Problem Faced while searching Duplicate rows in MYSQL - mysql

Question

I have been trying to find duplicate and got confused by 3 lines of codes. Can you please help me understand it?
Please find DESC of my table :
| Field | Type | NULL | Key | Extra |
| id | int(11) | NO | PRI | auto_increment |
| name | varchar(50) | YES | | |
| membership | enum('Silver','Gold','Diamond') | YES | | |
| interest |set('Movie','Music','Concert') | YES | | |
This is the output for all the present data in table.
Select * from clients
+----+--------+------------+---------------+
| id | name | membership | interest |
+----+--------+------------+---------------+
| 1 | Sourav | Silver | Movie,Concert |
| 2 | Yash | Diamond | Music |
| 3 | Yash | Diamond | Music |
| 4 | Yash | Diamond | Music |
| 5 | Yash | Diamond | Music |
| 6 | Yash | Diamond | Music |
| 7 | Yash | Diamond | Music |
| 8 | Yash | Diamond | Music |
| 9 | Yash | Diamond | Music |
| 10 | Yash | Diamond | Music |
| 11 | Yash | Diamond | Music |
| 12 | Yash | Diamond | Music |
| 13 | Yash | Diamond | Music |
| 14 | Yash | Diamond | Music |
| 15 | Sneha | Silver | Concert |
+----+--------+------------+---------------+
15 rows in set (0.001 sec)
Confusion arose in below three queries for why do they return 3 completely different outputs?
1.
SELECT id, name, membership, interest, count(*)
FROM clients
GROUP BY name, membership, interest
HAVING count(*) > 1;
//This one being the correct line of code
+----+------+------------+----------+----------+
| id | name | membership | interest | count(*) |
+----+------+------------+----------+----------+
| 2 | Yash | Diamond | Music | 13 |
+----+------+------------+----------+----------+
SELECT name, membership, interest, count()
FROM clients
HAVING count() > 1;
//Returns first entry with count as 15
+--------+------------+---------------+----------+
| name | membership | interest | count(*) |
+--------+------------+---------------+----------+
| Sourav | Silver | Movie,Concert | 15 |
+--------+------------+---------------+----------+
SELECT id, name, membership, interest, count(*)
FROM clients
GROUP BY name, membership, interest;
//Returns Unique Entries only while eliminating the duplicate entries. However, Sneha (last entry) was on top for some reason
+----+--------+------------+---------------+----------+
| id | name | membership | interest | count(*) |
+----+--------+------------+---------------+----------+
| 15 | Sneha | Silver | Concert | 1 |
| 1 | Sourav | Silver | Movie,Concert | 1 |
| 2 | Yash | Diamond | Music | 13 |
+----+--------+------------+---------------+----------+
What necessary changes happened when I used statement because given same syntax without statement clause, it was working completely different.
In addition to that the last statement query was intriguing me as even though I did not apply any clause for Unique data, it delivered the same to me.
Finally, why did the last entry appeared at top in the 3. query?

Answer 1

Correct query finding dupes in your case should be (one possibility as might be other solution as well),
select min(id) as id,
name,
membership,
interest,
count(*) as cnt
from clients
group by name,membership,interest
having count(*) > 1;
https://dbfiddle.uk/HsV-S897
As per your tries.
The first query should fail with sql_mode only_full_group_by enabled which should be enabled .
In the second query HAVING without GROUP BY acts like WHERE, still the second query will fail with sql_mode only_full_group_by enabled.
The third query is correctly formed but you need order by count(*) desc to get Yash first.