MySQL Joining 3 Quiz Tables with Calculated Totals By Attribute/User - mysql

Question

I have the following tables which I am trying to combine into one query (with the ultimate goal being to output a CSV file from PHP):
users
id, name, email
eg. data
1, John, email#email.com
2, Jane, email#email.com
questions - a static list of questions, each question is of one of 3 types stored as "attribute," which is a single letter [A, B, C]
id, text, attribute
eg. data
1, How cool are dogs?, A
2, How cool are cats?, B
3, How cool are fish?, A
4, How cool are mice?, C
5, How cool are birds?, B
users_questions - where answer is an integer [1-5]
id, user_id, question_id, answer
eg. data
1, 1, 1, 2
2, 1, 2, 5
3, 1, 3, 1
4, 1, 4, 1
5, 1, 5, 4
6, 2, 1, 4
7, 2, 2, 1
8, 2, 3, 3
9, 2, 4, 2
10, 2, 5, 2
Desired results:
I'm trying to combine all of this data as one query with the questions for each user totalled grouped by attribute so the output format is something like:
users.id, users.name, users.email , A_question_total, B_question_total, C_question_total
1, John , email#email.com, 3, 9, 1
2, Jane , email#email.com, 7, 3, 2
What I have currently:
I've tried the queries below which all give me almost what I need:
I'm able to select everything joined together, but this duplicates users and questions and doesn't give me the question totals by user/attribute:
Select * FROM users
JOIN users_questions ON users.id = users_questions.user_id
JOIN questions ON questions.id = users_questions.question_id;
I can also select all the question totals by user/attribute, but then I'd have to grab the users separately and then connect them together in PHP.
SELECT questions.attribute, users_questions.user_id, SUM(users_questions.answer) AS `total` FROM `questions` LEFT OUTER JOIN `users_questions` ON questions.id = users_questions.`question_id` GROUP BY users_questions.user_id, questions.attribute;
I'm wondering if there is a way with complex joining, grouping, subqueries, etc. to be able to do this all in one query. I'm struggling as to how to combine the above two queries and also convert what is essentially separate calculated total "rows" in columns for each user.
Here's the sql dump of the example data:
DROP TABLE IF EXISTS `questions`;
CREATE TABLE `questions` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`text` varchar(30) DEFAULT NULL,
`attribute` enum('A','B','C') DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `questions` (`id`, `text`, `attribute`)
VALUES
(1,'How cool are dogs?','A'),
(2,'How cool are cats?','B'),
(3,'How cool are fish?','A'),
(4,'How cool are mice?','C'),
(5,'How cool are birds?','B');
DROP TABLE IF EXISTS `users`;
CREATE TABLE `users` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(10) DEFAULT NULL,
`email` varchar(20) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `users` (`id`, `name`, `email`)
VALUES
(1,'John','email#email.com'),
(2,'Jane','email#email.com');
DROP TABLE IF EXISTS `users_questions`;
CREATE TABLE `users_questions` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) unsigned DEFAULT NULL,
`question_id` int(11) unsigned DEFAULT NULL,
`answer` tinyint(1) unsigned DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `users_questions` (`id`, `user_id`, `question_id`, `answer`)
VALUES
(1,1,1,2),
(2,1,2,5),
(3,1,3,1),
(4,1,4,1),
(5,1,5,4),
(6,2,1,4),
(7,2,2,1),
(8,2,3,3),
(9,2,4,2),
(10,2,5,2);
Any help is appreciated!

Answer 1

I think this query (SQLFiddle) will solve your problem:
Select u.id, u.name, u.email,
SUM(case q.attribute when 'A' then uq.answer else 0 end) as A_question_total,
SUM(case q.attribute when 'B' then uq.answer else 0 end) as B_question_total,
SUM(case q.attribute when 'C' then uq.answer else 0 end) as C_question_total
FROM users u
JOIN users_questions uq ON u.id = uq.user_id
JOIN questions q ON q.id = uq.question_id
group by u.id
Output:
id name email A_question_total B_question_total C_question_total
1 John email#email.com 3 9 1
2 Jane email#email.com 7 3 2