MySQL - The most occuring for the specific day? - mysql

I'm stuck on this problem.
Basically I need to find out for each department how to figure out which days had the most sales made in them. The results display the department number and the date of the day and a department number can appear several times in the results if there were several days that have equally made the most sales.
This is what I have so far:
SELECT departmentNo, sDate FROM Department
HAVING MAX(sDate)
ORDER BY departmentNo, sDate;
I tried using the max function to find which dates occurred most. But it only returns one row of values. To clarify more, the dates that has the most sales should appear with the corresponding column called departmentNo. Also, if two dates for department A has equal amount of most sales then department A would appear twice with both dates showing too.
NOTE: only dates with the most sales should appear and the departmentNo.
I've started mySQL for few weeks now but still struggling to grasp the likes of subqueries and store functions. But i'll learn from experiences. Thank you in advance.
UPDATED:
Results I should get:
DepartmentNo Column 1: 1 | Date Column 2: 15/08/2000
DepartmentNo Column 1: 2 | Date Column 2: 01/10/2012
DepartmentNo Column 1: 3 | Date Column 2: 01/06/1999
DepartmentNo Column 1: 4 | Date Column 2: 08/03/2002
DepartmentNo Column 1: nth | Date Column 2: nth date
These are the data:
INSERT INTO Department VALUES ('1','tv','2012-05-20','13:20:01','19:40:23','2');
INSERT INTO Department VALUES ('2','radio','2012-07-22','09:32:23','14:18:51','4');
INSERT INTO Department VALUES ('3','tv','2012-09-14','15:15:43','23:45:38','3');
INSERT INTO Department VALUES ('2','tv','2012-06-18','06:20:29','09:57:37','1');
INSERT INTO Department VALUES ('1','radio','2012-06-18','11:34:07','15:41:09','2');
INSERT INTO Department VALUES ('2','batteries','2012-06-18','16:20:01','23:40:23','3');
INSERT INTO Department VALUES ('2','remote','2012-06-18','13:20:41','19:40:23','4');
INSERT INTO Department VALUES ('1','computer','2012-06-18','13:20:54','19:40:23','4');
INSERT INTO Department VALUES ('2','dishwasher','2011-06-18','13:20:23','19:40:23','4');
INSERT INTO Department VALUES ('3','lawnmower','2011-06-18','13:20:57','20:40:23','4');
INSERT INTO Department VALUES ('3','lawnmower','2011-06-18','11:20:57','20:40:23','4');
INSERT INTO Department VALUES ('1','mobile','2012-05-18','13:20:31','19:40:23','4');
INSERT INTO Department VALUES ('1','mouse','2012-05-18','13:20:34','19:40:23','4');
INSERT INTO Department VALUES ('1','radio','2012-05-18','13:20:12','19:40:23','4');
INSERT INTO Department VALUES ('2','lawnmowerphones','2012-05-18','13:20:54','19:40:23','4');
INSERT INTO Department VALUES ('2','tv','2012-05-12','06:20:29','09:57:37','1');
INSERT INTO Department VALUES ('2','radio','2011-05-23','11:34:07','15:41:09','2');
INSERT INTO Department VALUES ('1','batteries','2011-05-21','16:20:01','23:40:23','3');
INSERT INTO Department VALUES ('2','remote','2011-05-01','13:20:41','19:40:23','4');
INSERT INTO Department VALUES ('3','mobile','2011-05-09','13:20:31','19:40:23','4');
For department1 the date 2012-05-18 would appear because that date occurred the most. And for every department, it should only show the one with the most sales, and if same amount of sales appears on the same date then both will appear, e.g. Department 1 will appear twice with both the dates of max sales.

I've tested the following query based on the table and two columns you've provided along with sample data. So, let me describe it for you. The inner-most "PREQUERY" is doing a count by department and date. The results of this will be pre-ordered by Department first, THEN the highest count in DESCENDING ORDER (so highest sales count is listed FIRST), it doesn't matter what date the count happened.
Next, by utilizing MySQL #variables, I'm pre-declaring two to be used in the query. #variables are like inline programming with MySQL. They can be declared once and then changed as applied to each record being processed. So, I'm defaulting to a bogus department value and a zero sales count.
Now, I'm grabbing the results of the PreQuery (Dept, #Sales and Date), but now, adding a test. If it is the FIRST ENTRY for a given department, use that record's "NumberOfSales" and put into the #maxSales variable and store as a final column name "MaxSaleCnt". The next column name uses the #lastDept and is set to whatever the current record's Department # is. So it can be compared to the next record.
If the next record is the same department, then it just keeps whatever the #maxSales value was from the previous, thus keeping the same first count(*) result for ALL entries on each respective department.
Now, the closure. I've added a HAVING clause (not a WHERE as that restricts what records get tested, but HAVING processes AFTER the records are part of the PROCESSED set. So now, it would have all 5 columns. I am saying ONLY KEEP those records where the final NumberOfSales for the record MATCHES the MaxSaleCnt for the department. If one, two or more dates, no problem it returns them all per respective department.
So, one department could have 5 dates with 10 sales each, and another department has 2 dates with only 3 sales each, and another with only 1 date with 6 sales.
select
Final.DepartmentNo,
Final.NumberOfSales,
Final.sDate
from
(select
PreQuery.DepartmentNo,
PreQuery.NumberOfSales,
PreQuery.sDate,
#maxSales := if( PreQuery.DepartmentNo = #lastDept, #maxSales, PreQuery.NumberOfSales ) MaxSaleCnt,
#lastDept := PreQuery.DepartmentNo
from
( select
D.DepartmentNo,
D.sDate,
count(*) as NumberOfSales
from
Department D
group by
D.DepartmentNo,
D.sDate
order by
D.DepartmentNo,
NumberOfSales DESC ) PreQuery,
( select #lastDept := '~',
#maxSales := 0 ) sqlvars
having
NumberOfSales = MaxSaleCnt ) Final
To clarify the "#" and "~" per you final comment. The "#" indicates a local variable to the program (or in this case and in-line sql variable) that can be used in the query. The '~' is nothing more than a simple string that probability would never exist that of any of your departments, so when it is compared to the first qualified record, does an IF( '~' = YourFirstDepartmentNumber, then use this answer, otherwise use this answer).
Now, how do the above work. Lets say the following is the results of your data returned by the inner-most query, grouped and ordered by the most sales at the top going down... SLIGHTLY altered from your data, lets just assume the following to simulate multiple dates on Dept 2 that have the same sales quantity...
Row# DeptNo Sales Date # Sales
1 1 2012-05-18 3
2 1 2012-06-18 2
3 1 2012-05-20 1
4 2 2012-06-18 4
5 2 2011-05-23 4
6 2 2012-05-18 2
7 2 2012-05-12 1
8 3 2011-06-18 2
9 3 2012-09-14 1
Keep track of the actual rows. The innermost query that finishes as alias "PreQuery" returns all the rows in the order you see here. Then, that is joined (implied) with the declarations of the # sqlvariables (special to MySQL, other sql engines dont do this) and starts their values with the lastDept = '~' and the maxSales = 0 (via assignment with #someVariable := result of this side ).
Now, think of the above being handled as a
DO WHILE WE HAVE RECORDS LEFT
Get the department #, Number of Sales and sDate from the record.
IF the PreQuery Record's Department # = whatever is in the #lastDept
set MaxSales = whatever is ALREADY established as max sales for this dept
This basically keeps the MaxSales the same value for ALL in the same Dept #
ELSE
set MaxSales = the # of sales since this is a new department number and is the highest count
END IF
NOW, set #lastDept = the department you just processed to it
can be compared when you get to the next record.
Skip to the next record to be processed and go back to the start of this loop
END DO WHILE LOOP
Now, the reason you need to have the #MaxSales and THEN the #LastDept as returned columns is they must be computed for each record to be used to compare to the NEXT record. This technique can be used for MANY application purposes. If you click on my name, look at my tags and click on the MySQL tag, it will show you the many MySQL answers I've responded to. Many of them do utilize # sqlvariables. In addition, there are many other people who are very good at working queries, so dont just look in one place. As for any question, if you find a good answer that you find helpful, even if you didn't post the question, clicking on an up-arrow next to the answer helps others indicate what really helped them understand and get resolution to questions -- again, even if its not your question. Good luck on your MySQL growth.

I think this can be achieved with a single query, but my experiences for similar functionality have involved either WITH (as defined in SQL'99) using either Oracle or MSSQL.
The best (only?) way to approach a problem like this is to break in into smaller components. (I don't think your provided statement provides all columns, so I'm going to have to make a few assumptions.)
First, how many sales were made for each day for each group:
SELECT department, COUNT(1) AS dept_count, sale_date
FROM orders
GROUP BY department, sale_date
Next, what's the most sales for each department
SELECT tmp.department, MAX(tmp.dept_count)
FROM (
SELECT department, COUNT(1) AS dept_count
FROM orders
GROUP BY department
) AS tmp
GROUP BY tmp.department
Finally, putting the two together:
SELECT a.department, a.dept_count, b.sale_date
FROM (
SELECT tmp.department, MAX(tmp.dept_count) AS max_dept_count
FROM (
SELECT department, COUNT(1) AS dept_count
FROM orders
GROUP BY department
) AS tmp
GROUP BY tmp.department
) AS a
JOIN (
SELECT department, COUNT(1) AS dept_count, sale_date
FROM orders
GROUP BY department, sale_date
) AS b
ON a.department = b.department
AND a.max_dept_count = b.dept_count

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FROM Leader a
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SELECT leader_id
, AVG(score) AverageScore
FROM Shift
INNER JOIN
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AND shift_time = visit_time --depends on how you are recording time what this really needs to be
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id
name
leaders_timetabke (can be multiple per leader)
id,
leader_id
shift_datetime (I assume it stores date and hour here, minutes and seconds are always 0
survey_scores
id,
visit_datetime
score
SELECT l.id, l.name, AVG(s.score) FROM leaders l
INNER JOIN leaders_timetable lt ON lt.leader_id = l.id
INNER JOIN survey_scores s ON lt.shift_datetime=DATE_FORMAT('Y-m-d H:00:00', s.visit_datetime)
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(1....1....1)
(2....1....2)
(3....1....3)
(4....2....2)
(5....2....3)
(6....2....5)
(7....3....2)
(8....3....3)
(9....3....4)
(10...3....5)
(11...4....3)
(12...4....4)
(13...4....5)
etc
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2002-2003
2003-2004
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HAVING COUNT(CSYWFY.S2_Customer_ID) > 14
HTH

Complex MySQL COUNT query

Evening folks,
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Thanks!
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COUNT(*) AS `total_count`
FROM `income`
GROUP BY `reference_id`