Pick records since last particular day of week - mysql

I want to pick and SUM values since last wednesday until NOW() in mysql. How can I do that?
Sorry for incomplete question, by the last Wednesday I did not mean to hard-code the date, rather I want my program to run that query, so it cannot hard-code--- Needs a flexible solution. Please help...

select date_sub(now(), interval dayofweek(date_sub(now(), interval 4 day)) day);
This works on any day of the week and always returns the Wednesday which has most recently passed. On a Wednesday itself, it returns the previous Wednesday. The next day, it returns yesterday

Genesis is right (He's very right, use his suggestion), but as an intellectual exercise: This is the best pure MySQL I could think of:
SELECT * FROM TABLE
WHERE
DATE_COLUMN > DATE_SUB( NOW(), INTERVAL DAYOFWEEK(NOW()) + 3 DAY);
NOW - DAYOFWEEK => this past Saturday. Weds. is three days before that.

SELECT SUM(value) FROM table WHERE date > '2011-07-20'
You should calculate your date from your programming language (fastest solution)

Related

MySQL Select From Prior Week, not just 7 days

I am attempting to get a query that selects the week prior (Sun-Sat). I've fought with this query and the closest I can get is the last 7 days, using the following:
SELECT *
FROM dates
WHERE date BETWEEN CURDATE()-INTERVAL 1 WEEK AND CURDATE();
I'm really unsure how to proceed from here. It seems as if I need to create some kind of relation between CURDATE() and the Saturday before maybe?
Any help is appreciated.
You are after the Week of the year.
Look at the Week Function: WEEK(date[,mode])
http://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_week
The mode describes how you define the week (which is the start of the week)
WEEK(date,3) is for a week that starts Monday.
SELECT *
FROM dates
WHERE
-- Last Week
WEEK(date,3) = WEEK(CURDATE(),3)-1
AND YEAR(date)= YEAR(CURDATE()) ;
Don't forget the year. Week is just a number between 1 and 52. So the Year is important!
The code above is not correct. it will fail on the last week of the year!

Rewritten mySQL Statement

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;
}

Counting days by non standard days end

I am trying to count events that happen in a day, which in itself is easy:
SELECT date(time),COUNT(DISTINCT comment)
FROM data
GROUP BY date(time)
But I need a day end to be set at 4pm. So all comments before 4pm is day N, and news after 4pm are day N+1. This isnt a global thing, so I'd rather not change time zone of entire program and etc. Maybe I can somehow apply timezone to this query or pad time?
SELECT
DATE_ADD(date(time),
INTERVAL case when time(time)>='16:00:00' then 1 else 0 end DAY) as day,
COUNT(DISTINCT comment)
FROM data
GROUP BY day
Have you tried this:
SELECT date(time + INTERVAL 8 HOUR) as day, COUNT(DISTINCT comment)
FROM data
GROUP BY day
You want your days to end at 4PM, or 8 hours early so, we add 8 hours to time, forcing the date to roll over at 4PM.

How do I select two weeks ago in MYSQL?

I have a report that is driven by a sql query that looks like this:
SELECT batch_log.userid,
batches.operation_id,
SUM(TIME_TO_SEC(ramses.batch_log.time_elapsed)),
SUM(ramses.tasks.estimated_nonrecurring + ramses.tasks.estimated_recurring),
DATE(start_time)
FROM batch_log
JOIN batches ON batch_log.batch_id=batches.id
JOIN ramses.tasks ON ramses.batch_log.batch_id=ramses.tasks.batch_id
JOIN protocase.tblusers on ramses.batch_log.userid = protocase.tblusers.userid
WHERE DATE(ramses.batch_log.start_time) > "2011-02-01"
AND ramses.batch_log.time_elapsed > "00:03:00"
AND DATE(ramses.batch_log.start_time) < now()
AND protocase.tblusers.active = 1
AND protocase.tblusers.userid NOT in ("ksnow","smanning", "dstapleton")
GROUP BY userid, batches.operation_id, date(start_time)
ORDER BY start_time, userid ASC
Since this is to be compared with the time from the current payperiod it causes an error.
Our pay periods start on a Sunday, the first pay period was 2011-02-01 and our last pay period started the 4th of this month. How do I put that into my where statement to strip the most recent pay period out of the query?
EDIT: So now I'm using date_sub(now(), INTERVAL 2 WEEK) but I really need a particular day of the week(SUNDAY) since it is wednesday it's chopping it off at wednesday.
You want to use DATE_SUB, and as an example.
Specifically:
select DATE_SUB(curdate(), INTERVAL 2 WEEK)
gets you two weeks ago. Insert the DATE_SUB ... part into your sql and you're good to go.
Edit per your comment:
Check out DAYOFWEEK:
and you can do something along the lines of:
DATE_SUB(DATE_SUB(curdate(), INTERVAL 2 WEEK), INTERVAL 2 + DAYOFWEEK(curdate()) DAY)
(I don't have a MySql instance to test it on .. but essentially subtract the number of days after Monday.)
Question isn't quite clear, especially after the edit - it isn't clear now is the "pay period" two weeks long or do you want just last two weeks back from last sunday? I assume that the period is two weeks... then you first need to know how many days the latest period (which you want to ignore, as it isn't over yet) has been going on. To get that number of days you can use expression like
DATEDIFF(today, FirstPeriod) % 14
where FirstPeriod is 2011-02-01. And now you strip that number of days from the current date in the query using date_sub(). The exact expression depends on how the period is defined but you should get the idea...

Mysql, php query WHERE statement, recordset expire in +1 day of post date?

I basically have a simple calendar I set up. As of now, it shows "future" events. And then the event expires that day... I would love to find a WHERE statement that I can use to have that "event" stay up for 1 day past the "post_date"
(so if I post it as Nov. 15th,) The event would show: Name of event - Nov. 15th
And It would stay active until +1 day from post_date? (Nov. 16th would be the expire date)
Here is what I have so far:
WHERE DATE(FROM_UNIXTIME(`date`)) >= DATE(NOW())
Thanks in advance...
WHERE post_date > DATE(NOW())-INTERVAL 1 DAY
and if you really want to keep your post_dates in UNIX timestamps:
WHERE FROM_UNIXTIME(post_date) > DATE(NOW())-INTERVAL 1 DAY
Change your where statement to:
WHERE DATE(FROM_UNIXTIME(`date`)) + INTERVAL 1 DAY >= CURDATE();
Also a good idea to to use real SQL dates instead of UNIX timestamps. There are functions to do calculations on them.