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As described in the title of my question, I would like to know if you have any way to get the path to a csv file, in case I need this path to automatically import the file.
This is my attempt to do this. form
<%= form.file_field :file %>
This input I’m using to get the file and the symbol :file as parameter to return the path.
users_controller
def create
@user = User.new(user_params)
binding.pry
require 'csv'
CSV.foreach(@user.file, col_sep: ',').with_index do |l, i|
unless (indice == 0)
User.create!(name: l[0], email: l[1], address: l[2])
end
end
respond_to do |format|
if @user.save
format.html { redirect_to @user, notice: 'User was successfully created.' }
format.json { render :show, status: :created, location: @user }
else
format.html { render :new }
format.json { render json: @user.errors, status: :unprocessable_entity }
end
end
end
I used Binding.Pry to show the result that is returning me, and also the reason for my question. When I go to check the parameters that came to the user_params this type of information is shown:
<ActionController::Parameters {"file"=>#<ActionDispatch::Http::UploadedFile:0xb454db98 @tempfile=#<Tempfile:/tmp/RackMultipart20190710-3063-2ah7nt.csv>, @original_filename="users.csv", @content_type="text/csv", @headers="Content-Disposition: form-data; name=\"user[file]\"; filename=\"users.csv\"\r\nContent-Type: text/csv\r\n">} permitted: true>
OK. Now I’ll show you what’s inside the parameter file when I execute @user.file on the Pry console.
"#<ActionDispatch::Http::UploadedFile:0xb454db98>"
All right, I’d like to know if you have a way to return the path to the file or convert this "#<ActionDispatch::Http::UploadedFile:0xb454db98>" on the way to the archive.