Ecto:如何访问预加载关联的字段
Ecto: How to access field of a preloaded association
我正在尝试获取与 friend_referral_code 没有关联的用户,或者他们的代码状态为 false 并且他们也没有使用任何代码。但是我无法访问预加载关联 friend_referral_code 的字段 status。以下是我的做法:
def get_first_free_invite_users() do
users =
list_users()
|> Repo.preload([:friend_referral_code])
|> Enum.filter(
fn u ->
u.friend_referral_code == [] or u.friend_referral_code["status"] == false and
Repo.all(from ref in FriendReferralCode,
where: ref.receiver_id == ^u.id and ref.status == true) == []
end)
users
end
它在 u.friend_referral_code.status == false
上抛出错误。我无法加入的原因是 FriendReferralCode 中可能没有特定用户的记录,我想捕获这些用户。这是关联的加载方式:
friend_referral_code: [
%MyApp.Accounts.FriendReferralCode{
__meta__: #Ecto.Schema.Metadata<:loaded, "friend_referral_code">,
challenge: #Ecto.Association.NotLoaded<association :challenge is not loaded>,
challenge_id: nil,
code: "RFQTS",
expiry: 1,
id: 16021,
inserted_at: ~N[2021-02-01 11:55:00],
order: 1,
prize: #Decimal<3>,
receiver: #Ecto.Association.NotLoaded<association :receiver is not loaded>,
receiver_id: 15002,
state: "create_competition",
status: false,
updated_at: ~N[2021-02-01 11:57:20],
user: #Ecto.Association.NotLoaded<association :user is not loaded>,
user_id: 15001
}
],
It throws error on u.friend_referral_code.status == false.
那是因为 u.friend_referral_code
在您的示例中是一个列表。您需要枚举 u.friend_referral_code
并检查每个枚举的状态。
尝试这样的事情。
def get_ref_codes_by_user(id) do
Repo.all(from ref in FriendReferralCode,
where: ref.receiver_id == ^u.id and ref.status == true)
end
def filter_referrals([], user_id, acc), do: acc
def filter_referrals([ref, rest], user_id, acc) do
is_used =
ref == [] || ref.status == false && get_ref_codes_by_user(id) == []
new_acc =
case is_used do
true -> acc
false -> [ref | acc]
end
filter_referrals(rest, user_id, new_acc)
end
def get_first_free_invite_users() do
list_users()
|> Repo.preload([:friend_referral_code])
|> Enum.map(
fn u ->
filter_referrals(u.friend_referral_code, u.id, [])
end)
end
这将使您更接近,并为您提供 FriendReferralCode
的列表。仔细检查这里的逻辑是否是您想要的。综上所述,您应该可以通过加入来做到这一点。查看有关联接类型的文档 here and this other SO answer。
尝试
Repo.all(from ref in FriendReferralCode,
right_join users u on ref.user_id = u.id
where: ref.status == true)
我正在尝试获取与 friend_referral_code 没有关联的用户,或者他们的代码状态为 false 并且他们也没有使用任何代码。但是我无法访问预加载关联 friend_referral_code 的字段 status。以下是我的做法:
def get_first_free_invite_users() do
users =
list_users()
|> Repo.preload([:friend_referral_code])
|> Enum.filter(
fn u ->
u.friend_referral_code == [] or u.friend_referral_code["status"] == false and
Repo.all(from ref in FriendReferralCode,
where: ref.receiver_id == ^u.id and ref.status == true) == []
end)
users
end
它在 u.friend_referral_code.status == false
上抛出错误。我无法加入的原因是 FriendReferralCode 中可能没有特定用户的记录,我想捕获这些用户。这是关联的加载方式:
friend_referral_code: [
%MyApp.Accounts.FriendReferralCode{
__meta__: #Ecto.Schema.Metadata<:loaded, "friend_referral_code">,
challenge: #Ecto.Association.NotLoaded<association :challenge is not loaded>,
challenge_id: nil,
code: "RFQTS",
expiry: 1,
id: 16021,
inserted_at: ~N[2021-02-01 11:55:00],
order: 1,
prize: #Decimal<3>,
receiver: #Ecto.Association.NotLoaded<association :receiver is not loaded>,
receiver_id: 15002,
state: "create_competition",
status: false,
updated_at: ~N[2021-02-01 11:57:20],
user: #Ecto.Association.NotLoaded<association :user is not loaded>,
user_id: 15001
}
],
It throws error on u.friend_referral_code.status == false.
那是因为 u.friend_referral_code
在您的示例中是一个列表。您需要枚举 u.friend_referral_code
并检查每个枚举的状态。
尝试这样的事情。
def get_ref_codes_by_user(id) do
Repo.all(from ref in FriendReferralCode,
where: ref.receiver_id == ^u.id and ref.status == true)
end
def filter_referrals([], user_id, acc), do: acc
def filter_referrals([ref, rest], user_id, acc) do
is_used =
ref == [] || ref.status == false && get_ref_codes_by_user(id) == []
new_acc =
case is_used do
true -> acc
false -> [ref | acc]
end
filter_referrals(rest, user_id, new_acc)
end
def get_first_free_invite_users() do
list_users()
|> Repo.preload([:friend_referral_code])
|> Enum.map(
fn u ->
filter_referrals(u.friend_referral_code, u.id, [])
end)
end
这将使您更接近,并为您提供 FriendReferralCode
的列表。仔细检查这里的逻辑是否是您想要的。综上所述,您应该可以通过加入来做到这一点。查看有关联接类型的文档 here and this other SO answer。
尝试
Repo.all(from ref in FriendReferralCode,
right_join users u on ref.user_id = u.id
where: ref.status == true)