Is there a way in a MySQL statement to order records (through a date stamp) by >= NOW() -1 so all records from the day before today to the future are selected?
Judging by the documentation for date/time functions, you should be able to do something like:
SELECT * FROM FOO
WHERE MY_DATE_FIELD >= NOW() - INTERVAL 1 DAY
Be aware that the result may be slightly different than you expect.
NOW()
returns a DATETIME
.
And INTERVAL
works as named, e.g. INTERVAL 1 DAY = 24 hours
.
So if your script is cron'd to run at 03:00
, it will miss the first three hours of records from the 'oldest' day
.
To get the whole day use CURDATE() - INTERVAL 1 DAY
. This will get back to the beginning of the previous day regardless of when the script is run.
Didn't see any answers correctly using DATE_ADD
or DATE_SUB
:
Subtract 1 day from NOW()
...WHERE DATE_FIELD >= DATE_SUB(NOW(), INTERVAL 1 DAY)
Add 1 day from NOW()
...WHERE DATE_FIELD >= DATE_ADD(NOW(), INTERVAL 1 DAY)
Sure you can:
SELECT * FROM table
WHERE DateStamp > DATE_ADD(NOW(), INTERVAL -1 DAY)
DATE_ADD
then -1 day
. –
Monicamonie when search field is timestamp and you want find records from 0 hours yesterday and 0 hour today use construction
MY_DATE_TIME_FIELD between makedate(year(now()), date_format(now(),'%j')-1) and makedate(year(now()), date_format(now(),'%j'))
instead
now() - interval 1 day
try this: SELECT * FROM FOO WHERE MY_DATE_FIELD >= NOW() - INTERVAL '1' DAY
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