如果使用GROUP BY，则MySQL计数

我正在尝试解决LeetCode问题（）：

我已在本地生成了

调查日志

：

mysql> select * from survey_log;
+------+--------+-------------+-----------+-------+-----------+
| uid  | action | question_id | answer_id | q_num | timestamp |
+------+--------+-------------+-----------+-------+-----------+
|    5 | show   |         285 |      NULL |     1 |       123 |
|    5 | answer |         285 |    124124 |     1 |       124 |
|    5 | show   |         369 |      NULL |     2 |       125 |
|    5 | skip   |         369 |      NULL |     2 |       126 |
+------+--------+-------------+-----------+-------+-----------+

我想使用这个辅助表：

mysql> select question_id, if(action='show', 1, 0) as is_show, if(action='answer', 1, 0) as is_answer from survey_log;
+-------------+---------+-----------+
| question_id | is_show | is_answer |
+-------------+---------+-----------+
|         285 |       1 |         0 |
|         285 |       0 |         1 |
|         369 |       1 |         0 |
|         369 |       0 |         0 |
+-------------+---------+-----------+

接下来，我要做的是获得每个

问题id

的

is\u show

和

is\u answer

列的总和。我认为这会起作用：

mysql> select question_id, count(if(action = 'show', 1, 0)) as show_count, count(if(action = 'answer', 1, 0)) as answer_count from survey_log group by question_id;
+-------------+------------+--------------+
| question_id | show_count | answer_count |
+-------------+------------+--------------+
|         285 |          2 |            2 |
|         369 |          2 |            2 |
+-------------+------------+--------------+

但是，这并没有给我预期的输出

+-------------+------------+--------------+
| question_id | show_count | answer_count |
+-------------+------------+--------------+
|         285 |          1 |            1 |
|         369 |          1 |            0 |
+-------------+------------+--------------+

最后一个查询有什么问题？我已经看过了，但似乎无法将其应用于这个问题。

只是计算非空值。因此

0

和

1

都将被计算在内。您需要做的是：

请注意，由于MySQL在数字上下文中将布尔值视为0（false）或1（true），因此可以删除查询的

if

部分：

select question_id, 
       sum(action = 'show') as show_count,
       sum(action = 'answer') as answer_count
from survey_log
group by question_id;

select question_id, 
       sum(action = 'show') as show_count,
       sum(action = 'answer') as answer_count
from survey_log
group by question_id;