How do I write this in Ecto Query?

Questions / Help
1

I have a designs table that has these field

design_id
design_type
user_id

and design table belongs to user table. So user has_many designs and designs belongs to user.

first I need to query on design_id and design_type and load all the designs.

Which I did something like this

from([a] in Design, where: a.design_type == ^type and a.design_id == ^id)

Where I will get results like this

[
%Design{
    design_id: 1,
    design_type: "document",
    user: #Ecto.Association.NotLoaded<association :user is not loaded>,
    user_id: 2,
  },
%Design{
    design_id: 1,
    design_type: "document",
    user: #Ecto.Association.NotLoaded<association :user is not loaded>,
    user_id: 3,
]

So, I want to get a list of users for the same design_id and design_type.

Also if I’m querying ` from([a] in Design, where: a.design_type == ^type and a.design_id == ^id) from elixir shell its giving me a list of designs.

But when I tried to use it in function

  def design_filters(%{"design_id" => id, "design_type" => type}) do
    from([t] in Design,
      where: t.design_id == ^id and t.design_type == ^type
    )
  end

I’m getting this error

** (FunctionClauseError) no function clause matching

2

Change [a] to just a?

3

Still the same error.

4

Also, my focus is on this right now. Do you have any suggestions here?

5

from the top of my head you would need something like:

Ecto:

from d in Design
 join: u in User, on: d.user_id == u.id,
 where: d.design_type_id = ^type and d.design_id = ^id,
 select: u

actually even more simple if you don’t need the user table only the id:

from d in Design, where: d.design_type_id = ^type and d.design_id = ^id, select: d.user_id
2 Likes
6

Thanks :slight_smile: