mysql 选择查询结果与 COUNT(*) 和 * 错误

Question

在 Mysql 中，我在 COUNT() 聚合函数的帮助下计算了分页的总记录，并且为了显示数据目的，我使用 * 而不是 COUNT() 但我得到了不同的总计数和返回行数。

查询 1：

SELECT
  COUNT(*)
FROM
  change_request
INNER JOIN task_main_management ON cr_project_id = project_id
LEFT JOIN all_module AS am
ON
  am.m_id = task_m_id
WHERE
  cr_tenent_id = '100' AND cr_entity_id = '100' AND cr_project_id = 'PROJ_310820200746367254' AND cr_type = '2' AND change_request.cr_id = '16' AND task_main_management.cr_id = 'CHRQ_020920200807413904'
GROUP BY
  change_request.cr_project_id

返回结果为2。

但是当我在选择中使用 * 而不是 COUNT(*) 时，它给了我单行。

Answer 1

这是预期的行为.. 给定

ariaDB [sandbox]> select * from t;
+------+--------------------+------------+------------+----------+
| id   | venue              | city       | date       | time     |
+------+--------------------+------------+------------+----------+
|    1 | Waldorf Hotel      | London     | 2020-01-01 | 07:00:00 |
|    2 | Waldorf Hotel      | London     | 2020-01-02 | 07:00:00 |
|    3 | Heathrow           | London     | 2020-01-02 | 14:00:00 |
|    4 | Lennon Airport     | Liverpool  | 2020-01-02 | 16:00:00 |
|    5 | Port of Liverpool  | Liverpool  | 2020-01-02 | 19:30:00 |
|    6 | Port of Liverpool  | Liverpool  | 2020-01-03 | 07:00:00 |
|    7 | Port of Liverpool  | Liverpool  | 2020-01-04 | 07:00:00 |
|    8 | Port of Liverpool  | Liverpool  | 2020-01-05 | 07:00:00 |
|    9 | Port of Liverpool  | Liverpool  | 2020-01-06 | 07:00:00 |
|   10 | Manchester Airport | Manchester | 2020-01-06 | 12:40:00 |
|   11 | Heathrow           | London     | 2020-01-06 | 14:40:00 |
+------+--------------------+------------+------------+----------+
11 rows in set (0.001 sec)

MariaDB [sandbox]>
MariaDB [sandbox]> SELECT * FROM T WHERE CITY = 'LONDON' GROUP BY CITY;
+------+---------------+--------+------------+----------+
| id   | venue         | city   | date       | time     |
+------+---------------+--------+------------+----------+
|    1 | Waldorf Hotel | London | 2020-01-01 | 07:00:00 |
+------+---------------+--------+------------+----------+
1 row in set (0.001 sec)

MariaDB [sandbox]>
MariaDB [sandbox]> SELECT COUNT(*)
    -> FROM T
    -> WHERE CITY = 'LONDON'
    -> GROUP BY CITY;
+----------+
| COUNT(*) |
+----------+
|        4 |
+----------+
1 row in set (0.001 sec)

没有聚合函数的 group by 将返回 1 行不确定的值 具有聚合函数但没有要聚合的列的 group by 将返回由 where 子句过滤的行的计数。

Answer 2

您的查询有一个GROUP BY。因此，COUNT(*) 是返回的 one 行中的值。如果使用聚合查询会更清楚：

SELECT change_request.cr_project_id, COUNT(*)

使用GROUP BY 和SELECT *，查询格式错误，应返回语法错误。不幸的是，旧版本的 MySQL 允许这种损坏的语法。

我不确定您真正想要什么，但您可能只是想删除 GROUP BY。

Answer 3

如果要计算结果集将包含多少条记录，则删除group by 子句并使用count(distinct ...)：

SELECT COUNT(DISTINCT  change_request.cr_project_id) cnt
FROM change_request
INNER JOIN task_main_management ON cr_project_id = project_id
LEFT JOIN all_module AS am ON am.m_id = task_m_id
WHERE
  cr_tenent_id = '100' 
  AND cr_entity_id = '100' 
  AND cr_project_id = 'PROJ_310820200746367254' 
  AND cr_type = '2' 
  AND change_request.cr_id = '16' 
  AND task_main_management.cr_id = 'CHRQ_020920200807413904'

请注意，对于您当前的 group by 子句，使用 select * 已损坏。您应该枚举SELECT 子句中的列，并可能对不是来自表change_request 的任何内容使用聚合函数。