I have a table that I'm going to simplify. Here's what it looks like:
tid session pos dateOn
-----------------------------------------------
1 23 0 12/24/2020 1:00:00
2 23 1 12/24/2020 1:01:23
3 12 0 12/24/2020 1:02:43
4 23 2 12/24/2020 1:04:01
5 23 3 12/24/2020 1:04:12
6 45 0 12/26/2020 4:23:15
This table tells me that there were 2 unique sessions 12/24/2020 and 1 on 12/26.
How do I write my SQL statement so that I get a result like this:
date recordCount
----------------------------
12/24/2020 2
12/26/2020 1
You should simply be able to convert to a date and aggregate:
select convert(date, dateon), count(distinct session)
from t
group by convert(date, dateon)
order by convert(date, dateon);
Related
I need to show the number of valid inspectors we have by month over the last five years. Inspectors are considered valid when the expiration date on their certification has not yet passed, recorded as the month end date. The below SQL code is text of the query to count valid inspectors for January 2017:
SELECT Count(*) AS RecordCount
FROM dbo_Insp_Type
WHERE (dbo_Insp_Type.CERT_EXP_DTE)>=#2/1/2017#);
Rather than designing 60 queries, one for each month, and compiling the results in a final table (or, err, query) are there other methods I can use that call for less manual input?
From this sample:
Id
CERT_EXP_DTE
1
2022-01-15
2
2022-01-23
3
2022-02-01
4
2022-02-03
5
2022-05-01
6
2022-06-06
7
2022-06-07
8
2022-07-21
9
2022-02-20
10
2021-11-05
11
2021-12-01
12
2021-12-24
this single query:
SELECT
Format([CERT_EXP_DTE],"yyyy/mm") AS YearMonth,
Count(*) AS AllInspectors,
Sum(Abs([CERT_EXP_DTE] >= DateSerial(Year([CERT_EXP_DTE]), Month([CERT_EXP_DTE]), 2))) AS ValidInspectors
FROM
dbo_Insp_Type
GROUP BY
Format([CERT_EXP_DTE],"yyyy/mm");
will return:
YearMonth
AllInspectors
ValidInspectors
2021-11
1
1
2021-12
2
1
2022-01
2
2
2022-02
3
2
2022-05
1
0
2022-06
2
2
2022-07
1
1
ID
Cert_Iss_Dte
Cert_Exp_Dte
1
1/15/2020
1/15/2022
2
1/23/2020
1/23/2022
3
2/1/2020
2/1/2022
4
2/3/2020
2/3/2022
5
5/1/2020
5/1/2022
6
6/6/2020
6/6/2022
7
6/7/2020
6/7/2022
8
7/21/2020
7/21/2022
9
2/20/2020
2/20/2022
10
11/5/2021
11/5/2023
11
12/1/2021
12/1/2023
12
12/24/2021
12/24/2023
A UNION query could calculate a record for each of 50 months but since you want 60, UNION is out.
Or a query with 60 calculated fields using IIf() and Count() referencing a textbox on form for start date:
SELECT Count(IIf(CERT_EXP_DTE>=Forms!formname!tbxDate,1,Null)) AS Dt1,
Count(IIf(CERT_EXP_DTE>=DateAdd("m",1,Forms!formname!tbxDate),1,Null) AS Dt2,
...
FROM dbo_Insp_Type
Using the above data, following is output for Feb and Mar 2022. I did a test with Cert_Iss_Dte included in criteria and it did not make a difference for this sample data.
Dt1
Dt2
10
8
Or a report with 60 textboxes and each calls a DCount() expression with criteria same as used in query.
Or a VBA procedure that writes data to a 'temp' table.
Wondering how to select from a table:
FIELDID personID purchaseID dateofPurchase
--------------------------------------------------
2 13 147 2014-03-21 00:00:00
3 15 165 2015-03-23 00:00:00
4 13 456 2018-03-24 00:00:00
5 1 133 2018-03-21 00:00:00
6 23 123 2013-03-22 00:00:00
7 25 456 2013-03-21 00:00:00
8 25 456 2013-03-23 00:00:00
9 22 456 2013-03-28 00:00:00
10 25 589 2013-03-21 00:00:00
11 82 147 1991-10-22 00:00:00
12 82 453 2003-03-22 00:00:00
I'd like to get a result table of two columns: weekday and the number of purchases of each weekday, but only count the distinct days of purchases if done by the same person on the same day - for example since personID 25 purchased two things on 2013-03-21, that should only count as one 'thursday' instead of 2.
Basically, if the personID and the dateofPurchase are the same for more than one row, only count it once is what I want.
Here is what I have currently: It does everything correctly except it will count the above scenario under the thursday twice, when I would only want to add one:
SELECT v.wkday as day, COUNT(*) as 'absences'
FROM dbo.AttendanceRecord pr CROSS APPLY
(VALUES (CASE WHEN DATEPART(WEEKDAY, date) IN (1, 7)
THEN 'Weekend'
ELSE DATENAME(WEEKDAY, date)
END)
) v(wkday)
GROUP BY v.wkday;
to clarify:
If an item is purchased for at least one puchaseID on a specific day they will be counted as purchased for that day, and do not need to be counted again for each new purchase ID on that day.
I think you want to count distinct persons, so that would be:
COUNT(DISTINCT personid) as absences
Note that single quotes are not appropriate around column aliases. If you need to escape them, use square braces.
EDIT:
If you want to count distinct person-days, then you can use:
COUNT(DISTINCT CONCAT(personid, ':', dateofpurchase) as absences
Is there a way to find the solution so that I need for 2 days, there are 2 UD's because there are June 24 2 times and for the rest there are single days.
I am showing the expected output here:
Primary key UD Date
-------------------------------------------
1 123 2015-06-24 00:00:00.000
6 456 2015-06-24 00:00:00.000
2 123 2015-06-25 00:00:00.000
3 658 2015-06-26 00:00:00.000
4 598 2015-06-27 00:00:00.000
5 156 2015-06-28 00:00:00.000
No of times Number of days
-----------------------------
4 1
2 2
The logic is 4 users are there who used the application on 1 day and there are 2 userd who used the application on 2 days
You can use two levels of aggregation:
select cnt, count(*)
from (select date, count(*) as cnt
from t
group by date
) d
group by cnt
order by cnt desc;
I have a database with the following data:
Group ID Time
1 1 16:00:00
1 2 16:02:00
1 3 16:03:00
2 4 16:09:00
2 5 16:10:00
2 6 16:14:00
I am trying to find the difference in times between the consecutive rows within each group. Using LAG() and DATEDIFF() (ie. https://stackoverflow.com/a/43055820), right now I have the following result set:
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 00:06:00
2 5 00:01:00
2 6 00:04:00
However I need the difference to reset when a new group is reached, as in below. Can anyone advise?
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 NULL
2 5 00:01:00
2 6 00:04:00
The code would look something like:
select t.*,
datediff(second, lag(time) over (partition by group order by id), time)
from t;
This returns the difference as a number of seconds, but you seem to know how to convert that to a time representation. You also seem to know that group is not acceptable as a column name, because it is a SQL keyword.
Based on the question, you have put group in the order by clause of the lag(), not the partition by.
I am using SQL Server 2008.
I want to write a query that gives me total activity for a number of given days. Specifically, I want to count total votes per day for the last seven days.
My table looks like this:
VoteID --- VoteDate -------------- Vote --- BikeID
1 2012-01-01 08:24:25 1 1234
2 2012-01-01 08:24:25 0 5678
3 2012-01-02 08:24:25 1 1289
4 2012-01-03 08:24:25 0 1234
5 2012-01-04 08:24:25 1 5645
6 2012-01-05 08:24:25 0 1213
7 2012-01-06 08:24:25 1 1234
8 2012-01-07 08:24:25 0 1125
I need my results to look like this
VoteDate ---- Total
2012-01-01 5
2012-01-02 6
2012-01-03 7
2012-01-04 1
2012-01-05 3
My thought is that I have to do something like this:
SELECT SUM(CASE WHEN Vote = 1 THEN 1 ELSE 0 END) AS Total
FROM Votes
GROUP BY VoteDate
This query doesn't work because it counts only votes that occurred (almost exactly) at the same time. Of course, I want to look only at a specific day. How do I make this happen?
Cast it as a date:
SELECT
cast(VoteDate as date) as VoteDate,
SUM(CASE WHEN Vote = 1 THEN 1 ELSE 0 END) AS Total
FROM Votes
WHERE VoteDate between dateadd(day, -7, GETDATE()) and GETDATE()
GROUP BY cast(VoteDate as date)
Your VoteDate column is a datetime, but you just want the date part of it. The easiest way to do that is to cast it as a date type. You can read more about SQL Server date types here.
And if your Vote column is either 1 or 0, you can just do sum(vote) as Total instead of doing the case statement.
SELECT SUM(Vote) As Total, YEAR(VoteDate),Month(VoteDate),Day(VoteDate)
FROM Votes
Group By YEAR(VoteDate),Month(VoteDate),Day(VoteDate)
Some SQL Server functions that may be of interest
Some MySQL functions that may be of interest