需要邏輯來找到 mysql 中同一表中另一列的最小值的一列的值

Question

我寫了下面的代碼來得到這樣的

SELECT `temp2`.`UserId` AS `UserId`,
              `temp2`.`ScheduleDate` AS `ScheduleDate`,
             `temp2`.`daystodue`
  FROM (
SELECT
      `temp1`.`UserId` AS `UserId`,
      `temp1`.`ScheduleDate` AS `ScheduleDate`,
      `temp1`.`DueDate` AS `DueDate`,
      `temp1`.`CompletedState` AS `CompletedState`,
      IF(((`temp1`.`CompletedState` = 0) AND ((`temp1`.`ScheduleDate` - CURDATE()) <= 0)), (TO_DAYS(`temp1`.`DueDate`) - TO_DAYS(CURDATE())), 0) AS `daystodue`
    FROM (SELECT
        `fd_dw`.`ComplianceFactTable`.`UserId` AS `UserId`,
        CAST(`fd_dw`.`ComplianceFactTable`.`CourseModule_ScheduleDateID` AS date) AS `ScheduleDate`,
        CAST(`fd_dw`.`ComplianceFactTable`.`CourseModule_dueDateID` AS date) AS `DueDate`,
        `fd_dw`.`ComplianceFactTable`.`CourseModuleComplete_completionstate` AS `CompletedState`
      FROM `fd_dw`.`ComplianceFactTable`
      WHERE ((`fd_dw`.`ComplianceFactTable`.`CourseModule_dueDateID` > 0)
      AND ((CAST(`fd_dw`.`ComplianceFactTable`.`CourseModule_dueDateID` AS date) - CURDATE()) > 0))) `temp1`) `temp2` 
  WHERE `temp2`.`UserId` IN (223699,223741,223780,223678,243988,380316,388737,121896,491562)

當我執行上述查詢時，我得到了這個 output

ID       Date      value
121896  2019-12-06  0
121896  2019-11-06  0
121896  2020-01-06  0
223678  2019-12-23  0
223678  2019-11-23  0
223678  2020-01-23  0
223678  2019-10-23  43
223699  2019-12-23  0
223699  2019-11-23  0
223699  2020-01-23  0
223699  2019-10-23  43
223741  2019-12-23  0
223741  2019-11-23  0
223741  2020-01-23  0
223741  2019-10-23  43
223780  2019-12-23  0
223780  2019-11-23  0
223780  2020-01-23  0
223780  2019-10-23  43
243988  2019-10-15  21
243988  2020-01-15  0
243988  2019-12-15  0
380316  2019-10-05  0
380316  2019-11-05  0
380316  2019-12-05  0
380316  2020-01-05  0
388737  2019-10-23  29
388737  2019-11-23  0
388737  2020-01-23  0
388737  2019-12-23  0
491562  2019-10-17  7
491562  2019-10-17  7
491562  2019-10-17  23
491562  2019-11-17  0
491562  2019-12-17  0
491562  2020-01-17  0
491562  2019-10-17  7
491562  2019-10-17  7

但我想要這樣的東西

121896  2019-11-06  0
223678  2019-10-23  43
223699  2019-10-23  43
223741  2019-10-23  43
223780  2019-10-23  43
243988  2019-10-15  21
380316  2019-10-05  0
388737  2019-10-23  29
491562  2019-10-17  7

Answer 1

我了解，對於每個 ID，您都希望以最短日期記錄。 當有多個具有最短日期的記錄時，您需要具有最短日期的記錄。

在 MySQL 8.0 中，您可以使用 window 函數：

select id, date, value
from (
    select 
        t.*,
        row_number() over(partition by id order by date, value) rn
    from mytable t
) t
where rn = 1

在早期版本中，您可以使用相關的 suquery 進行過濾：

select id, date, min(value) value
from mytable t
where t.date = (
    select t1.date from mytable t1 where t1.id = t.id order by t1.date, t1.value limit 1
)
group by id, date

或者：

select id, date, min(value) value
from mytable t
where t.date = (select min(t1.date) from mytable t1 where t1.id = t.id)
group by id, date

在這個關於 DB Fiddle 的演示中，所有 3 個查詢都返回：

| id     | date       | value |
| ------ | ---------- | ----- |
| 121896 | 2019-11-06 | 0     |
| 223678 | 2019-10-23 | 43    |
| 223699 | 2019-10-23 | 43    |
| 223741 | 2019-10-23 | 43    |
| 223780 | 2019-10-23 | 43    |
| 243988 | 2019-10-15 | 21    |
| 380316 | 2019-10-05 | 0     |
| 388737 | 2019-10-23 | 29    |
| 491562 | 2019-10-17 | 7     |

Answer 2

Select min(column1), min(column2) from yourtablename

Answer 3

如果您沒有 mysql 8 使用相關查詢

  SELECT *
  FROM yourTable t1
  WHERE date = (SELECT MIN(date)
                FROM yourTable t2
                WHERE t1.user_id = t2.user_id)

Answer 4

您預期的 output 與您的問題標題不同，因為它包含第一列的每個值，第二列（日期）中具有最小值的行。
使用NOT EXISTS ：

select t.* from tablename t
where not exists (
  select 1 from tablename
  where id = t.id and date < t.date
)