Struggling to structure an SQL query - mysql

Question

Currently I'm trying to run a query that returns the 'Trainer' username and 'Type' title for every trainer who has caught all unique 'Species' of a specific type. For example, if there are two distinct species who are type fighting, and Jenny has caught both of them, then she should be output as such:
---------------------
| Username | Type |
---------------------
| Jenny | Fighting |
---------------------
If she has also caught all distinct species of another type, another row would be output, but with a different type. This must account for the fact a species have two types and that a trainer may catch more than one of each species, so to determine if someone has caught all species of a type, both types are taken into account. Here is what the tables look like:
(Type) (Species) (Trainer) (Pokemon)
------------ ---------------------------- --------------- ------------------------
| id | title | | id | title | type1 | type2 | | id | username | | id | species | trainer |
------------ ---------------------------- --------------- ------------------------
I've also provided the schema and sample data for the query. With an expected result set underneath it. So far I've taken an approach to determining how many unique species of each type there are with the following query:
SELECT Type.id AS TypeID, Type.title, COUNT(Species.id) AS 'Number of Species of Type'
FROM Type, Species WHERE Type.id = Species.type1 OR Type.id = Species.type2 GROUP BY Type.id;
My next thought was to determine how many species of each type a trainer has caught, so that I can compare the two. But I'm stuck on how I could structure the query to do so. Views or common table expressions also cannot be used. Any suggestions or ideas would be appreciated.
create database pokemon;
use pokemon;
CREATE TABLE IF NOT EXISTS `Type` (
`id` TINYINT NOT NULL,
`title` VARCHAR(50) NOT NULL,
PRIMARY KEY (`id`));
CREATE TABLE IF NOT EXISTS `Species` (
`id` TINYINT UNSIGNED NOT NULL,
`title` VARCHAR(50) NOT NULL,
`type1` TINYINT NOT NULL,
`type2` TINYINT NULL,
PRIMARY KEY (`id`),
FOREIGN KEY (`type1`) REFERENCES `Type` (`id`),
FOREIGN KEY (`type2`) REFERENCES `Type` (`id`));
CREATE TABLE IF NOT EXISTS `Trainer` (
`id` INT NOT NULL,
`username` VARCHAR(50) NOT NULL,
PRIMARY KEY (`id`));
CREATE TABLE IF NOT EXISTS `Pokemon` (
`id` BIGINT NOT NULL,
`species` TINYINT UNSIGNED NOT NULL,
`trainer` INT NULL,
PRIMARY KEY (`id`),
FOREIGN KEY (`trainer`) REFERENCES `Trainer` (`id`),
FOREIGN KEY (`species`) REFERENCES `Species` (`id`));
insert into Type values (1,'Normal');
insert into Type values (2,'Fighting');
insert into Type values (3,'Flying');
insert into Type values (4,'Poison');
insert into Type values (5,'Ground');
insert into Type values (6,'Rock');
insert into Type values (7,'Bug');
insert into Type values (8,'Ghost');
insert into Type values (9,'Steel');
insert into Type values (10,'Fire');
insert into Type values (11,'Water');
insert into Type values (12,'Grass');
insert into Type values (13,'Electric');
insert into Type values (14,'Psychic');
insert into Type values (15,'Ice');
insert into Type values (16,'Dragon');
insert into Type values (17,'Dark');
insert into Type values (18,'Fairy');
insert into Species values (100,'Voltorb',13,null);
insert into Species values (101,'Electrode',13,null);
insert into Species values (102,'Exeggcute',12,14);
insert into Species values (103,'Exeggutor',12,14);
insert into Species values (104,'Cubone',5,null);
insert into Species values (105,'Marowak',5,null);
insert into Species values (106,'Hitmonlee',2,null);
insert into Species values (107,'Hitmonchan',2,null);
insert into Species values (108,'Lickitung',1,null);
insert into Species values (109,'Koffing',4,null);
insert into Species values (110,'Weezing',4,null);
insert into Species values (111,'Rhyhorn',5,6);
insert into Species values (112,'Rhydon',5,6);
insert into Species values (113,'Chansey',1,null);
insert into Species values (114,'Tangela',12,null);
insert into Species values (115,'Kangaskhan',1,null);
insert into Species values (116,'Horsea',11,null);
insert into Species values (117,'Seadra',11,null);
insert into Species values (118,'Goldeen',11,null);
insert into Species values (119,'Seaking',11,null);
insert into Species values (120,'Staryu',11,null);
insert into Species values (121,'Starmie',11,14);
insert into Trainer values (1,'Ash');
insert into Trainer values (2,'Brock');
insert into Trainer values (3,'Misty');
insert into Trainer values (4,'Jenny');
insert into Trainer values (5,'Luna');
insert into Pokemon values (1,109,1);
insert into Pokemon values (2,110,1);
insert into Pokemon values (3,115,1);
insert into Pokemon values (4,113,1);
insert into Pokemon values (5,108,1);
insert into Pokemon values (6,117,1);
insert into Pokemon values (7,102,2);
insert into Pokemon values (8,103,2);
insert into Pokemon values (9,121,2);
insert into Pokemon values (10,104,2);
insert into Pokemon values (11,111,3);
insert into Pokemon values (12,112,3);
insert into Pokemon values (13,121,3);
insert into Pokemon values (14,106,4);
insert into Pokemon values (15,107,4);
insert into Pokemon values (16,110,4);
Expected result set for the sample data is:
---------------------
| Username | Type |
| --------------------|
| Ash | Poison |
| --------------------|
| Ash | Normal |
| --------------------|
| Brock | Psychic |
| --------------------|
| Misty | Rock |
| --------------------|
| Jenny | Fighting |
--------------------

Answer 1

WITH
cte1 AS ( SELECT Species.id sid, t1.id, t1.title ttitle
FROM Species
JOIN Type t1 ON Species.type1 = t1.id
UNION ALL
SELECT Species.id sid, t2.id, t2.title ttitle
FROM Species
JOIN Type t2 ON Species.type2 = t2.id ),
cte2 AS ( SELECT ttitle, COUNT(id) ids
FROM cte1
GROUP BY ttitle ),
cte3 AS ( SELECT Trainer.username, cte1.ttitle, COUNT(*) ids
FROM Pokemon
JOIN Trainer ON Pokemon.trainer = Trainer.id
JOIN cte1 ON Pokemon.species = cte1.sid
GROUP BY Trainer.username, cte1.ttitle )
SELECT cte3.username, cte2.ttitle type
FROM cte2
JOIN cte3 USING (ttitle, ids);
The same without CTE, for MySQL 5.x
SELECT cte3.username, cte2.ttitle type
FROM ( SELECT ttitle, COUNT(id) ids
FROM ( SELECT Species.id sid, t1.id, t1.title ttitle
FROM Species
JOIN Type t1 ON Species.type1 = t1.id
UNION ALL
SELECT Species.id sid, t2.id, t2.title ttitle
FROM Species
JOIN Type t2 ON Species.type2 = t2.id) AS cte1
GROUP BY ttitle ) AS cte2
JOIN ( SELECT Trainer.username, cte1.ttitle, COUNT(*) ids
FROM Pokemon
JOIN Trainer ON Pokemon.trainer = Trainer.id
JOIN ( SELECT Species.id sid, t1.id, t1.title ttitle
FROM Species
JOIN Type t1 ON Species.type1 = t1.id
UNION ALL
SELECT Species.id sid, t2.id, t2.title ttitle
FROM Species
JOIN Type t2 ON Species.type2 = t2.id) AS cte1 ON Pokemon.species = cte1.sid
GROUP BY Trainer.username, cte1.ttitle) AS cte3 USING (ttitle, ids);
https://dbfiddle.uk/?rdbms=mysql_8.0&rdbms2=mysql_5.7&fiddle=0a33601724fa53d9ee4a583c983ff01a

Answer 2

If it is the case a trainer can take the same Specie more than once, and of course, that shouldn't count as two species taken you would just have to substitute table Pokémon by a version of the table with no duplicates, meaning:
(select distinct species, trainer from pokemon)
So the query would be:
select t.username, tp.title
from
(select p.trainer, t.id TypeID, count(species) Cuenta
from (select distinct species, trainer from pokemon) p, species s, type t
where p.species = s.id
and (s.type1 = t.id or s.type2 = t.id)
group by p.trainer, t.id) Tra
, (SELECT Type.id AS TypeID, Type.title, COUNT(Species.id) Cuenta
FROM Type, Species WHERE Type.id = Species.type1 OR Type.id = Species.type2
GROUP BY Type.id) ty
, trainer t, type tp
where ty.TypeID = Tra.TypeId
and ty.Cuenta = Tra.Cuenta
and t.id = Tra.Trainer
and tp.id = Tra.TypeId;
In order to test the result I added these three records of Ground type; Specie Cubone to Brock:
insert into Pokemon values (17,104,2);
insert into Pokemon values (18,104,2);
insert into Pokemon values (19,104,2);
And I observed that, with the previous query, Brock would appear as a certificate in Ground and with this new one it does not.
PS.: You can make same substitution in Akina's query and I am sure it will work just as well.

Answer 3

I really like Akina's solution. It is well organised and beautiful. Mine does not look so nice, but, just in case you want another approach:
select t.username, tp.title
from
(select p.trainer, t.id TypeID, count(species) Cuenta
from pokemon p, species s, type t
where p.species = s.id
and (s.type1 = t.id or s.type2 = t.id)
group by p.trainer, t.id) Tra
, (SELECT Type.id AS TypeID, Type.title, COUNT(Species.id) Cuenta
FROM Type, Species WHERE Type.id = Species.type1 OR Type.id = Species.type2
GROUP BY Type.id) ty
, trainer t, type tp
where ty.TypeID = Tra.TypeId
and ty.Cuenta = Tra.Cuenta
and t.id = Tra.Trainer
and tp.id = Tra.TypeId
Tra is about the courses that each trainer has.
ty is the query that you put in the question about how many species per type.