SQL Group by Day, Show Orders for Each Day

SQL group by day, with count

I like this approach in (MS SQL):

SELECT 
Convert(char(8), LogDate, 112),
count(distinct RefundId)
FROM RefundProcessing
GROUP BY Convert(char(8), LogDate, 112)

Top 3 services used for each day - SQL

You can use window functions with aggregation. Something like this:

SELECT b.*
FROM (SELECT date, products_name AS product,
COUNT(*) AS NumberOfBookings,
ROW_NUMBER() OVER (PARTITION BY date ORDER BY COUNT(*) DESC) as seqnum
FROM bookings
WHERE date >= '2021-06-01' AND date < '2021-07-01'
GROUP BY products_name, date
) b
WHERE seqnum <= 3
ORDER BY Date, seqnum;

Note: This uses standard date formats for the date constants.

Group by day and still show days without rows?

In order to accomplish this, you need to have a table (or derived table) which contains the dates that you can then join from, using a LEFT JOIN.

SQL operates on the concept of mathematical sets, and if you don't have a set of data, there is nothing to SELECT.

If you want more details, please comment accordingly.

SQL query for calculate cont of orders for each day

I think you want:

SELECT DATE(time_order), COUNT(*)
FROM orders
WHERE DATE(time_order) >= DATE(NOW()) - INTERVAL 7 DAY
GROUP BY DATE(time_order);

That is, your version is outputting the date for each time_order. However, the GROUP BY includes the time, so you are getting a row for each date/time.

MySQL Query GROUP BY day / month / year

GROUP BY YEAR(record_date), MONTH(record_date)

Check out the date and time functions in MySQL.



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