Accidentally the server date was set to a wrong value for 1 day. You have correctly set the date now. But you want to change all the date entries made on that day. Write a query to change the day by 1 of all dates in the query table on 31st Jan 2013.
I have written the following query
UPDATE query SET date= DATE_ADD(date, INTERVAL 1 DAY)
WHERE date="2013-01-31";
but it is not executing correctly
Your code is almost correct. But you need to do following change.
UPDATE query SET date= DATE_ADD(date, INTERVAL 1 DAY) WHERE date(date)="2013-01-31";
I hope this works for you.
Related
So this has been the biggest pita... My db is in UTC and I am trying to create a select statement that accurately displays the count of newly created customers today in my TZ. The first interval works as intended, but when the clock strikes 5pm PST (UTC -7) and the db switches over to the next day, the second part after the and clause fails, and I am left with an empty set. Tweeked the hell out of the second part but have had no success. Any help would be most appreciated!
SELECT count(*) FROM product36.associate where date(created_date - interval 7 hour) >= curdate() and date(created_date + interval 17 hour) >=curdate();
I to have this
Date
29/03/2018
+2
Due
31/03/2018
I am using this code , very simple and it works if i add seconds and hours, but if i go above 1 day the due column goes to 00:00:000
$sq2= "UPDATE Tickets SET Tickets.due= date(`date`)+20000;
( this currently adds 2 hours..)
how do i get past 1 day and have it work correctly?
If you want a day you can use date_add
UPDATE Tickets
SET Tickets.due= DATE(DATE_ADD(`date`, INTERVAL 1 DAY))
if you need minute hour and minute too the don't use date but
UPDATE Tickets
SET Tickets.due= DATE_ADD(`date`, INTERVAL 1 DAY)
I've a table in a db with some date field with format yyyy-mm-dd
I'm trying to perform a query that take just records with a interval of 3 month from today.
I've done like this
WHERE DATE_SUB(myTable.myField, INTERVAL 3 MONTH) = CURDATE()
and it works, but my second step is ignore years of my date field and from curdate().
I've tried EXTRACT or DATEFORMAT, but query doesn't work with those function.
How can I modify my query?
Thanks
The condition is wrong.
Try this instead:
...WHERE myTable.myField >= CURDATE() - INTERVAL 3 MONTH...
EDIT:
Based on your comment:
with my query i've got all record that have in dateField this date
'2016-12-07' (curdate() is today '2016-09-07') and it's fine. but i
want that query gives me also date that have 12 on month and 07 on
day, ignoring year. Eg. if i have '2016-12-07' and '2014-12-07', my
query must give me both records. it's a query that will run every day
...WHERE DATE_FORMAT(myTable.myField,'%m-%d') =
DATE_FORMAT((CURDATE() + INTERVAL 3 MONTH),'%m-%d')...
Use below condition which will find records with a interval of 3 month from today.
WHERE myTable.myField = DATE_SUB(CURDATE(), INTERVAL 3 MONTH)
SELECT DATE(STR_TO_DATE(CONCAT(CONCAT(YEAR('$uDate1'), week), ' Monday'), '%X%V %W') +
INTERVAL (7 - DAYOFWEEK(STR_TO_DATE(CONCAT(CONCAT(YEAR('$uDate1'), week), ' Monday'),
'%X%V %W'))) DAY)
as week_end_date
What this statement does is take the date I give it ($uDate1) and give me the week end date (Saturday) of that week. This works well and I am happy with it, kinda.
I was wondering if there were some things I missed that would either make this more efficient or even if I missed some shortcuts to this.
Any suggestions for me?
week >= WEEK('$uDate1') AND week <= WEEK('$uDate2')
This is in my WHERE clause. So basically if I use this...
DATE('$uDate1', INTERVAL 7 - DAYOFWEEK('$uDate1') DAY)
...then it returns the same day for all records. I need it to be able to go over a span of a few weeks.
I have a column in my database named 'week'. It simply stores an INT that corresponds to the week of the year. (ex. 21 for this week)
I then have two date picker boxes. The output gets the week end date based of each week that is BETWEEN and INCLUDES the days chosen.
5/10/2016 & 5/26/2016 outputs 5/14/2016, 5/21/2016, 5/28/2016
What gets exported to CSV file looks something like this..
WEEK END, LAST NAME, FIRST NAME, ...
5/10/2016, Smith, John, ...
5/26/2016, Jones, James, ...
It outputs anyone who had hours during the week, with the week end date.
SIDE NOTE: I do appreciate the comments and help. I don't want anyone to stress over this though! Just curious if better way. :)
I am not sure why your current SQL is so complicated.
You say it is just to take a date and give me the week end date (Saturday) of that week .
How you are doing this at the moment is:-
Yours is taking the year
Adding the week of the year (I assume - should be WEEK('$uDate1') I think)
Adding on the day as a string (so for example for today it would be 2016 21 Monday )
Changing that string back to a date a datetime value
Converting that datetime value back to a date.
Then taking the year again
Adding the week of the year again
Getting the day of the week of the resulting string. As you have concatenated Monday on to the date then the day of the week will always be 2.
Taking that resulting day of the week and subtracting it from 7. As the day of the week will always be 2 this will always result in 5
Adding on the day as a string (so for example for today it would be 2016 21 Monday ).
This value is then added on to the previously calculated date, taking the Monday date and adding 5 days.
My suggestion was to just use:-
DATE_ADD($uDate1, INTERVAL 7 - DAYOFWEEK($uDate1) DAY)
which is far simpler, and appears to cover your requirements.
EDIT
Looking at your edit you want a list of all the Saturdays for weeks all or partially between 2 passed dates.
If so I think the following will do it and hopefully be more efficient as there is no need to translate dates to and from string. Note it relies on your week table to add to the date, hence only copes with date ranges of up to that many weeks.
SELECT DATE_ADD(DATE_ADD('$uDate1', INTERVAL 7 - DAYOFWEEK('$uDate1') DAY), INTERVAL `week` WEEK) AS aDate
FROM `week`
HAVING aDate BETWEEN '$uDate1' AND DATE_ADD('$uDate2', INTERVAL 7 - DAYOFWEEK('$uDate2') DAY)
ORDER BY aDate
As I mentioned in comment you should move this transformation from mysql query to php code.
I see no reason to do this calculation on mysql side.
http://ideone.com/48zLvF
$week_day = intval(date('w',$uDate1));
if ($week_day<6) {
$end_of_week = $uDate1+(86400*(6-$week_day));
} else {
$end_of_week = $uDate1;
}
I have a column with timestamp, contain example value "2014-04-16 18:00:00","2014-04-17 18:00:00"....
Now, if I will call a page before "2014-04-17 12:00:00" I need this value-"2014-04-16 18:00:00"
And if I call my page after "2014-04-17 12:00:00" I need this value "2014-04-17 18:00:00".
I think my question is very complicated to understand, having complications in date & times, please check date & time properly.
I want to fetch this data from DB in mysql, The page I was saying is that where I'm going to add your mysql query.
Thanx in advance
Generalising what your asking for a bit the following will return dates from the previous day if it's before noon and dates from today if it's after noon:
SELECT date_column
FROM yourTable
WHERE DATE(DATE_SUB(NOW(), INTERVAL 12 HOUR)) = DATE(date_column);
Edit:
The WHERE clause First gets the current time (NOW()) and subtracts 12 hours. This wont affect the date unless the time is before 12. This means DATE_SUB(NOW(), INTERVAL 12 HOUR) gives us today if it's after noon and yesterday if it's before.
We then check if the date_column matches the date we've created (using the DATE function so that the time is ignored).
Adding some rows to the SELECT may help you see how these dates are built up.