Question

我在nil上获得了未定义的方法`name'：简单文件上传表单上的NilClass。

查看

<%= form_tag({action: :upload}, multipart: true) do %>
  <%= file_field_tag 'picture' %>
  <%= submit_tag "Upload & Produce Result" %>
<% end %>

控制器

def upload
  file_name = params['picture']
  if file_name[-4,4] != ".txt"
    respond_to do |format|
      flash[:notice] = "Please upload only txt file"
      format.html { redirect_to :action => :index }
    end
  end
  File.open(Rails.root.join('public', 'uploads', file_name.original_filename), 'w') do |file|
    file.write(file_name.read)
  end
end

路线

root :to => 'home#index'
post 'upload', to: 'home#upload'

发布参数

{"utf8"=>"✓",
 "authenticity_token"=>"HPM76LFj/3Pt7ur1EfMz4SR5T8A9gvoe+JRmneHbu5c=",
 "picture"=>#<ActionDispatch::Http::UploadedFile:0x007f7418370210 @tempfile=#  <Tempfile:/tmp/RackMultipart20130907-9689-viscwz>,
 @original_filename="data.txt",
 @content_type="text/plain",
 @headers="Content-Disposition: form-data; name=\"picture\";   filename=\"data.txt\"\r\nContent-Type: text/plain\r\n">,
 "commit"=>"Upload & Produce Result"}

即使我尝试了

raise params.to_yaml

它给出了同样的错误。

任何帮助都将不胜感激。

感谢。

Answer 1

你正在考虑将params ['picture']作为字符串，而你给出的Post参数显示params ['picture']是'ActionDispatch :: Http :: UploadedFile'对象

所以，您可以尝试用

替换params ['picture']

params[:picture].original_filename

您还可以获取文件上传的其余部分：params [：picture] .tempfile，params [：picture] .headers，params [：picture] .content_type

您也可以使用类似问题Reading in file contents rails

来引用这篇文章

Rails undefined方法`name'代表nil：FileUpload上的NilClass太多未知

1 个答案: