Question

我相对较新的rails无法获得以下代码。我正在尝试上传数据文件（Excel或csv），将其复制到临时文件夹并在Datafiles模型中创建记录，该模型包含基本文件信息，例如文件名，类型和日期。然后我想读取文件并使用数据在其他几个模型中创建或更新记录。如果一切顺利，请将文件移动到永久位置，并在数据文件记录中写入新路径。

控制器：

def new
  @datafile = Datafile.new
  respond_to do |format|
    format.html # new.html.erb
    format.xml  { render :xml => @datafile }
  end
end

def create
  @datafile = Datafile.new(params[:upload])
  @datafile.save!
  redirect_to datafile_path(@datafile), :notice => "Successfully imported datafile"

rescue => e
  logger.error( 'Upload failed. ' + e.to_s )
  flash[:error] = 'Upload failed. Please try again.'
  render :action => 'new'
end

查看：

<%= form_for @datafile, :html => {:multipart => true} do |f| %>
  <p>
    <%= f.label(:upload, "Select File:") %>
    <%= f.file_field :upload %>
  </p>
  <p> <%= f.submit "Import" %> </p>
<% end %>

型号：

require 'spreadsheet'

class Datafile < ActiveRecord::Base
  attr_accessor :upload
  attr_accessible :upload
  before_create :upload_file

  def upload_file
    begin
      File.open(Rails.root.join('uploads/temp', upload.original_filename), 'wb') do |file|
        file.write(upload.read)
        self.filename = upload.original_filename
        Spreadsheet.client_encoding = 'UTF-8'
        @book = Spreadsheet.open(file.path)
        self.import
      end          
    rescue => e
      @upload_path = Rails.root.join('uploads/temp', upload.original_filename)
      File.delete(@upload_path) if File::exists?(@upload_path)
      raise e
    end
  end

  def import
    case @book.worksheet(0).row(0)[0]

      when "WIP Report - Inception to Date"
      self.report_type = 'WIP'
      puts 'report_type assigned'
      self.import_wip

      else
      self.report_type = 'Unknown'
    end
  end  

  def import_wip
    self.end_date = @book.worksheet(0).row(0)[3]
    puts 'end_date assigned'
  end

  def formatted_end_date
   end_date.strftime("%d %b, %Y")
  end
end

然而，它失败并且rails服务器窗口显示

Started POST "/datafiles" for 127.0.0.1 at 2011-05-24 16:05:25 +0200
  Processing by DatafilesController#create as HTML
  Parameters: {"utf8"=>"✓", "datafile"=>{"upload"=>#<ActionDispatch::Http::UploadedFile:0xa0282d0 @original_filename="wip.xls", @content_type="application/vnd.ms-excel", @headers="Content-Disposition: form-data; name=\"datafile[upload]\"; filename=\"wip.xls\"\r\nContent-Type: application/vnd.ms-excel\r\n", @tempfile=#<File:/tmp/RackMultipart20110524-14236-1kcu3hm>>}, "commit"=>"Import"}
Upload failed. undefined method `original_filename' for nil:NilClass
Rendered datafiles/new.html.erb within layouts/application (54.5ms)
Completed 200 OK in 131ms (Views: 56.3ms | ActiveRecord: 0.0ms)

我有传递的rspec模型测试和保存后无法重定向的控制器测试。如果它有用，我可以发布它们。

我插入了raise @ datafile.to_yaml并在终端中获得了以下内容。

ERROR RuntimeError: --- !ruby/object:Datafile 
attributes: 
  filename: 
  report_type: 
  import_successful: 
  project: 
  begin_date: 
  end_date: 
  created_at: 
  updated_at: 
attributes_cache: {}

changed_attributes: {}

destroyed: false
marked_for_destruction: false
persisted: false
previously_changed: {}

readonly: false

我注意到：上传未列出。我可以从表单中设置模型实例变量吗？：upload是一个实例变量，而不是一个属性，因为我不想将上传的文件保存在数据库中（只是它的本地目录路径）。如果我不能在视图的表单中设置实例变量，有什么建议吗？将文件上传到控制器中的临时文件夹是否有意义（就MVC而言），然后通过传递临时文件的路径来创建模型记录？

Answer 1

你好我对Rails很陌生并且对此很感兴趣我找到了一个解决方案虽然它可能不是最好的。它确实有用。

在你的模型中创建一个名为import_upload的公共函数

def import_upload( upload ) 
  @uploaded_file = upload
end

现在在您的控制器中，您可以显式传递它。我不知道为什么如果你创建一个与file_field同名的attr_accsessor，这不会自动发生，但这是对我有用的解决方案。

def new
  foo = Foo.new( params[:foo] )
  foo.import_upload( params[:foo][:uploaded_file] ) #This is were the majic happens
  #Do your saving stuff and call it a day
end

无法将上传的文件传递给模型

1 个答案: