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我即將建立一個類似的機器爲我的食譜類。Rails簡單像按鈕與ajax jquery
class LikesController
def update
@like = Like.find(params[:id])
end
def destroy
@like = Like.find(params[:id])
@recipe = Recipe.find(params[:id])
@like.destroy
end
end
def show
@recipe = Recipe.find(params[:id])
end
食譜/ show.html.erb
<div id="like_form">
<% if like = current_user.likes.find_by_recipe_id(@recipe.id) %>
<%= form_for like, :html => { :method => :delete },:remote => true do |f| %>
<%= f.submit "Unlike" %>
<% end %>
<% else %>
<%= form_for current_user.likes.create(:recipe_id => @recipe.id), :remote => true do |f| %>
<%= f.hidden_field :recipe_id %>
<%= f.hidden_field :user_id %>
<%= f.submit "Like" %>
<% end %>
<% end %>
喜歡/ destroy.js.erb
$("#like_form").html("<%= escape_javascript(render('recipes/like')) %>")
喜歡/ update.js.erb
$("#like_form").html("<%= escape_javascript(render('recipes/unlike')) %>")
食譜/_unlike.html.erb
個<%= form_for @like, :html => { :method => :delete },
:remote => true do |f| %>
<%= f.submit "Unlike" %>
<% end %>
食譜/ _like.html.erb
<%= form_for current_user.likes.create(:recipe_id => @recipe.id), :remote => true do |f| %>
<%= f.hidden_field :recipe_id %>
<%= f.hidden_field :user_id %>
<%= f.submit "Like" %>
<% end %>
點擊 「不像」, 「贊」 按鈕應該被渲染後。但我不知道如何在_like.html部分中創建form_for參數。
我不知道如何獲取recipe.id的值。如何使_like.html.erb中的recipe.id可用?
ActionView::Template::Error (undefined method `id' for nil:NilClass):1:
<%= form_for current_user.likes.create(:recipe_id => @recipe.id),
:remote => true do |f| %>
2: <%= f.hidden_field :recipe_id %>
3: <%= f.hidden_field :user_id %>
4: <%= f.submit "Like" %>
app/views/recipes/_like.html.erb:1:in
_app_views_recipes__like_html_erb___1003281063_65806008'
app/views/likes/destroy.js.erb:1:in
_app_views_likes_destroy_js_erb___41116873_65874912'