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我有一个关于 Rails 的问题。我的数据库中有一个没有主键的表，其中包含这些列 - person_id、year、salary。我的问题是当我想编辑其中一份薪水时。我不明白该怎么做。那是我的代码：

工资控制员：

def index
    @salaries = Salary.all
end

def edit
    @salary = Salary.find_by_person_id_and_year(params[:person_id], params[:year])
    @people = Person.all
end

index.html.erb

<% @salaries.each do |salary| %>
    <b><%= salary.person.Name %></b>
    <b><%= salary.Year %></b>
    <b><%= salary.Amount %></b>
    <%= link_to "Edit", edit_salary_path(salary.person_id, salary.year)%>
<% end %>

所以，我将它添加到我的 routes.rb 中：

get 'salaries/index', :to => 'salaries#index',:as => 'salaries'
get 'salaries/:person_id/:year', :to => 'salaries#show', :as => 'salary'   
get 'salaries/new/:person_id/:year', :to => 'salaries#edit', :as => 'new_salary'
get 'salaries/edit/:person_id/:year', :to => 'salaries#edit', :as => 'edit_salary'
put 'salaries/:person_id/:year', :to => 'salaries#update'
post 'salaries/:person_id/:year', :to => 'salaries#create'
delete 'salaries/:person_id/:year', :to => 'salaries#destroy'

但是我的 _form.html.erb 中有这些错误消息：“undefined method `join' for nil:NilClass” on this line: <%= form_for(@salary) do |f| %>

def edit
    @person_id = params[:person_id]
    @year = params[:year]
    @salary = Salary.find_by_person_id_and_year(params[:person_id], params[:year])
    @people = People.all
end

def update
    @people = People.all
    @salary = Salary.new(params[:salary])
    if @salary.save
        flash[:success] = "Successfully edited salary"
        redirect_to salaries_path
    else
        @title = "Edit salary"
        render 'edit'
    end
end

_form.html.erb

<%= form_for(@salary, :url => salaries_path(:person_id => @person_id, :year => @year)) do |f|   %>
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2 回答 2

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Rails 路由会自动寻找 id，所以你需要自定义你的路由。

例如，不要resources :salaries在 config/routes.rb 中尝试类似：

get 'salaries/index', :to => 'salaries#index',:as => 'salaries'
get 'salaries/edit/:person_id/:year', :to => 'salaries#edit', :as => 'edit_salaries'
put 'salaries/:person_id/:year', :to => 'salaries#update'

使用您需要的其他路线，例如：

get 'salaries/index', :to => 'salaries#index',:as => 'salaries'
get 'salaries/:person_id/:year', :to => 'salaries#show', :as => 'salary'
get 'salaries/new/:person_id/:year', :to => 'salaries#edit', :as => 'new_salary'
get 'salaries/edit/:person_id/:year', :to => 'salaries#edit', :as => 'edit_salary'
put 'salaries/:person_id/:year', :to => 'salaries#update'
post 'salaries/:person_id/:year', :to => 'salaries#create'
delete 'salaries/:person_id/:year', :to => 'salaries#destroy'

和形式：

form_for(@salary, :url => salary_path(:person_id => X, :year => Y)) do |f|
于 2012-09-30T09:09:02.493 回答
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如果您可以在薪金表中添加额外的 ID 列，则可以使用标准的 rails 函数，例如

(in app/models/person.rb)

class Person
  has_many :salaries
  has_one :current_salary, :class_name => "Salary", 
          :conditions => lambda { {:year => Time.now.year} }
  # outer curlies are for the lambda, inner curlies are for it to return a hash
end

(in config/routes.rb)


resources :people do
  resources :salaries
end

然后通过ID使用普通访问

于 2012-09-30T08:44:49.043 回答