如何在同一个 select 语句中使用 count 和 group by

这将做你想要的（城镇列表，每个城镇的用户数量）：

select town, count(town) 
from user
group by town

在使用 GROUP BY 时，您可以使用大多数 aggregate functions：

（COUNT、MAX、COUNT DISTINCT 等）

更新（在更改问题和评论之后）

您可以为用户数声明一个变量并将其设置为用户数，然后使用它进行选择。

DECLARE @numOfUsers INT
SET @numOfUsers = SELECT COUNT(*) FROM user

SELECT DISTINCT town, @numOfUsers
FROM user

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您可以使用 COUNT(DISTINCT ...) ：

SELECT COUNT(DISTINCT town) 
FROM user

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另一种方式是：

/* Number of rows in a derived table called d1. */
select count(*) from
(
  /* Number of times each town appears in user. */
  select town, count(*)
  from user
  group by town
) d1

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十个未删除的答案；大多数人不做用户要求的事情。大多数答案误读了这个问题，认为每个城镇有 58 个用户，而不是总共 58 个用户。即使是少数正确的也不是最佳的。

mysql> flush status;
Query OK, 0 rows affected (0.00 sec)

SELECT  province, total_cities
    FROM       ( SELECT  DISTINCT province  FROM  canada ) AS provinces
    CROSS JOIN ( SELECT  COUNT(*) total_cities  FROM  canada ) AS tot;
+---------------------------+--------------+
| province                  | total_cities |
+---------------------------+--------------+
| Alberta                   |         5484 |
| British Columbia          |         5484 |
| Manitoba                  |         5484 |
| New Brunswick             |         5484 |
| Newfoundland and Labrador |         5484 |
| Northwest Territories     |         5484 |
| Nova Scotia               |         5484 |
| Nunavut                   |         5484 |
| Ontario                   |         5484 |
| Prince Edward Island      |         5484 |
| Quebec                    |         5484 |
| Saskatchewan              |         5484 |
| Yukon                     |         5484 |
+---------------------------+--------------+
13 rows in set (0.01 sec)

SHOW session status LIKE 'Handler%';

+----------------------------+-------+
| Variable_name              | Value |
+----------------------------+-------+
| Handler_commit             | 1     |
| Handler_delete             | 0     |
| Handler_discover           | 0     |
| Handler_external_lock      | 4     |
| Handler_mrr_init           | 0     |
| Handler_prepare            | 0     |
| Handler_read_first         | 3     |
| Handler_read_key           | 16    |
| Handler_read_last          | 1     |
| Handler_read_next          | 5484  |  -- One table scan to get COUNT(*)
| Handler_read_prev          | 0     |
| Handler_read_rnd           | 0     |
| Handler_read_rnd_next      | 15    |
| Handler_rollback           | 0     |
| Handler_savepoint          | 0     |
| Handler_savepoint_rollback | 0     |
| Handler_update             | 0     |
| Handler_write              | 14    |  -- leapfrog through index to find provinces  
+----------------------------+-------+

在OP的上下文中：

SELECT  town, total_users
    FROM       ( SELECT  DISTINCT town  FROM  canada ) AS towns
    CROSS JOIN ( SELECT  COUNT(*) total_users  FROM  canada ) AS tot;

由于 tot 中只有一行，因此 CROSS JOIN 不会像其他情况下那样庞大。

通常的模式是 COUNT(*) 而不是 COUNT(town)。后者意味着检查 town 是否不为空，这在这种情况下是不必要的。

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使用 Oracle，您可以使用分析函数：

select town, count(town), sum(count(town)) over () total_count from user
group by town

您的其他选择是使用子查询：

select town, count(town), (select count(town) from user) as total_count from user
group by town

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如果您想按计数订购（听起来很简单，但我在堆栈上找不到如何做到这一点的答案），您可以这样做：

        SELECT town, count(town) as total FROM user
        GROUP BY town ORDER BY total DESC

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您可以在 COUNT 中使用 DISTINCT，就像 Milkovsky 所说的那样

就我而言：

select COUNT(distinct user_id) from answers_votes where answer_id in (694,695);

这将拉取被认为与 user_id 相同的答案票数作为一个计数

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我知道这是 SQL Server 中的旧帖子：

select  isnull(town,'TOTAL') Town, count(*) cnt
from    user
group by town WITH ROLLUP

Town         cnt
Copenhagen   58
NewYork      58
Athens       58
TOTAL        174

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如果要选择城镇和总用户数，可以使用以下查询：

SELECT Town, (SELECT Count(*) FROM User) `Count` FROM user GROUP BY Town;

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如果您想使用带计数选项的全选查询，试试这个...

 select a.*, (Select count(b.name) from table_name as b where Condition) as totCount from table_name  as a where where Condition

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试试下面的代码：

select ccode, count(empno) 
from company_details 
group by ccode;

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