为什么不能连接 id 上行数不同的 2 个表？
why can't join 2 tables with different num of rows on id?
我需要将 table A 的时间除以 table B 以找到以任务的百分比表示的相对完成时间。
时间栏显示学生完成任务所需的时间。
如果总时间为2分钟，尝试时间为1分钟，则相对时间为50%
一个 table 有 30 行 - 这是一个显示总时间的派生 table，
另外 59 行 - 没有求和的每一步花费的时间，
公共字段名为 st_id (step_id)。
第table个甲：
+-------+---------+
| st_id | time |
+-------+---------+
| 10 | 0:00:28 |
| 11 | 0:01:51 |
| 12 | 0:10:09 |
| 13 | 0:03:44 |
| 14 | 0:16:44 |
| 15 | 0:19:12 |
| 16 | 0:03:43 |
| 17 | 0:07:56 |
| 18 | 0:06:19 |
| 19 | 0:31:14 |
| 20 | 0:11:50 |
| 81 | 0:14:12 |
| 82 | 0:18:59 |
| 83 | 0:30:24 |
| 84 | 0:11:50 |
| 85 | 0:40:35 |
| 86 | 1:28:57 |
| 87 | 0:24:10 |
| 88 | 0:51:53 |
| 100 | 0:05:31 |
| 101 | 0:13:41 |
| 103 | 0:11:50 |
| 104 | 0:22:18 |
| 105 | 0:35:18 |
| 106 | 0:55:02 |
| 107 | 0:42:02 |
| 108 | 0:08:52 |
| 109 | 1:07:07 |
| 110 | 0:11:50 |
| 111 | 0:11:55 |
+-------+---------+
另一个 table 有 59 行 - table B:
+------------+---------+---------------+
| student_id | st_id | time |
+------------+---------+---------------+
| 59 | 10 | 0:00:28 |
| 59 | 11 | 0:01:11 |
| 59 | 12 | 0:06:09 |
| 59 | 13 | 0:02:24 |
| 59 | 14 | 0:00:05 |
| 59 | 14 | 0:00:04 |
| 59 | 14 | 0:00:20 |
| 59 | 14 | 0:00:23 |
| 59 | 14 | 0:09:42 |
| 59 | 14 | 0:00:10 |
| 59 | 15 | 0:11:44 |
| 59 | 15 | 0:00:08 |
| 59 | 16 | 0:02:23 |
| 59 | 17 | 0:04:53 |
| 59 | 17 | 0:00:23 |
| 59 | 18 | 0:00:07 |
| 59 | 18 | 0:03:19 |
| 59 | 18 | 0:00:53 |
| 59 | 19 | 0:00:07 |
| 59 | 19 | 0:00:29 |
| 59 | 19 | 0:00:10 |
| 59 | 19 | 0:00:06 |
| 59 | 19 | 0:00:04 |
| 59 | 19 | 0:00:07 |
| 59 | 19 | 0:00:24 |
| 59 | 19 | 0:00:08 |
| 59 | 19 | 0:00:08 |
| 59 | 19 | 0:02:11 |
| 59 | 19 | 0:00:10 |
| 59 | 19 | 0:15:50 |
| 59 | 20 | 0:07:10 |
| 59 | 81 | 0:08:52 |
| 59 | 82 | 0:11:31 |
| 59 | 82 | 0:00:08 |
| 59 | 83 | 0:00:04 |
| 59 | 83 | 0:00:08 |
| 59 | 83 | 0:00:51 |
| 59 | 83 | 0:00:15 |
| 59 | 83 | 0:17:46 |
| 59 | 84 | 0:07:10 |
| 59 | 85 | 0:24:35 |
| 59 | 86 | 0:53:14 |
| 59 | 86 | 0:00:23 |
| 59 | 87 | 0:07:10 |
| 59 | 87 | 0:07:40 |
| 59 | 88 | 0:31:13 |
| 59 | 100 | 0:03:31 |
| 59 | 101 | 0:08:21 |
| 59 | 103 | 0:07:10 |
| 59 | 104 | 0:13:38 |
| 59 | 105 | 0:21:18 |
| 59 | 106 | 0:33:02 |
| 59 | 107 | 0:07:10 |
| 59 | 107 | 0:18:12 |
| 59 | 108 | 0:05:32 |
| 59 | 109 | 0:40:27 |
| 59 | 110 | 0:07:10 |
| 59 | 111 | 0:07:15 |
+------------+---------+---------------+
我试图通过 step_id 进行内部连接，但它给了我错误：
“派生 table 具有不同的行数”
但我只是想加入那些存在于 table 中的那些？这就是内连接的工作原理？
我在 with 子句中有两个 tables，它们提取 student_id = 59:
的尝试时间数据
with step_helper(st_id, t) as (
select step_id, sec_to_time(sum(t)) as total_time from (select step_id,
case when (submission_time - attempt_time) > 3600 then sec_to_time((select * from s1))
else sec_to_time((submission_time - attempt_time)) end as t
from step_student where student_id = 59)k
group by step_id),
time_attempt(st_id, t) as (
select student.student_id, step.step_id,
case when (submission_time - attempt_time) > 3600 then sec_to_time((select * from s1))
else sec_to_time((submission_time - attempt_time)) end as Время_попытки
from step_student inner join student on student.student_id = step_student.student_id
inner join step on step.step_id = step_student.step_id where student.student_id = 59)
上述 2 table 的内部连接不起作用：
SELECT *FROM time_attempt INNER JOIN step_helper ON time_attempt.st_id = step_helper.st_id
In definition of view, derived table or common table expression,
SELECT list and column names list have different column counts
问题就在这里time_attempt(st_id, t) as (。 time_attempt 的 SELECT 子查询中的列为 3，但您只列出 2。根据 MySQL documentation
The number of names in the list must be the same as the number of columns in the result set.
因此，您的选择是要么列出结果集中的所有三列 time_attempt(st_id, t, ttt) as (，要么根本不列出它们 time_attempt as (.
更新：
Here is an update fiddle 证明 return 错误与行差异无关。
我需要将 table A 的时间除以 table B 以找到以任务的百分比表示的相对完成时间。 时间栏显示学生完成任务所需的时间。
如果总时间为2分钟，尝试时间为1分钟，则相对时间为50%
一个 table 有 30 行 - 这是一个显示总时间的派生 table，
另外 59 行 - 没有求和的每一步花费的时间，
公共字段名为 st_id (step_id)。
第table个甲：
+-------+---------+
| st_id | time |
+-------+---------+
| 10 | 0:00:28 |
| 11 | 0:01:51 |
| 12 | 0:10:09 |
| 13 | 0:03:44 |
| 14 | 0:16:44 |
| 15 | 0:19:12 |
| 16 | 0:03:43 |
| 17 | 0:07:56 |
| 18 | 0:06:19 |
| 19 | 0:31:14 |
| 20 | 0:11:50 |
| 81 | 0:14:12 |
| 82 | 0:18:59 |
| 83 | 0:30:24 |
| 84 | 0:11:50 |
| 85 | 0:40:35 |
| 86 | 1:28:57 |
| 87 | 0:24:10 |
| 88 | 0:51:53 |
| 100 | 0:05:31 |
| 101 | 0:13:41 |
| 103 | 0:11:50 |
| 104 | 0:22:18 |
| 105 | 0:35:18 |
| 106 | 0:55:02 |
| 107 | 0:42:02 |
| 108 | 0:08:52 |
| 109 | 1:07:07 |
| 110 | 0:11:50 |
| 111 | 0:11:55 |
+-------+---------+
另一个 table 有 59 行 - table B:
+------------+---------+---------------+
| student_id | st_id | time |
+------------+---------+---------------+
| 59 | 10 | 0:00:28 |
| 59 | 11 | 0:01:11 |
| 59 | 12 | 0:06:09 |
| 59 | 13 | 0:02:24 |
| 59 | 14 | 0:00:05 |
| 59 | 14 | 0:00:04 |
| 59 | 14 | 0:00:20 |
| 59 | 14 | 0:00:23 |
| 59 | 14 | 0:09:42 |
| 59 | 14 | 0:00:10 |
| 59 | 15 | 0:11:44 |
| 59 | 15 | 0:00:08 |
| 59 | 16 | 0:02:23 |
| 59 | 17 | 0:04:53 |
| 59 | 17 | 0:00:23 |
| 59 | 18 | 0:00:07 |
| 59 | 18 | 0:03:19 |
| 59 | 18 | 0:00:53 |
| 59 | 19 | 0:00:07 |
| 59 | 19 | 0:00:29 |
| 59 | 19 | 0:00:10 |
| 59 | 19 | 0:00:06 |
| 59 | 19 | 0:00:04 |
| 59 | 19 | 0:00:07 |
| 59 | 19 | 0:00:24 |
| 59 | 19 | 0:00:08 |
| 59 | 19 | 0:00:08 |
| 59 | 19 | 0:02:11 |
| 59 | 19 | 0:00:10 |
| 59 | 19 | 0:15:50 |
| 59 | 20 | 0:07:10 |
| 59 | 81 | 0:08:52 |
| 59 | 82 | 0:11:31 |
| 59 | 82 | 0:00:08 |
| 59 | 83 | 0:00:04 |
| 59 | 83 | 0:00:08 |
| 59 | 83 | 0:00:51 |
| 59 | 83 | 0:00:15 |
| 59 | 83 | 0:17:46 |
| 59 | 84 | 0:07:10 |
| 59 | 85 | 0:24:35 |
| 59 | 86 | 0:53:14 |
| 59 | 86 | 0:00:23 |
| 59 | 87 | 0:07:10 |
| 59 | 87 | 0:07:40 |
| 59 | 88 | 0:31:13 |
| 59 | 100 | 0:03:31 |
| 59 | 101 | 0:08:21 |
| 59 | 103 | 0:07:10 |
| 59 | 104 | 0:13:38 |
| 59 | 105 | 0:21:18 |
| 59 | 106 | 0:33:02 |
| 59 | 107 | 0:07:10 |
| 59 | 107 | 0:18:12 |
| 59 | 108 | 0:05:32 |
| 59 | 109 | 0:40:27 |
| 59 | 110 | 0:07:10 |
| 59 | 111 | 0:07:15 |
+------------+---------+---------------+
我试图通过 step_id 进行内部连接，但它给了我错误： “派生 table 具有不同的行数”
但我只是想加入那些存在于 table 中的那些？这就是内连接的工作原理？
我在 with 子句中有两个 tables，它们提取 student_id = 59:
的尝试时间数据with step_helper(st_id, t) as (
select step_id, sec_to_time(sum(t)) as total_time from (select step_id,
case when (submission_time - attempt_time) > 3600 then sec_to_time((select * from s1))
else sec_to_time((submission_time - attempt_time)) end as t
from step_student where student_id = 59)k
group by step_id),
time_attempt(st_id, t) as (
select student.student_id, step.step_id,
case when (submission_time - attempt_time) > 3600 then sec_to_time((select * from s1))
else sec_to_time((submission_time - attempt_time)) end as Время_попытки
from step_student inner join student on student.student_id = step_student.student_id
inner join step on step.step_id = step_student.step_id where student.student_id = 59)
上述 2 table 的内部连接不起作用：
SELECT *FROM time_attempt INNER JOIN step_helper ON time_attempt.st_id = step_helper.st_id
In definition of view, derived table or common table expression, SELECT list and column names list have different column counts
问题就在这里time_attempt(st_id, t) as (。 time_attempt 的 SELECT 子查询中的列为 3，但您只列出 2。根据 MySQL documentation
The number of names in the list must be the same as the number of columns in the result set.
因此，您的选择是要么列出结果集中的所有三列 time_attempt(st_id, t, ttt) as (，要么根本不列出它们 time_attempt as (.
更新：
Here is an update fiddle 证明 return 错误与行差异无关。