Sql Postgres 在结果表中组合 COUNTS 和 GROUP BY

https://dbfiddle.uk/z01O9eNR个

我有下表:

CREATE TABLE fruits (
    id INT PRIMARY KEY,
    fruit_name VARCHAR,
    purchase_date DATE
);

以下为表格中的数据:

+------------------+
| Results Table 1a |
+----+------------+---------------+
| id | fruit_name | purchase_date |
+----+------------+---------------+
|  1 | apple      |   2021-02-02  |
|  2 | pear       |   2021-02-02  |
|  3 | orange     |   2021-02-02  |
|  4 | apple      |   2021-02-02  |
|  5 | grapes     |   2021-02-04  |
|  6 | grapes     |   2021-02-05  |
+----+------------+---------------+

如何创建一条不仅返回表内容的SELECT语句，如下所示:

SELECT *
FROM fruits;

但也给了我一个额外的专栏，告诉我how many purchases were made the same date.我想要一个如下所示的结果表:

+------------------+
| Results Table 1b |
+----+------------+---------------+--------------------+
| id | fruit_name | purchase_date | same_day_purchases |
+----+------------+---------------+--------------------+
|  1 | apple      |   2021-02-02  |                  4 |
|  2 | pear       |   2021-02-02  |                  4 |
|  3 | orange     |   2021-02-02  |                  4 |
|  4 | apple      |   2021-02-02  |                  4 |
|  5 | grapes     |   2021-02-04  |                  1 |
|  6 | grapes     |   2021-02-05  |                  1 |
+----+------------+---------------+--------------------+

例如:在2021-02-02天有4笔购买，所以对于那一天的每purchase_date笔交易，same_day_purchases应该是4笔.

我可以使用GROUP BY创建单独的语句:

SELECT purchase_date, COUNT(*) AS same_date_purchases
FROM fruits
GROUP BY purchase_date;

但我正在寻找一种方法来声明启用Results Table 1b的结果表.