MySQL-如何計算當前行中先前出現的次數

Question

帶有下表：

+----------+------+
| Platform | day  |
+----------+------+
| 1670254  |    0 |
| 1670254  |    0 |
| 1665087  |    0 |
| 1670254  |    0 |
| 1670254  |    1 |
| 1670254  |    1 |
| 1670254  |    1 |
| 1670254  |    1 |
| 1670254  |    1 |
| 1665160  |    1 |
| 1670254  |    1 |
| 1670254  |    2 |
| 1670254  |    2 |
| 1670254  |    3 |
| 1670254  |    4 |
| 1667145  |    5 |

我想添加第三列，該列計算先前日期的出現次數。 例如，在第1天，平台1670254執行了3次先前的操作。

結果應顯示為：

+----------+------+-----------+
| Platform | day  |  Previous | 
+----------+------+-----------+
| 1670254  |    0 |     0     |
| 1670254  |    0 |     0     |
| 1665087  |    0 |     0     |
| 1670254  |    0 |     0     |
| 1670254  |    1 |     3     |
| 1670254  |    1 |     3     |
| 1670254  |    1 |     3     |
| 1670254  |    1 |     3     |
| 1670254  |    1 |     3     |
| 1665160  |    1 |     0     |
| 1670254  |    1 |     3     |
| 1670254  |    2 |     9     |
| 1670254  |    2 |     9     |
| 1670254  |    3 |    11     |
| 1670254  |    4 |    12     |
| 1667145  |    5 |     0     |

Answer 1

只需一個相關查詢即可獲得每個平台的前一天。

SQL演示

   SELECT *, (SELECT COUNT(Platform) 
              FROM YourTable t2 
              WHERE t1.Platform = t2.Platform
                AND t1.day > t2.day) as Previous
   FROM YourTable t1