计算日期之间的分钟数并获得前10名

Question

So I have a table that holds two different dates and I am selecting the minutes difference between:

    select customerID, customers.telNumber,
   sum(round((enddate - startdate) * 1440)) over (partition by telNumber) total_mins
    from table;

And after that I want to get only the top 5 that have the highest amount of minutes, something like

     rank() over (partition by total_mins order by total_mins)

How would one go about doing that?

Answer 1

Something like this should work for you:

SELECT * 
FROM (
  SELECT customerId, telNumber, rank() over (order by total_mins) rnk
  FROM (
    SELECT customerId,telNumber,
     sum(round((enddate - startdate) * 1440)) over (partition by telNumber) total_mins
    FROM YourTable
  ) t
) t
WHERE rnk <= 10

This will get you ties, so it could return more than 10 rows. If you only want to return 10 rows, use ROW_NUMBER() instead of RANK() .

SQL Fiddle Demo

Answer 2

I would add to sgeddes's example that the combination of rank() and row_number() is the best as rank() may return the same rank values for all or few rows. But row_number() will always be different. I'd use row_number() in Where clause, not rank().