MySQL SELECT 查看 returns 不同的结果
MySQL SELECT to VIEW returns different results
我整理了一个小的 sql 查询,它从一个 table 中提取数据并将其排序在新的列名下。 sql 看起来像这样:
SELECT course_id AS course, NOW() as datum,
(SELECT COUNT(*) FROM users_courses WHERE course_id = course) AS antal_registrerade,
(SELECT COUNT(*) FROM users_courses WHERE status = 1 AND course_id = course) AS antal_aktiva,
(SELECT COUNT(*) FROM users_courses WHERE status = 3 AND course_id = course) AS antal_avklarade
FROM users_courses GROUP BY course_id
以上查询returns如下:
| course | datum | antal_registrerade | antal_aktiva | antal_avklarade |
-----------------------------------------------------------------------------------------
| 31 | 2016-01-12 16:24:58 | 142 | 19 | 83 |
| 38 | 2016-01-12 16:24:58 | 826 | 45 | 49 |
| 39 | 2016-01-12 16:24:58 | 2 | 2 | NULL |
| 43 | 2016-01-12 16:24:58 | 169 | 29 | 32 |
| 44 | 2016-01-12 16:24:58 | 11 | 4 | 2 |
| 45 | 2016-01-12 16:24:58 | 67 | 8 | 7 |
| 46 | 2016-01-12 16:24:58 | 2 | 1 | 1 |
还好吧?就像我想要的那样。但是当我将此查询另存为视图并且 运行 结果不同时。我得到的每一行的数据都相同,但课程和数据列除外。
| course | datum | antal_registrerade | antal_aktiva | antal_avklarade |
-----------------------------------------------------------------------------------------
| 31 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 38 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 39 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 43 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 44 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 45 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 46 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
有人知道这是为什么吗?在已保存视图中找到的 sql 如下所示:
SELECT `database`.`users_courses`.`course_id` AS `course`,now() AS `datum`,
(SELECT COUNT(0) from `database`.`users_courses` where (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`)) AS `antal_registrerade`,
(SELECT COUNT(0) from `database`.`users_courses` where ((`database`.`users_courses`.`status` = 1) and (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`))) AS `antal_aktiva`,
(SELECT COUNT(0) from `database`.`users_courses` where ((`database`.`users_courses`.`status` = 3) and (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`))) AS `antal_avklarade`
FROM `database`.`users_courses`
GROUP BY `database`.`users_courses`.`course_id`
使用条件聚合来表达要简单得多:
SELECT course_id AS course, NOW() as datum,
COUNT(*) as antal_registrerade,
SUM(status = 1) as antal_aktiva,
SUM(status = 3) AS antal_avklarade
FROM users_courses
GROUP BY course_id;
这应该可以解决您的结果问题。
由于某种原因,保存的视图代码的关联子句不正确。我的猜测是您在 table 中没有 course
和 course_id
的两列,因此您的第一个查询不完全是进入视图的内容。无论如何,请使用更简单的查询来解决此问题。
我整理了一个小的 sql 查询,它从一个 table 中提取数据并将其排序在新的列名下。 sql 看起来像这样:
SELECT course_id AS course, NOW() as datum,
(SELECT COUNT(*) FROM users_courses WHERE course_id = course) AS antal_registrerade,
(SELECT COUNT(*) FROM users_courses WHERE status = 1 AND course_id = course) AS antal_aktiva,
(SELECT COUNT(*) FROM users_courses WHERE status = 3 AND course_id = course) AS antal_avklarade
FROM users_courses GROUP BY course_id
以上查询returns如下:
| course | datum | antal_registrerade | antal_aktiva | antal_avklarade |
-----------------------------------------------------------------------------------------
| 31 | 2016-01-12 16:24:58 | 142 | 19 | 83 |
| 38 | 2016-01-12 16:24:58 | 826 | 45 | 49 |
| 39 | 2016-01-12 16:24:58 | 2 | 2 | NULL |
| 43 | 2016-01-12 16:24:58 | 169 | 29 | 32 |
| 44 | 2016-01-12 16:24:58 | 11 | 4 | 2 |
| 45 | 2016-01-12 16:24:58 | 67 | 8 | 7 |
| 46 | 2016-01-12 16:24:58 | 2 | 1 | 1 |
还好吧?就像我想要的那样。但是当我将此查询另存为视图并且 运行 结果不同时。我得到的每一行的数据都相同,但课程和数据列除外。
| course | datum | antal_registrerade | antal_aktiva | antal_avklarade |
-----------------------------------------------------------------------------------------
| 31 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 38 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 39 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 43 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 44 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 45 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
| 46 | 2016-01-12 16:24:58 | 1219 | 108 | 174 |
有人知道这是为什么吗?在已保存视图中找到的 sql 如下所示:
SELECT `database`.`users_courses`.`course_id` AS `course`,now() AS `datum`,
(SELECT COUNT(0) from `database`.`users_courses` where (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`)) AS `antal_registrerade`,
(SELECT COUNT(0) from `database`.`users_courses` where ((`database`.`users_courses`.`status` = 1) and (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`))) AS `antal_aktiva`,
(SELECT COUNT(0) from `database`.`users_courses` where ((`database`.`users_courses`.`status` = 3) and (`database`.`users_courses`.`course_id` = `database`.`users_courses`.`course_id`))) AS `antal_avklarade`
FROM `database`.`users_courses`
GROUP BY `database`.`users_courses`.`course_id`
使用条件聚合来表达要简单得多:
SELECT course_id AS course, NOW() as datum,
COUNT(*) as antal_registrerade,
SUM(status = 1) as antal_aktiva,
SUM(status = 3) AS antal_avklarade
FROM users_courses
GROUP BY course_id;
这应该可以解决您的结果问题。
由于某种原因,保存的视图代码的关联子句不正确。我的猜测是您在 table 中没有 course
和 course_id
的两列,因此您的第一个查询不完全是进入视图的内容。无论如何,请使用更简单的查询来解决此问题。