SQL 查询以在每个月的第一天到最后一天的列中显示每日计数结果

Question

Need a Query to Display daily count of each item bought by customers in columns from 1st day of month to last day

Sample data table "Item"

+--------+--------+----------+---------------+
| Purchase Date | Item Code| Item Name| Price|
|--------+--------+----------+--------------+
| 01-JAN-20     | 11       | Apple   | 1    |
| 01-JAN-20     | 11       | Apple   | 1    |
| 02-JAN-20     | 12       | Orange  | 2    |
| 02-JAN-20     | 11       | Apple   | 1    |
| 03-JAN-20     | 12       | Orange  | 2    |
| 03-JAN-20     | 12       | Orange  | 2    |
| 04-JAN-20     | 12       | Orange  | 2    |
| 04-JAN-20     | 11       | Apple   | 1    |
+--------+--------+----------+--------------+

SQL Query should Display Daily Count using Item code and Result to be displayed as below table. Count daily with each day displayed in column base on the day eg If today is 4th of Jan then count tomorrow will create new column with count result and continues until last day of month or something similar.

+--------+--------+----------+---------------+
| Items  | Jan 01| Jan 02| Jan 03|Jan 04| etc
+--------+--------+----------+--------------+
| Apple  | 2     |   1   |   2   |   1  |
| Orange | 0     |   1   |   0   |   1  |
+--------+--------+----------+--------------+

Answer 1

If you know what dates you want, you can use conditional aggregation:

select item,
       sum(case when purchase_date = '2020-01-01' then 1 else 0 end) as jan_1,
       sum(case when purchase_date = '2020-01-02' then 1 else 0 end) as jan_2,
       sum(case when purchase_date = '2020-01-03' then 1 else 0 end) as jan_3,
       sum(case when purchase_date = '2020-01-04' then 1 else 0 end) as jan_4,
       . . .
from items
group by item;

Note that this assumes that purchase_date is really stored as an internal date format. So the comparison is a date constant -- however, that might differ among databases.

If you do not have a specific set of dates in mind, then you will need to use dynamic SQL.