Table:
DayOfWeek Enrollments
Monday 35
Monday 12
Saturday 25
Tuesday 15
Monday 9
Tuesday 15
Basically I'm trying to sum the total enrolments for each day.
so the Output will look like:
DayOfWeek Enrollments
Monday 56
Saturday 25
Tuesday 30
I've spent around 4 hours trying to work this out trying many many different ways but no luck.
The problem I'm having is i can count how many enrollments for each day but can't have it aligned with the correct day when i run the query e.g. I want The total to be on the same line as the day it was calculated from. (I hope that is clear enough)
Group by DayOfWeek, and ask for the sum of Enrollments within each group. The SQL will look like this.
SELECT DayOfWeek, Sum(Enrollments) AS SumOfEnrollments
FROM YourTable
GROUP BY DayOfWeek;
If you're using the Access query designer to create this, select your fields, then click the symbol for "Totals" query (Greek character sigma). In the "Total:" row of the design grid, select Group By and Sum for the appropriate fields.
Related
At my last meeting someone asked me if it was possible to hide people who where ill since a year from a dashboard. So I'm searching for the best way to actually KNOW who has been ill for 12 months.
I am working with a table with the number of days you've been absent for every kind of absence you could have, the number of days you should have been working that month, with a row per person, department and profession each month.
So it looks something like this :
PersonID
YEAR
MONTH
DEPARTMENT
PROFESSION
Absence1
Absence2
Absence3
WORKING DAYS OF THE MONTH...
11111
2021
07
HR
ASSISTANT
0
2
0
22
11111
2021
08
HR
ASSISTANT
0
0
0
22
==> So if I'm on a row of July 2021 I need to check the lines from June 2020 to June 2021.
My guess is that I need to add a column to this table who will say (with some kind of loop maybe) "if for the last 12 months (rows) the total number of days of absence equal the number of working days of the last 12 months then "ILL FOR A YEAR OR MORE" for each person (knowing that a person can work in more than one department or more than one profession so she'll have more than one row per month).
But I really have no idea how to actually write it in a script as I usually do very basic things. I'm using SQL SERVER and have 429 207 rows in the table. I'm thinking about doing it in the whole table and not only treating this month's rows because in the dashboard we show an historic.
Your table is heavily denormalized. If you want to represent all this information in the database, I would have expected the following tables, instead of just one:
Person
Department
Illness (list of illnesses)
IllnessAbsence (join table between Person and Illness)
Either way, you can get the information you need with something like this:
I've assumed you want the whole table, so you need a window function
We need to flip the logic on its head: exclude all rows which have no non-absence in the last 12 months
SELECT
PersonID,
YEAR,
MONTH,
DEPARTMENT,
PROFESSION,
ILLNESS1,
ILLNESS2,
ILLNESS3,
[WORKING DAYS OF THE MONTH]
FROM (
SELECT *,
NotIllLast12Months = COUNT(CASE WHEN DATEFROMPARTS(YEAR + 1, MONTH, 1) >= GETDATE()
AND ILLNESS1 + ILLNESS2 + ILLNESS3 = 0 THEN 1 END)
OVER (PARTITION BY ID)
FROM HETP_ABS
) abs
WHERE NotIllLast12Months > 0;
So I want to make a query to show me if a certain calendar week has all 7 Day.
It would be okay if it just returns the numbers 1-7.
The table that I have contains articles of the 3 month of 2020 but even so the first week just contains Wednesday to Sunday it still counts it as a calendar week.
With that select I would make pl/sql Script to check it and if yes something happens.
This is an example of the Table:
Date Articel_Id
14.10.2020 78
15.10.2020 80
16.10.2020 96
17.10.2020 100
18.10.2020 99
Can I Use to_char() to check if Calendar Week has all 7 Days ?
If yes, how ?
The challenging is actually defining the weeks. If you want to define them using the ISO standard, then aggregate:
select to_char(date, 'IYYYY-IW') as yyyyww,
count(distinct trunc(date)) as num_days
from t
group by to_char(date, 'IYYYY-IW')
order by yyyyww;
This counts the number of days per week. I'm not sure if you want to filter, have a flag, or what the result set should look like. For filtering, using a having clause, such as having count(distinct trunc(date)) = 7.
I am trying to produce a query in SQLite where I can determine the average sales made each weekday in the year.
As an example, I'd say like to say
"The average sales for Monday are $400.50 in 2017"
I have a sales table - each row represents a sale you made. You can have multiple sales for the same day. Columns that would be of interest here:
Id, SalesTotal, DayCreated, MonthCreated, YearCreated, CreationDate, PeriodOfTheDay
Day/Month/Year are integers that represent the day/month/year of the week. DateCreated is a unix timestamp that represents the date/time it was created too (and is obviously equal to day/month/year).
PeriodOfTheDay is 0, or 1 (day, or night). You can have multiple records for a given day (typically you can have at most 2 but some people like to add all of their sales in individually, so you could have 5 or more for a day).
Where I am stuck
Because you can have two records on the same day (i.e. a day sales, and a night sales, or multiple of each) I can't just group by day of the week (i.e. group all records by Saturday).
This is because the number of sales you made does not equal the number of days you worked (i.e. I could have worked 10 saturdays, but had 30 sales, so grouping by 'saturday' would produce 30 sales since 30 records exist for saturday (some just happen to share the same day)
Furthermore, if I group by daycreated,monthcreated,yearcreated it works in the sense it produces x rows (where x is the number of days you worked) however that now means I need to return this resultset to the back end and do a row count. I'd rather do this in the query so I can take the sales and divide it by the number of days you worked.
Would anyone be able to assist?
Thanks!
UPDATE
I think I got it - I would love someone to tell me if I'm right:
SELECT COUNT(DISTINCT CAST(( julianday((datetime(CreationDate / 1000, 'unixepoch', 'localtime'))) ) / 7 AS INT))
FROM Sales
WHERE strftime('%w', datetime(CreationDate / 1000, 'unixepoch'), 'localtime') = '6'
AND YearCreated = 2017
This would produce the number for saturday, and then I'd just put this in as an inner query, dividing the sale total by this number of days.
Buddy,
You can group your query by getting the day of week and week number of day created or creation date.
In MSSQL
DATEPART(WEEK,'2017-08-14') // Will give you week 33
DATEPART(WEEKDAY,'2017-08-14') // Will give you day 2
In MYSQL
WEEK('2017-08-14') // Will give you week 33
DAYOFWEEK('2017-08-14') // Will give you day 2
See this figures..
Day of Week
1-Sunday, 2- Monday, 3-Tuesday, 4-Wednesday, 5-Thursday, 6-Saturday
Week Number
1 - 53 Weeks in a year
This will be the key so that you will have a separate Saturday's in every month.
Hope this can help in building your query.
I'm currently running the following sql statement in JasperReports Server to bring back my data using derived tables.
Select count(createddate) as ModulesCreatedDuringPastWeek,
count(updateddate) as ModulesUpdatedDuringPastWeek,
createddate,
updateddate
from merchendisingmodule
group by merchendisingmodule.createddate, merchendisingmodule.updateddate
However when grouping my data, I am only able to do it in Year, quarter, month and day. However for my report I'm needing the data to be group weeks, and so I was wondering what I will need to add to my code to do this.
DATEADD(D,-DATEPART(weekday,createddate)+1,createddate)
I use this method to prevent issues around the year transitions (week 53 in first days of januari and also in the last days of december, will group days together that are 360 days apart).
I use the first day of the week, instead of week numbers. I can use these dates to group by.
Also this will ensure that every week is 7 days long, instead of the last week of the year being only 3 or 4 days long.
Btw, in this example the first day of the week is sunday.
If your dates include time, use:
CAST(FLOOR(CAST(createddate AS FLOAT)) AS DATETIME)
instead of createddate in the above SYNTAX
Greetings SQL gurus,
I don't know if you can help me, but I will try. I have several large databases grouped by year (each year in a different database). I want to be able to compare values from a particular week from one year to the next. For example, "show me week 17 of 2008 vs. week 17 of 2002."
I have the following definition of weeks that ideally I would use:
Only 52 weeks each year and 7 days a week (that only takes 364 days),
The first day of the first week starts from January 2nd - which means we do not use January 1st data, and
In leap year, the first day of the first week ALSO starts from the January 2nd plus we skip Feb. 29.
Any ideas?
Thanks in advance.
Best to avoid creating a table because then you have to update and maintain it to get your queries to work.
DatePart('ww',[myDate]) will give you the week number. You may run into some issues though deciding which week belongs to which year - for example if Jan 1 2003 is on Wednesday does the week belong as week 52 in 2002 or week 1 in 2003? Your accounting department will have a day of the week that is your end of week (usually Sat). I usually just pick the year that has the most days in it. DatePart will always count the first week as 1 and in the case of the example above the last week as 53. You may not care that much either way. You can create queries for each year
SELECT DatePart('ww',[myDate]) as WeekNumber,myYearTable.* as WeekNumber
FROM myYearTable
and then join the queries to get your data. You'll loose a couple days at the end of the year if one table has 52 weeks and one has 53 (most will show as 53). Or you can do it by your weekending day - this always gives you Saturday which would push a late week into the following year.
(7-Weekday([myDate]))+[myDate]
then
DatePart('ww',(7-Weekday([myDate]))+[myDate])
Hope that helps
To get the week number
'to get the week number in the year
select datepart( week, datefield)
'to get the week number in the month
select (datepart(dd,datefield) -1 ) / 7 + 1
You don't need to complicate things thinking about leap years, etc. Just compare weeks mon to sun
SInce you havea a specifc defintion of when the week starts that is differnt that the standard used by the db, I think a weeks table is the solution to your problem. For each year create a table that defines the dates contained in each week and the week number. Then by joining to that table as well as the relevant other tables, you can ask for just the data for week 17.
Table structure
Date Week
20090102 1
20090103 1
etc.
I needed to create a query that shows BOTH year AND week numbers, like 2014-52. The year shows correct when you use the Datepart() formula to convert week 53 to week 52 in the previous year, but shows the wrong year for the week that was week 1 previously that should be week 52 now. It show that week as 2015-52 instead of 2014-52.
Furthermore, it sorts the data wrong if you only use only the week number, eg:
2014-1,2014-11,2014-2
To overcome this I created the following query to insert a 0 and also to check for days in week 1 that should still fall under week 52.
ActualWeek: IIf(DatePart("ww",[SomeDate],1,3)=52 And DatePart("ww",[SomeDate])=1, DatePart("yyyy",[SomeDate],1,3)-1,DatePart("yyyy",[SomeDate],1,3)) & "-" & IIf(DatePart("ww",[SomeDate],1,3)<10,"0" & DatePart("ww",[SomeDate],1,3),DatePart("ww",[SomeDate],1,3))