How to use count and group by at the same select statement

SqlCountGroup By

Sql Problem Overview


I have an sql select query that has a group by. I want to count all the records after the group by statement. Is there a way for this directly from sql? For example, having a table with users I want to select the different towns and the total number of users

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

I want to have a column with all the towns and another with the number of users in all rows.

An example of the result for having 3 towns and 58 users in total is :

Town         Count
Copenhagen   58
NewYork      58
Athens       58

Sql Solutions


Solution 1 - Sql

This will do what you want (list of towns, with the number of users in each):

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

You can use most aggregate functions when using GROUP BY:

(COUNT, MAX, COUNT DISTINCT etc.)

Update (following change to question and comments)

You can declare a variable for the number of users and set it to the number of users then select with that.

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

SELECT DISTINCT town, @numOfUsers
FROM user

Solution 2 - Sql

You can use COUNT(DISTINCT ...) :

SELECT COUNT(DISTINCT town) 
FROM user

Solution 3 - Sql

The other way is:

/* 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

Solution 4 - Sql

Ten non-deleted answers; most do not do what the user asked for. Most Answers mis-read the question as thinking that there are 58 users in each town instead of 58 in total. Even the few that are correct are not optimal.

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  
+----------------------------+-------+

In the OP's context:

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

Since there is only one row from tot, the CROSS JOIN is not as voluminous as it might otherwise be.

The usual pattern is COUNT(*) instead of COUNT(town). The latter implies checking town for being not null, which is unnecessary in this context.

Solution 5 - Sql

With Oracle you could use analytic functions:

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

Your other options is to use a subquery:

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

Solution 6 - Sql

If you want to order by count (sound simple but i can`t found an answer on stack of how to do that) you can do:

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

Solution 7 - Sql

You can use DISTINCT inside the COUNT like what milkovsky said

in my case:

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

This will pull the count of answer votes considered the same user_id as one count

Solution 8 - Sql

I know this is an old post, in SQL Server:

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

Town         cnt
Copenhagen   58
NewYork      58
Athens       58
TOTAL	     174

Solution 9 - Sql

If you want to select town and total user count, you can use this query below:

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

Solution 10 - Sql

if You Want to use Select All Query With Count Option, try this...

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

Solution 11 - Sql

Try the following code:

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

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionStavrosView Question on Stackoverflow
Solution 1 - SqlOdedView Answer on Stackoverflow
Solution 2 - SqlmilkovskyView Answer on Stackoverflow
Solution 3 - SqlZhenYu WangView Answer on Stackoverflow
Solution 4 - SqlRick JamesView Answer on Stackoverflow
Solution 5 - SqlTommiView Answer on Stackoverflow
Solution 6 - SqlRenato ProbstView Answer on Stackoverflow
Solution 7 - SqlJur PView Answer on Stackoverflow
Solution 8 - SqlMarcusView Answer on Stackoverflow
Solution 9 - SqlViolendy FirdausView Answer on Stackoverflow
Solution 10 - SqlPrakashView Answer on Stackoverflow
Solution 11 - SqlbalajibranView Answer on Stackoverflow