使用列名稱變量的導軌

Question

我有一個Like模型，它連接用戶喜歡的項目。另外標準liked狀態，項目可以是spinned & pinned，所有都是布爾字段。

就像型號

_________________________________________________ 
| item_id | user_id | liked | spinned | pinned | 
|---------|---------|---------|----------|--------| 
| 1 | 1 | true | true | false | 
| 2 | 1 | false | true | true | 
| 3 | 2 | true | false | false | 
------------------------------------------------- 

一個項目都得到通過同一AJAX調用更新的不同的狀態，在我likes controller調用同一action。如果還不存在，它會創建一個類似於所需的狀態爲真的狀態。

likes_controller

def index 
    # get the item id 
    action = params[:act] 
    mid = params[:mid] 

    # gets the desired like item to see if it already exists 
    like = Like.where(:item_id => mid, :user_id => current_user.id).first 

    case action 

    # like the item 
    when "like" 
     # if the item doesn't exist 
     if like.blank? 
     # create the like 
     Like.create(:user_id => current_user.id, :item_id => mid, :liked => true) 
     else 
     # else toggle the liked status 
     like.toggle! :liked 
     end 

    # pin the item 
    when "pin" 
     if like.blank? 
     # create the like 
     Like.create(:user_id => current_user.id, :item_id => mid, :pinned => true) 
     else 
     # else toggle the pinned status 
     like.toggle! :pinned 
     end 

    # spin the item 
    when "spin" 
     if like.blank? 
     # create the like 
     Like.create(:user_id => current_user.id, :item_id => mid, :spinned => true) 
     else 
     # else toggle the spinned status 
     like.toggle! :spinned 
     end 
    end 
end 

正如你看到的，在case代碼變得非常重複，我想知道如果我可以用我的action（「像」，「腳」，「旋轉」）變量更有效。

通過從我的另一個參數值中發送另一個參數值，我可以很容易地將action轉換爲相應的字段名稱（「like」=>「liked」，「pin」=>「pinned」，「spin」=>「spinned」 view。

我可以使用它只有一塊代碼而不是3個類似的代碼嗎？

即

action = "liked" # this would change depending on the desired action 

# if the item doesn't exist 
if like.blank? 
    # original 
    Like.create(:user_id => current_user.id, :item_id => mid, :liked => true) 
    # doesn't work, how would I do this appropriately? 
    Like.create("user_id = ?, item_id = ?, ? = true", current_user.id, mid, action) 
else 
    # this works!, what about SQL injection? How can I prevent this? 
    like.toggle!("#{action}") 
end 

Answer 1

事實證明使用create()和toggle()時，我可以簡單地用我的變量action而不是列選擇:column。

其結果是：

if like.blank? 
    # create the like 
    Like.create(:user_id => current_user, :item_id => mid, action => true) 
else 
    # else toggle the desired status 
    like.toggle!(action) 
end 

Answer 2

您可以使用select或新的，然後保存，如果事件的原子是不是主要問題。

like = Like.where(:item_id => mid, :user_id => current_user.id).first 
like ||= Like.new(:item_id => mid, :user_id => current_user.id, :pinned => false, :spinned => false, :liked => false) 
when "like" 
like.liked = !like.liked 
when "spin" 
like.spinned = !like.spinned 
when "pin" 
like.pinned = !like.pinned 
end 
like.save!