重複數據加入MySQL查詢

Question

可能是重複的問題... 這裏有兩個表tasks_0_tasks_staff和tasks_0_staff_time_management：

 
tasks_0_tasks_staff 

+----+---------+---------+ 
| id | task_id | user_id | 
+----+---------+---------+ 
| 1 |  1 |  348 | 
| 2 |  1 |  350 | 
| 3 |  2 |  350 | 
| 4 |  2 |  351 | 
| 5 |  2 |  357 | 
+----+---------+---------+ 

和

 
tasks_0_staff_time_management 
+----+---------+---------+-------------+-------+ 
| id | user_id | task_id | day_of_week | hours | 
+----+---------+---------+-------------+-------+ 
| 1 |  350 |  1 |   2 |  2 | 
| 2 |  350 |  1 |   3 |  3 | 
+----+---------+---------+-------------+-------+ 

當我執行以下查詢它顯示左連接表中的重複數據...我如何顯示空（這是實際的情況下）...

 
SELECT tasks_0_tasks_staff.`task_id`, 
tasks_0_staff_time_management.day_of_week, 
tasks_0_staff_time_management.hours 

FROM tasks_0_tasks_staff 

LEFT JOIN `tasks_0_staff_time_management` ON `tasks_0_staff_time_management`.`user_id`= `tasks_0_tasks_staff`.`user_id` 

WHERE `tasks_0_tasks_staff`.`user_id`= 350 

ORDER BY `tasks_0_tasks_staff`.`task_id` ; 

results: 
+---------+-------------+-------+ 
| task_id | day_of_week | hours | 
+---------+-------------+-------+ 
|  1 |   2 |  2 | 
|  1 |   3 |  3 | 
|  2 |   2 |  2 | 
|  2 |   3 |  3 | 
+---------+-------------+-------+ 

爲什麼結果最後兩行是顯示DAY_OF_WEEK和小時列值，因爲這些不存在於左連接表？這兩行怎麼顯示NULL值...？

我期待的結果是這樣的

 
+---------+-------------+-------+ 
| task_id | day_of_week | hours | 
+---------+-------------+-------+ 
|  1 |   2 |  2 | 
|  1 |   3 |  3 | 
|  2 |  NULL | NULL | 
|  2 |  NULL | NULL | 
+---------+-------------+-------+ 

Answer 1

你的代碼應該添加task_id作爲條件你的加入：

SELECT tasks_0_tasks_staff.`task_id`, 
    tasks_0_staff_time_management.day_of_week, 
    tasks_0_staff_time_management.hours 
FROM tasks_0_tasks_staff 
LEFT JOIN `tasks_0_staff_time_management` ON `tasks_0_staff_time_management`.`user_id`= `tasks_0_tasks_staff`.`user_id` AND `tasks_0_staff_time_management`.`task_id`= `tasks_0_tasks_staff`.`task_id` 
WHERE `tasks_0_tasks_staff`.`user_id`= 350 
ORDER BY `tasks_0_tasks_staff`.`task_id` ; 

還是有一些語義SQL：

SELECT tasks_0_tasks_staff.`task_id`, 
    tasks_0_staff_time_management.day_of_week, 
    tasks_0_staff_time_management.hours 
FROM tasks_0_tasks_staff 
LEFT JOIN `tasks_0_staff_time_management` USING (`task_id`, `user_id`) 
WHERE `tasks_0_tasks_staff`.`user_id`= 350 
ORDER BY `tasks_0_tasks_staff`.`task_id` ; 

這裏的概念SQL搗鼓一點證明：

http://sqlfiddle.com/#!9/2a2425/1

Answer 2

您需要在添加一個條件的地方，使其只匹配具有相同TASK_ID記錄。

請注意，重複的結果意味着不同的相關行是相同的，即您的情況爲而不是。

Answer 3

您應該使用「INNER JOIN」。

SELECT tasks_0_tasks_staff.`task_id`, 
tasks_0_staff_time_management.day_of_week, 
tasks_0_staff_time_management.hours 

FROM tasks_0_tasks_staff 

INNER JOIN `tasks_0_staff_time_management` ON `tasks_0_staff_time_management`.`user_id`= `tasks_0_tasks_staff`.`user_id` 

WHERE `tasks_0_tasks_staff`.`user_id`= 350 

ORDER BY `tasks_0_tasks_staff`.`task_id` ; 

Answer 4

你必須改變這樣的

SELECT tasks_0_tasks_staff.`task_id`, 
tasks_0_staff_time_management.day_of_week, 
tasks_0_staff_time_management.hours 

FROM tasks_0_tasks_staff 

LEFT JOIN `tasks_0_staff_time_management` ON `tasks_0_staff_time_management`.`user_id`= `tasks_0_tasks_staff`.`user_id` 

WHERE `tasks_0_tasks_staff`.`user_id`= 350 AND `tasks_0_tasks_staff`.`task_id` = tasks_0_staff_time_management.`task_id` 

ORDER BY `tasks_0_tasks_staff`.`task_id` ;