I need a query to get results from a table that has 2 columns
Column startdt (datetime), Column enddt (datetime)
there are some records with startdt 2013-07-19 and enddt 2013-07-29
I need to get the records with weekday = 1 (Tuesday)
the record with date 2013-07-19 is weekday 4 and ends 2013-07-29 which is 0
Actually i want to get the results that has for weekday Monday or another weekday.
You can check the above link for an example
http://sqlfiddle.com/#!2/a80ce/1
If you don't understand what i want to do let me explain. I have an event that starts July 15 and ends July 25. (Starts Monday and ends Thursday) The user selects one of the week days (Monday, Tuesday etc). If he select Tuesday then i want the query that will get all events that are active in Tuesday.
I already found the answer so if anyone want to check it
SELECT articleid,startdt,enddt,dayofweek(startdt), DATEDIFF(enddt,startdt) datedf
FROM events
WHERE (dayofweek(events.startdt) <= 3 AND dayofweek(events.enddt) >= 3)
OR DATEDIFF(enddt,startdt) >=6
(3 is the number of the weekday "Tuesday")
How about using the comments that other people gave you and use a query that combines both dayofweek and a simple greater/smaller/equal syntax as follows:
SELECT * FROM events where dayofweek(events.startdt) <= 6 AND dayofweek(events.enddt) >= 6
This gives the following results if the user specified a friday (= 6):
ARTICLEID STARTDT ENDDT
4 July, 12 2013 00:00:00+0000 July, 26 2013 00:00:00+0000
6 July, 16 2013 00:00:00+0000 July, 20 2013 00:00:00+0000
I do think that you are better of using dayofmonth however as this (maybe just to me) makes it clearer, possibly combining the use of both to ensure that it's active on a friday.
The OP indicates that events which are in the history should also be retrieved and as such the following query does what he wants:
SELECT * FROM events where dayofweek(events.startdt) <= 6 AND dayofweek(events.enddt) >= 6 OR DATEDIFF(enddt,startdt) >=6
How about this solution:
first you convert the day of week in format that 6 is Saturday and 7 is sunday(it's easier for me)
if(dayofweek(o.start_date) = 1, 7, dayofweek(o.start_date) -1)
after that you calc the days from the start_date needed to reach some of the weekdays
(7 - if(dayofweek(o.start_date) = 1, 7, dayofweek(o.start_date) -1)
finally you make sure that the difference in days between the two dates is no less that the above calculation
(7 - if(dayofweek(o.start_date) = 1, 7, dayofweek(o.start_date) -1) <= datediff(o.end_date, o.start_date)
Related
I am working in mySQL and I currently have a count of total orders by day, but I would like to add Saturday and Sunday orders to Monday then remove Saturday and Sunday values. I have done some research on this but I cannot seem to find anything similar to what I am trying to do.
My current data table looks like this:
Date | Daily Count
8-6-2020 25
8-7-2020 82
8-8-2020 24
8-9-2020 33
8-10-2020 18
8-11-2020 10
8-12-2020 25
8-13-2020 15
I need it to look something like this:
Date | Daily Count
8-6-2020 25
8-7-2020 82
8-10-2020 75
8-11-2020 10
8-12-2020 25
8-13-2020 15
In this one the Daily counts for the 8th and 9th are added to the 10th, then removed, because they are weekend days. Thank you in advance for your help!
Consider using a case expression to adjust the date:
select
case weekday(date)
when 5 then date + interval 2 day
when 6 then date + interval 1 day
else date
end as new_date,
sum(daily_count) as daily_count
from mytable
group by new_date
I have such data in my table:
I need to calculate "Paid" field of UserID which reoccurs in 7 day intervals. In this example I will SUM(Paid) for UserID "01" because it occurs 2 times in 7 days interval.
I can calculate it programmaticaly, but only in such date intervals (2016-01-01 - 2016-01-07; 2016-01-07 - 2016-01-13; etc.).
Maybe there is some possibility to perform this calculation at MySQL level in any 7 day intervals? For example: 2016-01-01 - 2016-01-07; 2016-01-02 - 2016-01-08; 2016-10-10 - 2016-10-16; etc.
I believe the method WEEK() returns the week number based on the calendar year, meaning that for 2016: 1st Jan, 2nd Jan and 3rd Jan would return 0, but 4th Jan would return 1, which to my understanding does not fit the requirements.
I would suggest:
SELECT `UserID`, SUM(`Paid`) FROM `table` GROUP BY DATEDIFF(`Date`, (SELECT MIN(`Date`) FROM `table`)) DIV 7, `UserID`
I need stats on orders, week by week, so I have done this:
SELECT YEAR(orders.date), WEEKOFYEAR(orders.date), COUNT(*)
FROM orders
GROUP BY YEAR(orders.date), WEEKOFYEAR(orders.date)
It worked for one year, but just now (new year) it does not count the last days of 53rd week (jan 1st, 2nd, 3rd). How can I update my Query in order to have the full last week (from Monday 2015-12-28 to Sunday 2016-01-03)?
You need to switch to YEARWEEK(orders.date,3) to get the ISO weeks as a single column. Using WEEK(orders.date,3) (which is exactly the same as WEEKOFYEAR) will return the correct week number, but YEAR(orders.date) will return either 2015 or 2016, splitting the week into four days in 2015 and and three days in 2016.
As Strawberry mentioned in the comments you're looking for the WEEK function. I just checked the documentation at the MySQL website.
Week(date [,mode])
This function returns the week number for date. The two-argument form of WEEK() enables you to specify whether the week starts on Sunday or Monday and whether the return value should be in the range from 0 to 53 or from 1 to 53. If the mode argument is omitted, the value of the default_week_format system variable is used
Here's an example
SELECT WEEK('2008-12-31',1);
=> 53
It should also be noted that this is not the same as the WEEKOFYEAR function.
Returns the calendar week of the date as a number in the range from 1 to 53. WEEKOFYEAR() is a compatibility function that is equivalent to WEEK(date,3).
We can see that the value of the mode parameter here is 3. Here is the table that shows the values for the modes
Mode First day of week Range Week 1 is the first week
0 Sunday 0-53 With a Sunday in this year
1 Monday 0-53 With 4 or more days this year
2 Sunday 1-53 With a Sunday in this year
3 Monday 1-53 With 4 or more days this year
4 Sunday 0-53 With a Sunday in this year
5 Monday 0-53 With 4 or more days this year
6 Sunday 1-53 With a Sunday in this year
7 Monday 1-53 With 4 or more days this year
Source
https://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_week
As I understand you, you want calculate total orders for one week only once not twice for various years.
I've written SQL query for SQL Server, but I think you can easy rewrite it on mysql.
DECLARE #test_table TABLE(created_date DATETIME, VALUE INT)
INSERT INTO #test_table
SELECT N'20151231', 10
UNION ALL
SELECT N'20151228', 20
UNION ALL
SELECT N'20160101', 40
UNION ALL
SELECT N'20160104', 5
UNION ALL
SELECT N'20160107', 2
SELECT first_day_week_number,
SUM(value) AS total
FROM
(SELECT *,
DATEPART(ww,
DATEADD(dd, -DATEPART(dw, created_date)+2, created_date)) AS first_day_week_number
FROM #test_table ) AS T
GROUP BY first_day_week_number
To avoid problems with years you could calculate first week day of order date and then get week number from it.
I use mySql 5 and IIS.
I have products, that have a start date field and an end date field.
I need to run a query that will take user entered Start and End dates, and output the number of days that the product ran within the date range.
Example:
Offer1 - July 1 2011 thru July 31 2011
Query - July 1 2011 thru Sept 15 2011
Results = 31
Example:
Offer1 - July 1 2011 thru July 31 2011
Query - July 1 2011 thru July 15 2011
Results = 15
If your products have a start_date and an end_date and your query has a qstart_date and a qend_date, then we want the number of days between:
GREATEST(start_date, qstart_date)
and
LEAST(end_date,qend_date)
. In MySQL I think this looks like
1 + DATEDIFF ( 'd' , GREATEST(start_date, qstart_date) , LEAST(end_date,qend_date) )
And you'll want to ignore negative numbers, replacing them with "0".
Does anyone know a way to convert a month, year, and day into the day's name for any year? Example:
function convert(day, year, month)
...
return "Monday"
end
Thanks in advance!
You can use the following method:
This method uses codes for different
months and years to speed up the
calculation of the day of the week.
You might even be able to memorize the
codes. We'll use December 16, 2482 as
an example.
Take the last 2 digits of the year. In
our example, this is 82.
Divide by 4, and drop any remainder.
82 / 4 = 20, remainder 2, so we think
"20."
Add the day of the month. In our
example, 20 + 16 = 36.
Add the month's key value, from the
following table. Jan Feb Mar Apr May
June July Aug Sept Oct Nov Dec 1 4 4
0 2 5 0 3 6 1 4 6
The month for our example is December,
with a key value of 6. 36 + 6 = 42.
If your date is in January or February
of a leap year, subtract 1. We're
using December, so we don't have to
worry about this step.
Add the century code from the
following table. (These codes are for
the Gregorian calendar. The rule's
slightly simpler for Julian dates.)
1700s 1800s 1900s 2000s 4 2 0 6
Our example year is 2482, and the
2400s aren't in the table. Luckily,
the Gregorian calendar repeats every
four hundred years. All we have to do
is add or subtract 400 until we have a
date that is in the table. 2482 - 400
= 2082, so we look at the table for the 2000s, and get the code 6. Now we
add this to our running total: 42 + 6
= 48.
Add the last two digits of the year.
48 + 82 = 130.
Divide by 7 and take the remainder.
This time, 1 means Sunday, 2 means
Monday, and so on. A remainder of 0
means Saturday.
How to calculate the day of the week
A quickl google for "day of week from date algorithm" showed up this Wikipedia article
But depending on the dates you need to work with, beware the strange history of Gregorian calendar adoption