SQL simple selection of rows according to their time - sql

I have a table with measures and the time this measures have been taken in the following form: MM/DD/YYYY HH:MI:SS AM. I have measures over many days starting at the same time every day.The datas are minute by minute so basically the seconds are always = 0. I want to select only the measures for the first 5 minutes of each day. I would have used the where statement but the condition would only be on the minutes and note the date is there a way to do this?
Thanks

You could try something like this:
SELECT * FROM SomeTable
WHERE
DATEPART(hh, timestamp_col) = 0 AND -- filter for first hour of the day
DATEPART(mm, timestamp_col) <= 5 -- filter for the first five minutes
Careful! 0 means midnight. If your "first hour" of the day is actually 8 or 9 AM then you should replace the 0 with an 8 or 9.

Related

How to calculate the time difference in SQL with DATEDIFF?

I am using the DATEDIFF function to calculate the difference between my two timestamps.
payment_time = 2021-10-29 07:06:32.097332
trigger_time = 2021-10-10 14:11:13
What I have written is : date_diff('minute',payment_time,trigger_time) <= 15
I basically want the count of users who paid within 15 mins of the triggered time
thus I have also done count(s.user_id) as count
However it returns count as 1 even in the above case since the minutes are within 15 but the dates 10th October and 29th October are 19 days apart and hence it should return 0 or not count this row in my query.
How do I compare the dates in my both columns and then count users who have paid within 15 mins?
This also works to calculate minutes between to timestamps (it first finds the interval (subtraction), and then converts that to seconds (extracting EPOCH), and divides by 60:
extract(epoch from (payment_time-trigger_time))/60
In PostgreSQL, I prefer to subtract the two timestamps from each other, and extract the epoch from the resulting interval:
Like here:
WITH
indata(payment_time,trigger_time) AS (
SELECT TIMESTAMP '2021-10-29 07:06:32.097332',TIMESTAMP '2021-10-10 14:11:13'
UNION ALL SELECT TIMESTAMP '2021-10-29 00:00:14' ,TIMESTAMP '2021-10-29 00:00:00'
)
SELECT
EXTRACT(EPOCH FROM payment_time-trigger_time) AS epdiff
, (EXTRACT(EPOCH FROM payment_time-trigger_time) <= 15) AS filter_matches
FROM indata;
-- out epdiff | filter_matches
-- out ----------------+----------------
-- out 1616119.097332 | false
-- out 14.000000 | true

Sql query to return data if AVG(7 days record count) > (Today's record count)

I want to write a SQL query in Oracle database for:
A priceindex(field name) have around 120(say) records each day and I have to display the priceindex name and today's date, if the avg of last 7 days record count is greater than Todays record count for the priceindex(group by priceindex).
Basically, There will be 56 priceindex and each should have around 120 records each day and is dump to database each day from external site. So want to make sure all records are downloaded to the database everyday.
Except for the clarification I requested in a Comment to your question (having to do with "how can we know today's final count, when today is not over yet), the problem can be solved along the following lines. Not tested since you didn't provide sample data.
From your table, select only the rows where the relevant DATE is between "today" - 7 and "today" (so there are really EIGHT days: the seven days preceding today, and today). Then group by PRICEINDEX. Count total rows for each group, and count rows just for "today". The rows for "today" should be less than 1/8 times the total count (this is easy algebra: this is equivalent to being less than 1/7 times the count of OTHER days).
Such conditions, at the group level, must be in the HAVING clause.
select priceindex
from your_table
where datefield >= trunc(sysdate) - 7 and datefield < trunc(sysdate) + 1
group by priceindex
having count(case when datefield >= trunc(sysdate) then 1 end) < 1/8 * count(*)
;
EDIT The OP clarified that the query runs every day at midnight; this means that "today" should actually mean "yesterday" (the day that just ended). In Oracle, and probably in all of computing, midnight belongs to the day that BEGINS at midnight, not the one that ends at midnight. The time-of-day at midnight is 00:00:00 (beginning of the new day), not 24:00:00.
So, the query above will have to be changed slightly:
select priceindex
from your_table
where datefield >= trunc(sysdate) - 8 and datefield < trunc(sysdate)
group by priceindex
having count(case when datefield >= trunc(sysdate) - 1 then 1 end)
< 1/8 * count(*)
;

Optimization: How to get TimeId from time for each minute in a week?

I am creating a table which will have 2 columns:
Day_time (time from 1978-01-01 00:00:00 Sunday, till 1978-01-07 23:59:00.0 Saturday, Granularity: Minute)
Time_id (a unique id for each minute), to be populated
I have column one populated. I want to populate column two.
How I am doing it right now:
EXTRACT(dayofweek FROM day_time) * 10000 + DATEDIFF('minutes', TRUNC(day_time), day_time)
I basically want a function where I pass any date and it tells me where I am in a week. So, I need a function, just like the function above. Just more optimized, where I give a date and get a unique ID. The unique ID should repeat weekly.
Example: ID for Jan 1, 2015 00:00:00 will be same as Jan 8, 2015 00:00:00.
Why 1978-01-01? cuz it starts from a Sunday.
Why 10,000? cuz the number of minutes in a day are in four digits.
You can do it all in one fell swoop, without needing to extract the date separately:
SELECT DATEDIFF('minutes', date_trunc('week',day_time), day_time) which I'd expect to be marginally faster.
Another approach that I'd expect to be significantly faster would be converting the timestamp to epoch, dividing by 60 to get minutes from epoch and then taking the value modulus of 10,080 (for 60 * 24 * 7 minutes in a week).
SELECT (extract(epoch from day_time) / 60) % 10080
If you don't care about the size of the weekly index, you could also do:
SELECT (extract(epoch from day_time)) % 604800 and skip the division step altogether, which should make it faster still.

Updating dates with random time

I have a datetime column, all of them at 12:00 am. Is there a way to update them with random hours, minutes to nearest 1/2 hour while keeping the same date(day) value?
Update Activities set ActivityDate = ....
Here's one option using dateadd:
update Activities
set ActivityDate = DateAdd(minute,
30 * (abs(checksum(NewId())) % 47), ActivityDate);
SQL Fiddle Demo
And here's a good post about generating random numbers. Using that, multiple by 30 minutes to get to the nearest half hour.
Note, this uses % 47 since there are 1440 minutes in a day -- that divides into 48 potential half hour segments in that same day.

date_trunc 5 minute interval in PostgreSQL [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the fastest way to truncate timestamps to 5 minutes in Postgres?
Postgresql SQL GROUP BY time interval with arbitrary accuracy (down to milli seconds)
I want to aggregate data at 5 minute intervals in PostgreSQL. If I use the date_trunc() function, I can aggregate data at an hourly, monthly, daily, weekly, etc. interval but not a specific interval like 5 minute or 5 days.
select date_trunc('hour', date1), count(*) from table1 group by 1;
How can we achieve this in PostgreSQL?
SELECT date_trunc('hour', date1) AS hour_stump
, (extract(minute FROM date1)::int / 5) AS min5_slot
, count(*)
FROM table1
GROUP BY 1, 2
ORDER BY 1, 2;
You could GROUP BY two columns: a timestamp truncated to the hour and a 5-minute-slot.
The example produces slots 0 - 11. Add 1 if you prefer 1 - 12.
I cast the result of extract() to integer, so the division / 5 truncates fractional digits. The result:
minute 0 - 4 -> slot 0
minute 5 - 9 -> slot 1
etc.
This query only returns values for those 5-minute slots where values are found. If you want a value for every slot or if you want a running sum over 5-minute slots, consider this related answer:
PostgreSQL: running count of rows for a query 'by minute'
Here's a simple query you can either wrap in a function or cut and paste all over the place:
select now()::timestamp(0), (extract(epoch from now()::timestamptz(0)-date_trunc('d',now()))::int)/60;
It'll give you the current time, and a number from 0 to the n-1 where n=60 here. To make it every 5 minutes, make that number 300 and so on. It groups by the seconds since the start of the day. To make it group by seconds since year begin, hour begin, or whatever else, change the 'd' in the date_trunc.