我需要一个表单，用户可以从列表中选择两个玩家（两个单独的选择字段）。 后来，在控制器中我需要找到那两个Player实例。 我知道在表单中传递对象的唯一方法是使用它的id，但我无法让用户选择原始数字 - 我希望能够向玩家展示＆＃39;选择框中的名称，但以某种方式传递id。有可能吗？
答案 0 :(得分：4)
您需要使用Rails options_for_select helper。此助手将在您的模型中向视图显示人类可读字段，同时通过 v3.hs:104:63:
Couldn't match expected type ‘[Rating] -> Bool’
with actual type ‘Bool’
In the first argument of ‘(.)’, namely
‘((map fst ratings) /= newUser)’
In the first argument of ‘filter’, namely
‘(((map fst ratings) /= newUser) . ratings)’
In the second argument of ‘(.)’, namely
‘filter (((map fst ratings) /= newUser) . ratings)’
v3.hs:104:72:
Couldn't match expected type ‘[(Char, b0)]’
with actual type ‘Film -> [Rating]’
Probable cause: ‘ratings’ is applied to too few arguments
In the second argument of ‘map’, namely ‘ratings’
In the first argument of ‘(/=)’, namely ‘(map fst ratings)’
v3.hs:104:93:
Couldn't match type ‘(String, Int)’ with ‘Film’
Expected type: Rating -> [Rating]
Actual type: Film -> [Rating]
In the second argument of ‘(.)’, namely ‘ratings’
In the first argument of ‘filter’, namely
‘(((map fst ratings) /= newUser) . ratings)’
哈希从同一模型返回id字段。
params您需要确保使用<%= f.select :user_id, options_for_select(@users.collect { | user | [user.name, user.id] }, @user.id), {}, {} %>
以与视图对应的控制器方法定义@users和@user。您显然需要在User模型中同时使用options_for_select和id字段，否则，此帮助程序将失败。