Question

我正在关注可排序列表railscast并且一切正常，除非我排序，每次都将相同（原始）顺序传递给update方法，因此它不会更新真正的顺序。

以下是代码：

在视图中：

%tbody#lineup{"data-update-url" => sort_lineups_url}
      - @lineup.pieces.each_with_index do |piece, index|
        = content_tag_for :tr, piece do
          = render 'piece', :piece => piece, :index => index

和pieces.js.coffee：

jQuery ->
  $('#lineup').sortable(
    axis: 'y'
    update: ->
      $.post($(this).data('update-url'), $(this).sortable('serialize'))
  );

排序并显示lineupscontroller的方法：

  def show
    @lineup = Lineup.find_by_user_id(current_user.id)
    @pieces = @lineup.pieces.order("position")
  end

  def sort
    params[:piece].each_with_index do |id, index|
      current_user.lineup.pieces.update_all({position: index+1}, {id: id})
    end
    render nothing: true
  end

和更新请求：

Started POST "/lineups/sort" for 127.0.0.1 at 2012-06-24 20:06:14 -0700
Processing by LineupsController#sort as */*
  Parameters: {"piece"=>["8", "2", "1", "4", "3", "7"]}
  User Load (0.4ms)  SELECT `users`.* FROM `users` WHERE `users`.`id` = 2 LIMIT 1
  Lineup Load (0.6ms)  SELECT `lineups`.* FROM `lineups` WHERE `lineups`.`user_id` = 2 LIMIT 1
  SQL (2.9ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 1 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 8
  SQL (0.6ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 2 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 2
  SQL (0.6ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 3 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 1
  SQL (0.5ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 4 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 4
  SQL (0.6ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 5 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 3
  SQL (0.5ms)  UPDATE `pieces` INNER JOIN `piece_lineups` ON `pieces`.`id` = `piece_lineups`.`piece_id` SET `position` = 6 WHERE `piece_lineups`.`lineup_id` = 3 AND `pieces`.`id` = 7
  Rendered text template (0.0ms)

所以每次我拖放列表，它都会再次发送相同的订单。我错误的是，参数没有通过正确的，更新的订单？在2757ms完成200 OK（浏览次数：1.0ms | ActiveRecord：6.7ms）

%tr{:style => 'height: 60px;'}
  %td{:style => 'text-align: center;'}
    %h5= index + 1
  %td{:style => 'padding-left: 10px;'}
    %h5= "#{piece.name} - #{piece.id}"
  %td{:style => 'text-align: center;'}
    %h5= piece.category
  %td{:style => 'padding-left: 10px;'}
    %h5= piece.description
  %td{:style => 'padding-left: 10px;'}
    %h5= @lineup.get_piece_status(piece.id)
  %td.pop{:style => 'text-align: center;', :rel => 'popover',:'data-title' => piece.name, :'data-placement' => 'bottom', :'data-content' => image_tag(piece.photo.url(:small))}
    = image_tag(piece.photo.url(:tiny))
  %td{:style => 'text-align: center;'}
    - unless ['Preparing', 'Shipped', 'Arrived'].include?(@lineup.get_piece_status(piece.id))
      = form_for(@lineup, :html => {:id => "remove#{piece.id}", :style => "margin-bottom: 0px;"}) do |f|
        = f.hidden_field(:piece_id, :value => piece.id)
        = link_to 'x', '#', :onclick=>"document.getElementById('remove#{piece.id}').submit(); return false;"

使用jquery ui的可排序列表未传递更新的列表顺序

0 个答案: