9

Ruby 很新，我已经被困了好几个小时。到处搜索，找不到答案。

所以我将引导日期选择器用于 rails gem。

因为我更改了日期选择器的日期格式，所以它不会存储在数据库中。猜测这是因为 simple_form 输入被用作字符串以避免 simple_form 应用的默认日期选择输入。

我的问题是：如何在将“06/18/2013”​​之类的字符串保存到数据库之前将其修改/转换为日期？这最好由控制器处理吗？

我的控制器：

# PUT /events/1
# PUT /events/1.json
def update
  @event = Event.find(params[:id])

  # Ugghhhh I need help
  # @event.event_date = Date.parse(params[:event_date]).to_date

  respond_to do |format|
    if @event.update_attributes(params[:event])
      format.html { redirect_to @event, :notice => 'Event was successfully updated.' }
      format.json { head :no_content }
    else
      format.html { render :action => "edit" }
      format.json { render :json => @event.errors, :status => :unprocessable_entity }
    end
  end
end

数据库架构：

  create_table "events", :force => true do |t|
    t.string   "event_name"
    t.string   "event_location"
    t.date     "event_date"
    t.time     "event_time"
    t.text     "event_content"
    t.datetime "created_at",                       :null => false
    t.datetime "updated_at",                       :null => false
    t.integer  "user_id"
    t.boolean  "approved",       :default => true
    t.integer  "category_id"
  end

这是服务器日志：

  Started PUT "/events/1" for 127.0.0.1 at 2013-06-14 02:37:15 -0700
  Processing by EventsController#update as HTML
    Parameters: {"utf8"=>"✓", "authenticity_token"=>"RYMdJ0lFmvG0+nVIsTtJXu5fyD/L3/WAKsk9FX6WWgo=", "event"=>{"user_id"=>"1", "category_id"=>"3", "event_name"=>"A Event Name", "event_location"=>"Event Location", "event_date"=>"06/13/2013", "event_time(1i)"=>"2000", "event_time(2i)"=>"1", "event_time(3i)"=>"1", "event_time(4i)"=>"02", "event_time(5i)"=>"18", "event_content"=>"First night"}, "commit"=>"Update Event", "id"=>"1"}
    User Load (0.2ms)  SELECT "users".* FROM "users" WHERE "users"."id" = 1 LIMIT 1
    Event Load (0.1ms)  SELECT "events".* FROM "events" WHERE "events"."id" = ? LIMIT 1  [["id", "1"]]
     (0.0ms)  begin transaction
     (0.0ms)  commit transaction
  Redirected to http://localhost:3000/events/1
  Completed 302 Found in 4ms (ActiveRecord: 0.3ms)
4

2 回答 2

16

您可以在控制器中使用 strptime。

DateTime.strptime("06/18/2013", "%m/%d/%Y")
于 2013-06-14T09:47:31.407 回答
0

正确的应该是

DateTime.strptime("02/06/2014", "%d/%m/%Y").strftime("%Y-%m-%d")

于 2014-07-17T06:32:48.767 回答