Question

我有以下问题

SELECT * from map_user_sticker WHERE map_user_sticker.user_id = 7;
+----+---------+------------+
| id | user_id | sticker_id |
+----+---------+------------+
| 35 |       7 |         55 |
|  3 |       7 |         30 |
| 32 |       7 |         49 |
| 33 |       7 |         52 |
| 34 |       7 |         43 |
| 36 |       7 |         50 |
+----+---------+------------+
6 rows in set (0.00 sec)


SELECT * FROM sticker;
+----+--------------------------------------------------+
| id | word                                             |
+----+--------------------------------------------------+
| 40 | I love Sonal                                     |
| 41 | Add User to Database                             |
| 39 | This is a dream Project                          |
| 33 | Narendra Sisodiya                                |
| 34 | Sourabh Parmar                                   |
| 30 | Sonal Sisodiya                                   |
| 42 | I love India                                     |
| 43 | I love Linux                                     |
| 44 | I hate Congress                                  |
| 45 | I love jQuery                                    |
| 48 | Modi will be the PM                              |
| 47 | Ramdev Baba is my Super Hero                     |
| 49 | हिन्दी से प्यार है |
| 50 | Linux is better then Windows                     |
| 52 | I am from Sehore                                 |
| 55 | I have 2 little kids - sadu and sonu             |
+----+--------------------------------------------------+
16 rows in set (0.00 sec)

我基本上想要生成像

这样的结果

+----+--------------------------------------------------+----------+
| id | word                                             |  present |
+----+--------------------------------------------------+----------+
| 40 | I love Sonal                                     |  0       |
| 41 | Add User to Database                             |  0       |
| 39 | This is a dream Project                          |  0       |
| 33 | Narendra Sisodiya                                |  0       |
| 34 | Sourabh Parmar                                   |  0       |
| 30 | Sonal Sisodiya                                   |  1       |
| 42 | I love India                                     |  0       |
| 43 | I love Linux                                     |  1       |
| 44 | I hate Congress                                  |  0       |
| 45 | I love jQuery                                    |  0       |
| 48 | Modi will be the PM                              |  0       |
| 47 | Ramdev Baba is my Super Hero                     |  0       |
| 49 | हिन्दी से प्यार है                                      |   1      |
| 50 | Linux is better then Windows                     |   1      |
| 52 | I am from Sehore                                 |   1      |
| 55 | I have 2 little kids - sadu and sonu             |   1      |
+----+--------------------------------------------------+----------+
16 rows in set (0.00 sec)

请帮帮我!! 我想根据以下条件生成一个名为“present”的额外列如果来自查询2的来自query1 == id的sticker_id，则present = 1 否则出现= 0

Answer 1

这里是我想要的答案！

答案1 -

SELECT s.* , IF(FIND_IN_SET(7,(GROUP_CONCAT(m.user_id))) != 0 , 1,0) as present, COUNT(m.user_id) as totalUsers 
FROM sticker s   
LEFT JOIN map_user_sticker m  
ON s.id = m.sticker_id 
GROUP BY id;

答案2 -

SELECT id,word, (case when 7 in (select user_id from map_user_sticker WHERE map_user_sticker.sticker_id=sticker.id) then 1 else 0 end) as present 
from sticker;

我不知道哪一个更快更好

Answer 2

试试这个：

select id,word,
case(when id in (select sticker_id from map_user_sticker) then 1 else 0 end) 
as present from sticker;

Answer 3

希望，这会有所帮助：

SELECT sticker.id, sticker.word, (IF(sticker_id IS NULL, 0, 1)) AS present FROM sticker LEFT JOIN map_user_sticker ON (sticker.id = map_user_sticker.sticker_id)

Answer 4

尝试此查询 -

SELECT s.*, IF(COUNT(m.sticker_id) > 0, 1, 0) present FROM sticker s
  LEFT JOIN map_user_sticker m
    ON s.id = m.sticker_id
GROUP BY
  s.id;

这是你想要的吗？

SELECT s.*, IF(m.sticker_id IS NULL, 0, 1) present FROM sticker s
  LEFT JOIN map_user_sticker m
    ON s.id = m.sticker_id
WHERE m.user_id = 7;

MySQL根据另一个查询在结果上添加额外的列

4 个答案: