How can I subtract time in MySQL? For example, today is 16 March; I want to subtract 15 days to reach 1 March. Are there any methods that can be used to subtract 15 days from the current date?
SELECT DATE(NOW()-INTERVAL 15 DAY)
For a list of units see http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_date-add
Not entirely related to this question but is related to the title:
SELECT SUBTIME("10:24:21", "5"); -- subtracts 5 seconds. (returns "10:24:16")
SELECT SUBTIME("10:24:21", "01:00:00"); -- subtracts one hour. (returns "09:24:21")
Documentation: MySQL SUBTIME function
Use:
SELECT NOW() - INTERVAL 15 DAY
to keep the datetime precision.
You can use this :
SELECT DATE(NOW()-INTERVAL 15 DAY);
for when you want to subtract the number of days.
In order to subtract the time instead, say 15 minutes, the following should work:
SELECT(DATE_SUB(NOW(), INTERVAL '15:0' MINUTE_SECOND));
Adding the reference link again :- https://dev.mysql.com/doc/refman/8.0/en/date-and-time-functions.html#function_date-add.
Yes its possible using date function in Mysql
select distinct
lastname,
changedat, date_add(changedat, interval -15 day) as newdate
from employee_audit;
lastname and changedat is field name and employee_audit is table name.
I have subtract 15 days from my date - check image please. thanks
Related
I'm trying to subtract now() - the created datetime field from 30 days to get the days remaining as a datetime field, mysql gives me an error for this sort of thing.
SELECT id, created, INTERVAL 30 DAY - CURRENT_DATE - created as timeleft FROM tablename
Use the Date add function to subtract 30 days like this.
SELECT DATE_ADD(current_date, interval -30 day);
Return the difference in days between two date values:
SELECT DATEDIFF(current_date, "2017-06-15");
You can use the combination of these functions to achieve the desired result.
In mysql you should translate the datetime to unix_timestamp to calculate the difference between the two days
SELECT id, created, 30 - (unix_timestamp(CURRENT_DATE) - unix_timestamp(created )) / 3600/24 as timeleft FROM tablename
I've got a column time type in mysql, i want to add 24 hours to this hour, i try with code below:
SELECT SUBSTRING(CAST(DATE_ADD(STR_TO_DATE('23:00', '%k:%i'), INTERVAL (TIME_TO_SEC('24:00') / 60) MINUTE) AS CHAR(8)), 1,5)
I need to return 23:00 again (because i add 24 hours but that 23:00 is of the next day) but this code return me 47:00.
Some help?
I think you're looking for ADDTIME
SELECT ADDTIME(‘23:00’,’24:00’)
https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_addtime
You need to first convert the column value to a datetime, add the hours, and then convert the result back to time:
select cast(date_add(cast(time_column as datetime), interval 24 hour ) as time)
from yourtable;
Example:
select cast(date_add(cast(cast('23:00' as time) as datetime), interval 24 hour ) as time)
The TIME type allows you to store up to (but not including) 839 hours (positive and negative). That's great if you need to store duration, but not so much if you want to store time of day. If you want the latter you should consider the DATETIME type instead.
This query is selecting rows applied last 30 days:
SELECT `amount` FROM `mg_inputs` WHERE `amount`<0 AND `product`='144' AND DATE(firstedit) BETWEEN CURDATE() - INTERVAL 30 DAY AND CURDATE()
How about queries selecting rows applied last 30 days, but 1 month ago (between: today-30days and today-60days)? Same question goes for "2 months ago". Nothing is working for me at all (SQL is returning errors).
One important thing to note here is that not all months are 30 days, so instead of using INTERVAL DAY use INTERVAL MONTH.
Next, you don't need to use the subtraction sign for dates, you can use the DATE_SUB() function which will do what you need.
Last, keeping those things in mind, you can use the BETWEEN operator to check for rows within a date range. So, for example, if you want all rows from one month ago, try this:
SELECT *
FROM myTable
WHERE dateColumn BETWEEN DATE_SUB(CURDATE(), INTERVAL 2 MONTH) AND DATE_SUB(CURDATE(), INTERVAL 1 MONTH);
You should note that for the BETWEEN operator to work properly, the older date must appear first. Here is an SQL Fiddle example that demonstrates that.
I have a table from which I'm trying to extracted summed timediff information grouped by days. I don't really know if this is possible
Table columns: mode_type, start_time.
A record exists in this table for each time an employee starts or stops a timer. mode_type = 1 for start, mode_type = 0 for stop.
I'd like to return a sum of the seconds used for each day in the last 30 days.
E.g:
date, seconds_used
02/04/2014, 25
03/04/2014, 12415
04/04/2014, 925
Currently I can return a list of seconds used per mode_type and date but this required later calc in PHP.
SELECT
mode_type,
Sum(Unix_Timestamp(start_time)) AS time,
start_time
FROM
activations
WHERE
start_time < Date(Now() + INTERVAL 1 MONTH)
GROUP BY
mode_type, Day(start_time)
ORDER BY
start_time
I'm stuck... is this possible or do I need to do revert to calculating the diff in PHP post request?
Thanks in advance.
Can you try with this:
SELECT DATE(start_time) AS startdate, TIME_TO_SEC(TIMEDIFF(NOW(),start_time)) AS secs
FROM activations
GROUP BY
startdate
I am trying to get one week earlier then current week of the year but my sql query is returning null. here is my query
select date_sub(yearweek('2014-01-01'),INTERVAL 1 week)
what is wrong with this query
If you want to get YEARWEEK of week prior to date, you can do this:
Note: YEARWEEK results in 6-digit number, first 4 digits are week year, trailing 2 digits are week number.
SELECT YEARWEEK('2014-01-01' - INTERVAL 1 WEEK)
If you need to get a date that is one week before a given date, then:
SELECT '2014-01-01' - INTERVAL 1 WEEK
Try this
select date_sub(date('2014-01-01'),INTERVAL 1 week)
Try this:-
DATE_SUB(date('2014-01-01'), INTERVAL 7 DAY)
or
SELECT '2014-01-01' - INTERVAL 1 WEEK
The problem is that DATE_SUB takes a date as the first arguement, but year week returns yyyyww i.e. not a date. So this:
SELECT yearweek('2014-01-01');
Returns 201352, this is then implicitly casted to a date, since it is not a date the result is null. You can replicate this by doing:
SELECT DATE(yearweek('2014-01-01'));
So if you subtract a week from NULL the result is also NULL.
The fix is to subtract the week first, then get the year week:
SELECT yearweek(date_sub('2014-01-01', INTERVAL 1 WEEK));
And the result is 201351.
Are you looking for week number??? If yes then plz try this if it will work for you
Select DatePart(Week, Date add(day,-7,convert(date time,'01-jan-2014')))
Pleas let me know if you are looking for something else.
SELECT * FROM [table] WHERE WEEKOFYEAR(date) = WEEKOFYEAR(NOW()) - 1;