How is it possible to compare a TIMESTAMP column by date or time?
I try to retrieve records that difference of TIMESTAMP column and NOW are a week or 2 hours or less 30 minutes.
SELECT FROM Tcase WHERE (date_time-datetime('now'))<7day
SELECT FROM Tcase WHERE (date_time-datetime('now'))<1hour
Example
SELECT * FROM Tcase
WHERE datetime(date_time,'-7 day', '+5 hour', '+10 minute') >= date('now')
Edit
SELECT * FROM Tcase
date('now') > datetime(date_time,'+7 day')
OR date('now') < datetime(date_time,'-7 day')
See SQLite date functions
SQLFiddle example
Related
How to get date time difference in PostgreSQL
I am using below syntax
select id, A_column,B_column,
(SELECT count(*) AS count_days_no_weekend
FROM generate_series(B_column ::timestamp , A_column ::timestamp, interval '1 day') the_day
WHERE extract('ISODOW' FROM the_day) < 5) * 24 + DATE_PART('hour', B_column::timestamp-A_column ::timestamp ) as hrs
FROM table req where id='123';
If A_column=2020-05-20 00:00:00 and B_column=2020-05-15 00:00:00 I want to get 72(in hours).
Is there any possibility to skip weekends(Saturday and Sunday) in first one, it means to get the result as 72 hours(exclude weekend hours)
i am getting 0
But i need to get 72 hours
And if If A_column=2020-08-15 12:00:00 and B_column=2020-08-15 00:00:00 I want to get 12(in hours).
One option uses a lateral join and generate_series() to enumerate each and every hour between the two timestamps, while filtering out week-ends:
select t.a_column, t.b_column, h.count_hours_no_weekend
from mytable t
cross join lateral (
select count(*) count_hours_no_weekend
from generate_series(t.b_column::timestamp, t.a_column::timestamp, interval '1 hour') s(col)
where extract('isodow' from s.col) < 5
) h
where id = 123
I would attack this by calculating the weekend hours to let the database deal with daylight savings time. I would then subtract the intervening weekend hours from the difference between the two date values.
with weekend_days as (
select *, date_part('isodow', ddate) as dow
from table1
cross join lateral
generate_series(
date_trunc('day', b_column),
date_trunc('day', a_column),
interval '1 day') as gs(ddate)
where date_part('isodow', ddate) in (6, 7)
), weekend_time as (
select id,
sum(
least(ddate + interval '1 day', a_column) -
greatest(ddate, b_column)
) as we_ival
from weekend_days
group by id
)
select t.id,
a_column - b_column as raw_difference,
coalesce(we_ival, interval '0') as adjustment,
a_column - b_column -
coalesce(we_ival, interval '0') as adj_difference
from weekend_time w
left join table1 t on t.id = w.id;
Working fiddle.
I'm having an issue generating a series of dates and then returning the COUNT of rows matching that each date in the series.
SELECT generate_series(current_date - interval '30 days', current_date, '1 day':: interval) AS i, COUNT(*)
FROM download
WHERE product_uuid = 'someUUID'
AND created_at = i
GROUP BY created_at::date
ORDER BY created_at::date ASC
I want the output to be the number of rows that match the current date in the series.
05-05-2018, 35
05-06-2018, 23
05-07-2018, 0
05-08-2018, 10
...
The schema has the following columns: id, product_uuid, created_at. Any help would be greatly appreciated. I can add more detail if needed.
Put the table generating function in the from and use a join:
SELECT g.dte, COUNT(d.product_uuid)
FROM generate_series(current_date - interval '30 days', current_date, '1 day':: interval
) gs(dte) left join
download d
on d.product_uuid = 'someUUID' AND
d.created_at::date = g.dte
GROUP BY g.dte
ORDER BY g.dte;
I have input: '2017-02-02 11:00:00' and '2017-02-13 15:00:00'
I want to know the difference in hours between these days, but there is a twist: I need to know the hours for each day.
the result would look something like
'2017-02-02', '13'
'2017-02-03', '24'
...
'2017-02-13', '15'
If it was only hours then I would use DATEDIFF, but assuming the twist I have no idea how to start. I would appreciate any idea.
ps: there can be different inputs, for example '2017-02-02 11:00:00' and '2017-02-02 15:00:00'
You can use generate_series to generate all the datetimes with an interval of 1 hour between the start and end dates. Then you can group by the date part to get the number of hours worked each day.
select dt_time::date,count(*) as hours_by_day
from (select generate_series('2017-02-02 11:00:00','2017-02-13 15:00:00',interval '1 hour') as dt_time
) x
group by dt_time::date
order by 1
One way uses generate_series() and some arithmetic:
with dates as (
select gs.dte, $date1 as dte1, $date2 as dte2
from generate_series(date_trunc('day', $date1), date_trunc('day', $date2), interval '1 day') gs(dte)
)
select gs.dte,
(case when dte1 > gs.dte and dte2 < gs.dte then 24
when date_trunc('day', dte2) = gs.dte then 24 - extract(hour from dte1)
else extract(hour from dte2)
end) as hours
from dates;
I am trying to get the max row from the sum of daily counts in a table. I have looked at several posts that look similar, however it doesn't seem to work. I have tried to follow
Get MAX row for GROUP in MySQL
but it doesn't work in Postgres. Here's what I have
select source, SUM(steps) as daily_steps, to_char("endTime"::date, 'MM/DD/YYYY') as step_date
from activities
where user_id = 1
and "endTime" <= CURRENT_TIMESTAMP + INTERVAL '1 day'
and "endTime" >= CURRENT_TIMESTAMP - INTERVAL '7 days'
group by source, to_char("endTime"::date, 'MM/DD/YYYY')
This returns the following
source, daily_steps, step_date
"walking";750;"11/17/2015"
"walking";821;"11/22/2015"
"walking";106;"11/20/2015"
"running";234;"11/21/2015"
"running";600;"11/24/2015"
I would like the result to return only the rows that have the max value for daily_steps by source. The result should look like
source, daily_steps, step_date
"walking";821;"11/22/2015"
"running";600;"11/24/2015"
Postgres offers the convenient distinct on syntax:
select distinct on (a.source) a.*
from (select source, SUM(steps) as daily_steps, to_char("endTime"::date, 'MM/DD/YYYY') as step_date
from activities a
where user_id = 1 and
"endTime" <= CURRENT_TIMESTAMP + INTERVAL '1 day' and
"endTime" >= CURRENT_TIMESTAMP - INTERVAL '7 days'
group by source, to_char("endTime"::date, 'MM/DD/YYYY')
) a
order by a.source, daily_steps desc;
I am looking for a where clause that can be used to retrieve records for the last 24 hours?
In MySQL:
SELECT *
FROM mytable
WHERE record_date >= NOW() - INTERVAL 1 DAY
In SQL Server:
SELECT *
FROM mytable
WHERE record_date >= DATEADD(day, -1, GETDATE())
In Oracle:
SELECT *
FROM mytable
WHERE record_date >= SYSDATE - 1
In PostgreSQL:
SELECT *
FROM mytable
WHERE record_date >= NOW() - '1 day'::INTERVAL
In Redshift:
SELECT *
FROM mytable
WHERE record_date >= GETDATE() - '1 day'::INTERVAL
In SQLite:
SELECT *
FROM mytable
WHERE record_date >= datetime('now','-1 day')
In MS Access:
SELECT *
FROM mytable
WHERE record_date >= (Now - 1)
SELECT *
FROM table_name
WHERE table_name.the_date > DATE_SUB(CURDATE(), INTERVAL 1 DAY)
MySQL :
SELECT *
FROM table_name
WHERE table_name.the_date > DATE_SUB(NOW(), INTERVAL 24 HOUR)
The INTERVAL can be in YEAR, MONTH, DAY, HOUR, MINUTE, SECOND
For example, In the last 10 minutes
SELECT *
FROM table_name
WHERE table_name.the_date > DATE_SUB(NOW(), INTERVAL 10 MINUTE)
Which SQL was not specified, SQL 2005 / 2008
SELECT yourfields from yourTable WHERE yourfieldWithDate > dateadd(dd,-1,getdate())
If you are on the 2008 increased accuracy date types, then use the new sysdatetime() function instead, equally if using UTC times internally swap to the UTC calls.
in postgres, assuming your field type is a timestamp:
select * from table where date_field > (now() - interval '24 hour');
If the timestamp considered is a UNIX timestamp
You need to first convert UNIX timestamp (e.g 1462567865) to mysql timestamp or data
SELECT * FROM `orders` WHERE FROM_UNIXTIME(order_ts) > DATE_SUB(CURDATE(), INTERVAL 1 DAY)
select ...
from ...
where YourDateColumn >= getdate()-1
SELECT *
FROM tableName
WHERE datecolumn >= dateadd(hour,-24,getdate())
Hello i now it past a lot of time from the original post but i got a similar problem and i want to share.
I got a datetime field with this format YYYY-MM-DD hh:mm:ss, and i want to access a whole day, so here is my solution.
The function DATE(), in MySQL: Extract the date part of a date or datetime expression.
SELECT * FROM `your_table` WHERE DATE(`your_datatime_field`)='2017-10-09'
with this i get all the row register in this day.
I hope its help anyone.
In SQL Server (For last 24 hours):
SELECT *
FROM mytable
WHERE order_date > DateAdd(DAY, -1, GETDATE()) and order_date<=GETDATE()
In Oracle (For last 24 hours):
SELECT *
FROM my_table
WHERE date_column >= SYSDATE - 24/24
In case, for any reason, you have rows with future dates, you can use between, like this:
SELECT *
FROM my_table
WHERE date_column BETWEEN (SYSDATE - 24/24) AND SYSDATE