;
Also, I am using it in a shell script if I can make a variable in the script and put it the query
To get data from the start of the current month until today:
SELECT TRUNC(timestamp) AS day,
COUNT(*)
FROM table1
WHERE timestamp >= TRUNC(SYSDATE, 'MM')
AND timestamp < TRUNC(SYSDATE) + INTERVAL '1' DAY
GROUP BY TRUNC(timestamp)
ORDER BY day
To get data from the same day last month until today:
SELECT TRUNC(timestamp) AS day,
COUNT(*)
FROM table1
WHERE timestamp >= ADD_MONTHS(TRUNC(SYSDATE), -1)
AND timestamp < TRUNC(SYSDATE) + INTERVAL '1' DAY
GROUP BY TRUNC(timestamp)
ORDER BY day
db<>fiddle here
I am using Oracle SQL developer. I am trying to select rows where max date is less than December 31st of the previous year. I tried this query below but kept getting an error.
Select ORG_ID, STS_DTE
from Table1
Group By RC_ORG_ID
Having MAX(STS_DTE)< '31-Dec-' || extract (year from SYSDATE-1)
I would use trunc():
having MAX(STS_DTE) < trunc(sysdate, 'YYYY') - interval '1 day'
I'm not sure if you want sysdate - interval '1' day.
I am currently trying to figure out the equivalent functions in PostgresSQL for the below MariaDB functions:
SELECT
CONCAT(DATE_ADD(DATE_ADD(LAST_DAY('2021-02-15 00:00:00'),INTERVAL 1 DAY),INTERVAL -1 MONTH), ' ', '00:00:00') AS first_day_of_the_month,
CONCAT(LAST_DAY('2021-02-15 23:59:59'), ' ' ,'23:59:59') AS last_day_of_the_month;
Expected Result:
first_day_of_the_month | last_day_of_the_month
------------------------|-------------------------------
2021-02-01 00:00:00 | 2021-02-28 23:59:59
I want to get the last day and the first day of the month of a certain date including the time 00:00:00 for the first day and 23:549:59 for the last day.
So far I am comming close to the expected result with these queries:
SELECT
(date_trunc('MONTH', '2021-02-15 00:00:00'::TIMESTAMP) + INTERVAL '1 MONTH - 1 day')::TIMESTAMP
+ '1 days'::INTERVAL - '1 months'::INTERVAL AS first_day_of_the_month,
(date_trunc('MONTH', '2021-02-15 23:59:59'::TIMESTAMP) + INTERVAL '1 MONTH - 1 day')::TIMESTAMP AS last_day_of_the_month;
However, as you can see in the DB-Fiddle for the last_day_of_the_month I am getting 00:00:00 instead of 23:59:59.
How do I need to change the query to get the correct timestamp?
demo:db<>fiddle
/* Option 1: Certain Date */
SELECT
date_trunc('month', timestamp '2021-02-15 00:00:00') AS first_timestamp,
date_trunc('month', timestamp '2021-02-15 23:59:59') + interval '1 month - 1 second' AS last_timestamp;
/* Option 2: Current_date */
SELECT
date_trunc('month', current_date)::timestamp AS first_timestamp,
date_trunc('month', current_date)::timestamp + interval '1 month - 1 second' AS last_timestamp;
First timestamp of month
date_trunc('month', ...) normalizes the date to the first possible timestamp of the month. So it returns 00:00:00 of the first day of the month
Last timestamp of month
If you add a month to the previous result, you get the first day/timestamp (00:00:00) of the next month. If you subtract a day from this, you get the last day of the current month. Of course, you can subtract just a second to get 23:59:59 of the last day of the month as you expected.
For the first day of the month:
date_trunc('month', the_date_column)
For the last day:
date_trunc('month', the_date_column) + interval '1 month' - interval '1 day'
Or if you prefer:
date_trunc('month', date) + interval '1 month - 1 day'
If you really one one second from midnight on the last day, replace day with second in the logic. I don't recommend that. In fact, if you want to define a range of date, define an open range and just use the first day of the next month. Then you can learn about tsrange which builds this into the language.
How do I get the last day of the month to be 11:59:59 pm? Right now it is showing 05/31/2019 12:00 AM. If I just do sysdate it just shows the time I ran it. Thanks.
LAST_DAY(TRUNC(SYSDATE,'MM'))
Try adding one day to the last day of the current month, then subtracting one minute:
SELECT
LAST_DAY(TRUNC(SYSDATE,'MM')) + INTERVAL '1' DAY - INTERVAL '1' SECOND
FROM dual;
Or, if you want to also see the time component, use CURRENT_TIMESTAMP in place of SYSDATE:
SELECT
LAST_DAY(TRUNC(CURRENT_TIMESTAMP,'MM')) + INTERVAL '1' DAY - INTERVAL '1' SECOND
FROM dual;
This outputted:
30.06.2019 23:59:59
The oldfashioned way would be to subtract a second (that's what 1 / (24*60*60) represents as there are 24 hours in a day, 60 minutes in an hour and 60 seconds in a minute) from the first day of the next month (and that's what trunc(add_months(sysdate, 1)) does):
SQL> select sysdate today,
2 trunc(add_months(sysdate, 1)) - 1/(24*60*60) result
3 from dual;
TODAY RESULT
---------------------- ----------------------
06/01/2019 07:52:40 AM 06/30/2019 11:59:59 PM
SQL>
I would use:
SELECT TRUNC(SYSDATE, 'MM') + INTERVAL '1' MONTH - INTERVAL '1' SECOND
FROM dual;
(This has one less step than Tim's solution.)
Or in the older-fashioned method:
SELECT ADD_MONTHS(TRUNC(SYSDATE, 'MM'), 1) - 1/(24*60*60)
If you convert this to a timestamp, you will see the time value:
SELECT CAST(TRUNC(sysdate, 'MM') + INTERVAL '1' MONTH - INTERVAL '1' SECOND as TIMESTAMP)
FROM dual;