Mysql – COUNT(), GROUP BY and GROUP_CONCAT() together on MySQL

countgroup bygroup-concatenationMySQL

My question is quite simple. I just want to count events from each client and present them in a comma separated single row.

I have this table:

+----+----------------+
| id | event          |
+----+----------------+
| 22 | a              |
| 23 | bb             |
| 24 | bb             |
| 25 | ccc            |
| 26 | ccc            |
| 27 | ccc            |
+----+----------------+
6 rows in set (0.01 sec)

So far I have this following query:

SELECT COUNT(event) AS total FROM test_table GROUP BY event;

Which gives me the following result:

+--------------+
| count(event) |
+--------------+
|            1 |
|            2 |
|            3 |
+--------------+

I'm trying to use GROUP_CONCAT() in order to show them in a single line, like this, but I tried all different approaches and I didn't get the desired result:

+--------------+
| result       |
+--------------+
| 1, 2, 3      |
+--------------+

Do you have a clue on how to do that?

Thanks in advance!

Best Answer

To create the result you wanted, I used a subquery and used GROUP_CONCAT() from there.

create table test
(
  id int,
  event varchar(3)
  );
  insert into test
  values
  (22,'a'),(23,'bb'),(24,'bb'),(25,'ccc'),(26,'ccc'),(27,'ccc')


select group_concat(total)
from
(
SELECT COUNT(event) AS total FROM test GROUP BY event
  )a

+---------------------+
|group_concat(total)  |
+---------------------+
|1,2,3                |
+---------------------+

DB Fiddle

YOUR ORIGINAL QUERY

select event_id, count(*) as num_bookmarks from event_connections
where connectiontype = 3 join event_id on events.id=event_connections.event_id group by event_id;

PROPER SYNTAX

select event_id, count(*) as num_bookmarks
from event_connections join events on events.id=event_connections.event_id
where event_connections.connectiontype = 3
group by event_id;

You just had the WHERE in the wrong place

SUGGESTIONS

If all you care about is the event_id and the counts, do this

select event_id, count(*) as num_bookmarks
from event_connections
where connectiontype = 3
group by event_id;

No need to join.

If you want the event title, then you must join

select events.title, count(*) as num_bookmarks
from event_connections join events on events.id=event_connections.event_id
where event_connections.connectiontype = 3
group by events.title;

GIVE IT A TRY !!!

Mysql – Getting the count of same value in consecutive rows that are ordered by different field

I have a solution that does not need GROUP_CONCAT

PROPOSED SOLUTION

SET @x = 0;
SET @name = '';
SET @result = '';
SELECT name,consecutive FROM
(SELECT
    name,
    (@nametag:=MD5(CONCAT(name,':',result))),
    (@x:=IF(@name=@nametag,@x+1,1)) consecutive,
    (@name:=@nametag) inc
FROM finals ORDER BY name,id) A
WHERE consecutive >= 3;

PROPOSED SOLUTION EXECUTED

mysql> SELECT name,consecutive FROM
    -> (SELECT
    ->     name,
    ->     (@nametag:=MD5(CONCAT(name,':',result))),
    ->     (@x:=IF(@name=@nametag,@x+1,1)) consecutive,
    ->     (@name:=@nametag) inc
    -> FROM finals ORDER BY name,id) A
    -> WHERE consecutive >= 3;
+------+-------------+
| name | consecutive |
+------+-------------+
| Kyle |           3 |
+------+-------------+
1 row in set (0.02 sec)

mysql>

SUBQUERY'S OUTPUT

mysql> SELECT
    ->     name,
    ->     (@nametag:=MD5(CONCAT(name,':',result))),
    ->     (@x:=IF(@name=@nametag,@x+1,1)) consecutive,
    ->     (@name:=@nametag) inc
    -> FROM finals ORDER BY name,id;
+------+------------------------------------------+-------------+----------------------------------+
| name | (@nametag:=MD5(CONCAT(name,':',result))) | consecutive | inc                              |
+------+------------------------------------------+-------------+----------------------------------+
| John | 84cc30b986fe149dfb765dd09fad8a60         |           1 | 84cc30b986fe149dfb765dd09fad8a60 |
| John | 534b3d163a04b74a72c6dbe68db1c01e         |           1 | 534b3d163a04b74a72c6dbe68db1c01e |
| John | 84cc30b986fe149dfb765dd09fad8a60         |           1 | 84cc30b986fe149dfb765dd09fad8a60 |
| John | 84cc30b986fe149dfb765dd09fad8a60         |           2 | 84cc30b986fe149dfb765dd09fad8a60 |
| Kyle | 30fac0873cf25ad17b38bc37bda4b850         |           1 | 30fac0873cf25ad17b38bc37bda4b850 |
| Kyle | 30fac0873cf25ad17b38bc37bda4b850         |           2 | 30fac0873cf25ad17b38bc37bda4b850 |
| Kyle | 30fac0873cf25ad17b38bc37bda4b850         |           3 | 30fac0873cf25ad17b38bc37bda4b850 |
| Kyle | 4a4e0aaa102c37f098bd6afd13ccfea0         |           1 | 4a4e0aaa102c37f098bd6afd13ccfea0 |
+------+------------------------------------------+-------------+----------------------------------+
8 rows in set (0.00 sec)

mysql>

GIVE IT A TRY !!!

I just answered a similar question 2 days ago (Logic Value of Assignment Expression)

Best Answer

Related Solutions

MySQL join for two queries

YOUR ORIGINAL QUERY

PROPER SYNTAX

SUGGESTIONS

GIVE IT A TRY !!!

Mysql – Getting the count of same value in consecutive rows that are ordered by different field

PROPOSED SOLUTION

PROPOSED SOLUTION EXECUTED

SUBQUERY'S OUTPUT

GIVE IT A TRY !!!

Related Question