一个桶不存在数据时，如何选择桶中的SQL数据？

Question

I'm trying to get a complete set of buckets for a given dataset, even if no records exist for some buckets .

For example, I want to display totals by day of week, with zero total for days with no records.

SELECT
WEEKDAY(transaction_date) AS day_of_week,
SUM(sales) AS total_sales
FROM table1
GROUP BY day_of_week

If I have sales every day, I'll get 7 rows in my result representing total sales on days 0-6.

If I don't have sales on Day 2, I get no result for Day 2.

What's the most efficient way to force a zero value for day 2?

Should I join to a temporary table or array of defined buckets? ['0','1','2','3','4','5','6']

Or is it better to insert zeros outside of MySQL, after I've done the query?

I am using MySQL, but this is a general SQL question.

Answer 1

Basically you want the answer to be 0 when the data is actually null for that bucket, therefore you want the max(null, 0). A max function wouldn't natively work with NULL in this way, however, you can use COALESCE to force it:

COALESCE(MAX(SUM(sales)),0)

as suggested by this answer

Answer 2

First off you need a calendar table; something like this or this . Or create calendar subset on the fly. I am not sure of the mySQL syntax, but here is what it would look like in SQL Server.

DECLARE
    @FromDate DATE
  , @ToDate   DATE

-- set these variables to appropriate values
SET @FromDate = '2020-03-01';
SET @ToDate = '2020-03-31';

;WITH cteCalendar (MyDate) AS
(
    SELECT CONVERT(DATE, @FromDate) AS MyDate
    UNION ALL
    SELECT DATEADD(DAY, 1, MyDate)
    FROM   cteCalendar
    WHERE  DATEADD(DAY, 1, MyDate) <= @ToDate
)
SELECT WEEKDAY(cte.MyDate) AS day_of_week,
SUM(sales) AS total_sales
FROM cteCalendar cte
LEFT JOIN table1 t1 ON cte.MyDate = t1.transaction_date
GROUP BY day_of_week

Answer 3

In MySQL, you could simply use a derived table of numbers from 1 to 7 , left join it with the table, then aggregate:

select d.day_of_week, sum(sales) AS total_sales
from (
    select 1 day_of_week union all select 2 union all select 3 union all select 4
    union all select 5 union all select 6 union all select 7
) d
left join table1 t1 on weekday(t1.transaction_date) = d.day_of_week
group by day_of_week

Very recent versions have the values(row...) syntax, which shorten the query:

select d.day_of_week, sum(sales) AS total_sales
from (values row(1), row(2), row(3), row(4), row(5), row(6), row(7)) d(day_of_week)
left join table1 t1 on weekday(t1.transaction_date) = d.day_of_week
group by day_of_week