方法find_by_user_id_and_friend_id似乎並沒有工作 - Rails的5

我不確定爲什麼方法 Friendship.find_by_user_id_and_friend_id(user, friend) & Friendship.find_by_user_id_and_friend_id(friend, user)給出了一個零，可以一個好心給我講解一下方法find_by_user_id_and_friend_id似乎並沒有工作 - Rails的5

爲何方法Friendship.find_by_user_id_and_friend_id(user, friend)給予零和方法 Friendship.find_by_user_id_and_friend_id(user.id, friend.id)不不是？基本上可以解釋一個區別我

2.3.0 :065 > user 
=> #<User id: 1, email: "[email protected]", created_at: "2016-11-22 15:56:19", updated_at: "2016-12-06 11:39:29", firstname: "richill", lastname: "artloe"> 

2.3.0 :068 > friend 
=> #<User id: 2, email: "[email protected]", created_at: "2016-11-22 16:19:25", updated_at: "2016-11-22 16:19:25", firstname: "emma", lastname: "watson"> 

Friendship.create(user: user, friend: friend, status: 'pending') 
Friendship.create(user: friend, friend: user, status: 'requested') 

2.3.0 :078 > Friendship.find_by_user_id_and_friend_id(user, friend) 
    Friendship Load (0.2ms) SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = ? AND "friendships"."friend_id" = ? LIMIT ? [["user_id", nil], ["friend_id", nil], ["LIMIT", 1]] 
=> nil 
2.3.0 :079 > 

2.3.0 :079 > Friendship.find_by_user_id_and_friend_id(friend, user) 
    Friendship Load (0.2ms) SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = ? AND "friendships"."friend_id" = ? LIMIT ? [["user_id", nil], ["friend_id", nil], ["LIMIT", 1]] 
=> nil 
2.3.0 :080 >

我相信我猜想得到類似下面的結果：

>> Friendship.find_by_user_id_and_friend_id(user, friend) 
=> #<Friendship:0x2bf74ec @attributes={"status"=>"pending", "accepted_at"=>nil, 
"id"=>"1", "user_id"=>"1", "position"=>nil, "created_at"=>"2007-01-03 18:34:09", 
"friend_id"=>"1198"}> 

>> Friendship.find_by_user_id_and_friend_id(friend, user) 
=> #<Friendship:0x490a7a0 @attributes={"status"=>"requested", "accepted_at"=>nil 
, "id"=>"2", "user_id"=>"1198", "position"=>nil, "created_at"=>"2007-01-03 18:34 
:20", "friend_id"=>"1"}>

很奇怪。

Friendship.find_by_user_id_and_friend_id(user, friend) 
=> nil

人能解釋這是爲什麼我

的區別是什麼

：當放置這樣它的工作原理，當放置這樣

Friendship.find_by_user_id_and_friend_id(user.id, friend.id) 
=> #<Friendship id: 1, user_id: 1, friend_id: 2, status: "pending", created_at: "2016-12-06 11:55:06", updated_at: "2016-12-06 11:55:06">

但不工作

2.3.0 :016 > Friendship.find_by_user_id_and_friend_id(user, friend) 
    Friendship Load (0.2ms) SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = ? AND "friendships"."friend_id" = ? LIMIT ? [["user_id", nil], ["friend_id", nil], ["LIMIT", 1]] 
=> nil 
2.3.0 :017 > Friendship.find_by_user_id_and_friend_id(user.id, friend.id) 
    Friendship Load (0.2ms) SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = ? AND "friendships"."friend_id" = ? LIMIT ? [["user_id", 1], ["friend_id", 2], ["LIMIT", 1]] 
=> #<Friendship id: 1, user_id: 1, friend_id: 2, status: "pending", created_at: "2016-12-06 11:55:06", updated_at: "2016-12-06 11:55:06"> 
2.3.0 :018 >

來源

2016-12-06 ARTLoe

嘗試傳遞'user.id'和'friend.id'雖然我不認爲會有所作爲。也可以使用'創造！'如果您在控制檯中是如此，如果不建立友誼，你會得到豁免。也避免採用這些動態方法。開始使用'find_by（user_id：user.id，friend_id：friend.id）' – jvnill

，find_by_id方法與使用where方法相同，只不過它只返回找到的第一條記錄。（See the source code）

由於該方法正在搜索的ID，我相信你需要通過ID，而不是直接的對象的實例爲它工作。這與使用.find方法時相同。

來源

2016-12-06 14:28:04 cdimitroulas

方法find_by_user_id_and_friend_id似乎並沒有工作 - Rails的5

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