Question

tl，dr ：我如何通过嵌套属性确保我的has_many的顺序与构建中设置的属性值一致，并始终分配相同的嵌套参数哈希键号（0 ，1，等）并始终以相同的顺序出现在表单中？

希望我能描述一下这样才有意义。我有一个小型原型应用程序模拟两个帐户，源帐户和目标帐户之间的简单银行转帐。我有一个Transfer类，一个Account类和一个TransferAccounts类，它是Transfer和Account之间many_to_many关联的直接连接。

以下是new中的TransfersController操作：

def new
  @transfer = Transfer.new
  @transfer.transfer_accounts.build(account_transfer_role: 'source').build_account
  @transfer.transfer_accounts.build(account_transfer_role: 'destination').build_account
  bank_selections
  account_selections
end

强大的参数：

def transfer_params
  params.require(:transfer).
    permit(:name, :description,
           transfer_accounts_attributes:
             [:id, :account_id, :account_transfer_role,
              account_attributes:
                [:id, :bank_id, :name, :description, :user_name,
                 :password, :routing_number, :account_number
                ]
             ])
end

因此，与transfer_accounts相关联的两个transfer均具有account_transfer_role属性，其中一个设置为 source ，另一个设置为< EM>目的地

现在，填写表单并提交时，参数在控制台中如下所示：

Parameters: {"utf8"=>"✓", "authenticity_token"=>"xxxxxxxxxxxxxxxxxx==",
"transfer"=>{"name"=>"Test Transfer 2", "description"=>"Second test transfer",
"transfer_accounts_attributes"=>{"0"=>{"account_transfer_role"=>"source", "account_id"=>"1",
"account_attributes"=>{"bank_id"=>"1", "name"=>"George's Checking",
"description"=>"George's personal checking account", "user_name"=>"georgeckg",
"password"=>"[FILTERED]", "account_number"=>"111111111", "routing_number"=>"101010101",
"id"=>"3"}, "id"=>"3"}, "1"=>{"account_transfer_role"=>"destination", "account_id"=>"2",
"account_attributes"=>{"bank_id"=>"2", "name"=>"George's Savings",
"description"=>"George's personal savings account", "user_name"=>"georgesav",
"password"=>"[FILTERED]", "account_number"=>"111101111", "routing_number"=>"100100100",
"id"=>"4"}, "id"=>"4"}}}, "commit"=>"Update Transfer", "id"=>"2"}

如您所见，transfer_account哈希中的每个transfer_account_attributes都有一个ID键， 0 或 1 （例如..."0"=>{"account_transfer_role"=>"source"...）。现在，我一直在假设（我认为可能会回来咬我，并且它已经），因为它们是在new动作中构建的顺序，所以源 transfer_account始终具有 0 的ID密钥，目标 transfer_account的ID密钥始终为 1 ，这使我在控制器的其他地方使用这些id键，就好像 0 代表源而 1 代表目的地

所有似乎都工作正常，直到我尝试创建新帐户或使用现有帐户的不同排列，创建新的或编辑现有的转移，突然显示表单目的地列出第一< / strong>和 source second ，之前没有发生过，导致条目现在目的地与 0 1 关联的em>和 source ，打破了上面提到的控制器中的代码。

为了更清楚，以下是表格：

#transfer_form = simple_form_for @transfer do |t| .form-inputs = t.input :name, label: 'Transfer Name' = t.input :description, required: false, label: 'Transfer Description' = t.simple_fields_for :transfer_accounts do |ta| - role = ta.object.account_transfer_role.titleize = ta.input :account_transfer_role, as: :hidden = ta.input :account_id, collection: @valid_accounts, include_blank: 'Select account...', label: "#{ role } Account", error: 'Account selection is required.' .account_fields{id: "#{ role.downcase }_account_fields"} = ta.simple_fields_for :account do |a| = a.input :bank_id, collection: @valid_banks, include_blank: 'Select bank...', label: "#{ role } Bank", error: 'Bank selection is required.', class: "#{ role.downcase }_account_input_field" = a.input :name, label: "#{ role } Account Name", class: "#{ role.downcase }_account_input_field" = a.input :description, required: false, label: "#{ role } Account Description", class: "#{ role.downcase }_account_input_field" = a.input :user_name, label: "#{ role } Account User Name", class: "#{ role.downcase }_account_input_field" = a.input :password, label: "#{ role } Account Password", class: "#{ role.downcase }_account_input_field" = a.input :account_number, label: "#{ role } Account Number", class: "#{ role.downcase }_account_input_field" = a.input :routing_number, label: "#{ role } Account Routing Number", class: "#{ role.downcase }_account_input_field" = t.submit

如何确保 源始终是 ，因此始终与ID键 0 和 目标总是第二个 ，始终与ID键 1 相关联？

Answer 1

答案似乎就像更改:transfer_accounts模型中的Transfer关联行一样简单：

has_many  :transfer_accounts, inverse_of: :transfer

到此：

has_many  :transfer_accounts, -> { order('account_transfer_role DESC') }, inverse_of: :transfer

如果有人认为它没有按我认为的那样做，请告诉我，因为目前它似乎正在解决我的问题。

Rails 4 - has_many的顺序，通过表单中的嵌套属性

1 个答案: