Objection JS - MySQL insert and update choices from User - mysql

Question

I'm facing an issue with how to set up my MySQL selected relationship table as you can see below.
&
Our users have a profile page, where they can answer to 3 questions from a dropdown menu with multiple choices.
My question is how to add their 3 choices in my MySQL table selected in a single query and where I can update their choices if needed. I can't use ON DUPLICATE KEY since my user_id key is only an index because each user will appear 3 times in selected since there are 3 questions per user.
On their profile page, when our users hit the Submit button I would like our selected table to insert and/or update the choice_id for that user.
I'm using objection JS to set up my models. Here are my User and Selected models
USER MODEL
class User extends Model {
static get tableName() {
return 'users';
}
static get relationMappings() {
return {
choice: {
relation: Model.HasManyRelation,
modelClass: Selected,
join: {
from: 'users.id',
to: 'selected.user_id',
},
},
user: {
relation: Model.ManyToManyRelation,
modelClass: Choices,
join: {
from: 'choices.id',
through: {
from: 'selected.user_id',
to: 'selected.choice_id',
},
to: 'users.id',
},
},
};
}
SELECTED MODEL
class Selected extends Model {
static get tableName() {
return 'selected';
}
static get jsonSchema() {
return {
type: 'object',
properties: {
id: {
type: 'integer',
},
choice_id: {
type: 'integer',
},
user_id: {
type: 'integer',
},
},
};
}
}
Any help to check out if the relationship table is correctly set up or help with the MySQL query would be much appreciated.
Thanks!
EDIT This is where I'm at. Missing the update function for existing answers.
insert into selected(choice_id, user_id)
select choices.Id, users.Id from choices join users
on choices.Id in (1, 6, 10) and users.Id = 91
WHERE NOT EXISTS (
SELECT user_id FROM selected WHERE user_id = 91
) LIMIT 3;

Answer 1

This is the kind of insert you want to execute:
insert into selected(Choice_id, User_id)
select User.Id, Choices.Id
from User
join Choices
on User.Id = 12 and Choices.Id in (4, 8, 15);
Instead of the hard-coded values above, you can use your actual values on the server-side. If you are using PHP, then check what you have in $_POST and $_SESSION and build your query, but make sure you do not allow SQL injection to occur.
EDIT
Sorry for the late reply. You can do the insertion with the criteria of non-existence, like this:
insert into selected(Choice_id, User_id)
select User.Id, Choices.Id
from User
join Choices
on User.Id = 12 and Choices.Id in (4, 8, 15)
where not exists (
select 1
from selected
where Choice_id = Choices.Id and User_id = User.id
);
However, you intend to perform an update if the user chooses a different answer of a question. You can do it like this
update selected
join Choices
on selected.Choice_Id = Choices.Id and Selected.User_id = 12
join Questions
on Choices.Question_id = Questions.Id and Questions.Id = 4
set Choices = 5;
Assuming that you perform updates like above before you perform the insert-select, you should get the result you prefer. Alternatively you could create a trigger before insert which would check whether the pair already exists and if so, it would update instead of insert.

Answer 2

In the end, I'm going to forget the update ability of this, since I believe my selected table wasn't built the 'right' way.
I should have only one row per user_id which lists the 3 choices (choice_id1, choice_id2 and choice_id3). Like that I could use ON DUPLICATE KEY UPDATE for the user_id value.
My workaround is to delete all rows for an user before inserting new values.
START TRANSACTION ;
DELETE FROM
selected
WHERE
user_id = 91;
INSERT INTO selected(choice_id, user_id)
select choices.Id, users.Id from choices join users
on choices.Id in (3, 5, 11) and users.Id = 91
COMMIT ;