How to query in mysql to find missing date from table
I have table like
uid date hours
4 06/07/2014 8:00
5 07/07/2014 6:00
6 11/07/2014 8:00
7 13/07/2014 7:00
I need to get missed dates like
08/07/2014
09/07/2014
10/07/2014
12/07/2014
please help me anyone...
thanks
Related
I am trying to get week numbers ( resetting at 1 for each month) as per ISO format for each month in 2019.For example I am interested in getting
All dates in July 2019: week 1 to 4,
All dates in Aug 2019 : week 1 to 4 and so on.
I first created the calculated field (Week_Number_ISO) to get the overall week number in year 2019.I used the following formula;
DATEPART('iso-week',[ Date]) which works as intended.
To get the monthly week number I used the following formula
INT((DATEPART('day',[Created Date])-DATEPART('iso-weekday',[Created Date])+7)/7)+1.
(Idea was to calculate the date of the first day of each week & then divide by 7 and take the integer part)
As per the ISO format, shouldn't July 29 to 31st be a part of week 4 for July?But the formula is showing it as week 5 for July 2019.I feel I am missing something in the formula or am missing something about ISO week number resetting at 1 for each month.
Can someone help me?
Here is an example of the dates in July 2019 and the associated week numbers.
Why would July 28th-July 31st 2019 be considered week 4?
I have sets of date/time range values, each set came from different column ("FROM" and "TO")
examples:
FROM: June 6, 2016 7:00 AM; TO: June 6, 2016 5:00 PM
FROM: June 6, 2016 8:00 PM; TO: June 7, 2016 6:00 AM
I want to get how many minutes of each set falls to 3:00 PM - 2:00 AM (next day) or "OFF PEAK HOURS", like this
for the example above, I want to get the number of minutes between 3PM - 2AM next day.
on the first row, from 7AM-5PM, two hours is included to my 3PM-2AM range criteria, which is 3PM-5PM (2 hours) . Thats why I got 120 mins.
on the second row, from 8PM-6AM next day, two hours is included to my 3PM-2AM range criteria, which is 8PM-2AM (6 hours). Thats why I got 360 mins.
see this graph for reference:
I want to count the number of minutes falls during "OFF PEAK HOURS"-shaded in orange/tan.
Is it doable in expression?
Thank you in advance.
Please note that I am using fetchXML for the query.
Supposing the table you put in the question is wrong and it is only an example, I think you can get the spent minutes between two dates by using this expression:
=DateDiff(DateInterval.Minute,
Fields!From.Value,
Fields!To.Value
)
Example:
=DateDiff(DateInterval.Minute,
CDATE("2016-06-06 07:00:00 AM"),
CDATE("2016-06-06 05:00:00 PM")
)
The above example returns 600. 10 hours by 60 minutes. DateDiff function returns the specified date/time interval of time between two specified dates.
Let me know if this helps.
Would it be possible to get same day of week last year using Excel? please below example:
Input: Monday 9 Nov 2015 | Output: 10 Nov 2014
Thanks
Simply subtract 52 full weeks with 7 days = 364 days. So if the date is in A1, the formula =A1-364 will get the date exactly 52 weeks before, which is the same day of week in the year before.
To show that it works even for leap years, try the following:
You see the formula date - 364 (=A2-364, =A3-364, ...) always gets the same day of week a year before. That is because it gets the day minus 52 full weeks (52 * 7 days) before. In leap years it gets a different day but the same day of week.
Try this:
=DATE(YEAR(A1)-1,MONTH(A1),DAY(A1))+WEEKDAY(A1)-WEEKDAY(DATE(YEAR(A1)-1,MONTH(A1),DAY(A1)))
It returns the closest date within a week. A1 is the cell with this year's date.
Postgresql 8.4.
I'm new to this concept so if people could teach me I'd appreciate it.
For Obamacare, anyone that works 30 hours per week or more must be offered the same healthcare as is offered to any other worker. We can't afford that so we have to limit work hours for temp and part-timers. This is affecting the whole country.
I need to calculate the hours worked (doesn't matter if overtime,
regular time, double time, etc) between two dates, say Jan 1, 2014,
and Nov 1, 2014 (Saturday) for each custom week (which beings on Sunday), not the week as defined by Postgresql (which begins on Monday).
Each of my custom work weeks begins on Sunday and ends on Saturday.
I don't know if I have to include weeks where
they did not work at all in the average, but let's assume I do. Zero hours that week would draw down the average.
Table name is 'employeetime', date field is 'employeetime.stopdate', hours worked per day is in the field 'employeetime.hours', employeeid field is 'employeetime.empid'.
I'd prefer to do this in one query per employee and I will execute the query once per employee as I loop through employees. If not I'm open to suggestions. But I'd like to understand the SQL presented in the answer.
Currently EXTRACT(week from '2014-01-01') calculates the start of the week as a Monday, so that doesn't work for me. Link here.
How would I do that without doing, say a separate query for each week, per person? We have 200 people to process.
Thank you.
I have set up a table to match your format:
select * from employeetime order by date;
id date hours
1 2014-11-06 10
1 2014-11-07 3
1 2014-11-08 5
1 2014-11-09 3
1 2014-11-10 5
You can get the week starting on Sunday by shifting. Note, here the 9th is a Sunday, so that is where we want the boundary.
select *, extract(week from date + '1 day'::interval) as week
from employeetime
order by week;
id date hours week
1 2014-11-07 3 45
1 2014-11-06 10 45
1 2014-11-08 5 45
1 2014-11-09 3 46
1 2014-11-10 5 46
And now the week shifts on Sunday rather than Monday. From here, the query to get hours by week/employee would be simple:
select id, sum(hours) as hours, extract(week from date + '1 day'::interval) as week
from employeetime
group by id, week
order by id, week;
id hours week
1 18 45
1 8 46
How to get the Date difference between two dates in postgresql? ( Excluding weekends Saturday and Sunday, and out of office hours). It should calculate the time only on weekdays during office hours( 9 - 6). My requirement is to calculate the time between 2 date columns in hours. Thanks in advance