Question

我有一个API，我在Base64中从Angular Client发送图像，我的方法创建的下一个代码：

if params[:profile][:picture]
        picture_params = params[:profile][:picture]


        #Create new Temporal File 
        temp_file = Tempfile.new("file_to_upload")
        temp_file.binmode
        temp_file.write(Base64.decode64(picture_params["body"]))

        #Create a new uploaded file
               uploaded_file = ActionDispatch::Http::UploadedFile.new(
                :tempfile => temp_file, 
                :original_filename => picture_params[:name]) 

        #After replace picture with the new uploaded file
        params[:profile][:picture] = uploaded_file
        Rails.logger.info "#{params[:profile]}"
    end

    @profile = @current_user.build_profile(profile_params)
    if @profile.save
        render "api/v1/profiles/show"
    else
        errors_array!(@profile.errors.full_messages, :unprocessable_entity)
    end

当我打印我的参数后，信息是零

"picture"=>#<ActionDispatch::Http::UploadedFile:0x007f872daac5b0 @tempfile=#<Tempfile:/var/folders/nx/30024wp17ggfgw542lyzs4zw0000gn/T/file_to_upload20160225-14952-by947r>, @original_filename=nil, @content_type=nil, @headers=nil>}

在尝试创建配置文件之后，我有下一个错误，我认为这是针对上一个问题

NoMethodError (undefined method `gsub' for nil:NilClass):
 app/controllers/api/v1/profiles_controller.rb:39:in `create'

我在Rails 4.2.5.1中使用paperclip gem上传文件

Answer 1

感谢您的观看，但有一个解决方案，我改变了方法：

    def set_selfies(selfie_params)
      picture_params = selfie_params
      encoded_picture = picture_params[:body]
      content_type = picture_params[:content_type]
      image = Paperclip.io_adapters.for("#{encoded_picture}")
      image.original_filename = picture_params[:name]
      image
    end

I Read this Blog: 这一切都对我有用，谢谢：D

如何在Rails 4中将所有属性设置为UploadedFile

1 个答案: