I have a set of files that are present in s3 and I have to zip them all and send the zipped file to the front end(ReactJS).
I am successfully able to create a folder in the tmp of the project and also zip them. Unfortunately, I get the error when I try to expand saying Unable to expand
Here is the code -
data = Zip::File.open(zip_file_name, ::Zip::File::CREATE) do |zipfile|
files.each do |file|
zipfile.add(file, file_path)
end
end
content_type "application/octet-stream"
header['Content-Disposition'] = "attachment; filename=abcd.zip"
env['api.format'] = :binary
File.open(zip_file_name, 'rb').read
Is there a way to solve the problem? Thanks
Related
I want to download uploaded files such as .zip, .xlsm, xlsx, pdf... ..., etc., that have been uploaded, I want to eventually compress them into a zip file for download.
However, if the uploaded files are not zipped, they will be downloaded with strange files stored in them.
In this case, .xlsm
Source Code
class DownloadZipsController < ::ApplicationController
def index
file_name = "#{Time.current.strftime("%Y%m%d_%H%M%S")}.zip"
zip_file = Tempfile.new(file_name)
Zip.unicode_names = true
Zip::OutputStream.open(zip_file.path) do |zip|
params[:product_ids].each do |product_id|
product = Product.find(product_id)
zip.put_next_entry("output.zip")
zip.print Net::HTTP.get URI.parse(product.submit_zip_file.url)
end
end
send_file zip_file.path,
type: "application/zip",
disposition: "attachment",
filename: file_name
zip_file.close
rescue => e
logger.debug e.backtrace
redirect_to request.referer, alert: e.message
end
end
Uploaded files are stored in AWS S3.
Is there any solution to this problem?
Office Documents such as .xlsm, .xlsx, .docx, and others are in fact zip files containing the document as a xml file plus additional resources.
The file tree you have shown in your screenshot shows the content of one such document if you interpret it as a zip file and unpack.
It appears that somewhere in your code, you have detected the document file as a zip file and interpreted it as such which resulted in its contents to be unpacked.
This is not apparent from the code you have posted though, so I would assume that you have some additional handling of zip files somewhere else (such as a function to download existing files which may then be send with a wrong content type to the browser, i.e as an application/zip rather than application/vnd.ms-excel.sheet.macroEnabled.12).
I've seen a lot of threads that open up a file from a directory and rename it. But in my case, the user is uploading a file to a POST endpoint. I'd like to rename that file before uploading to my blob storage.
Here's what I have so far:
def picture
#file = params[:file]
#new_file_name = generate__filename()
#this line didn't work
File.rename(#file, #new_file_name + File.extname(#file))
# begin upload here
end
I'm testing this by selecting the form-data value as file in Postman. How do I rename this file?
How can I download a file through a custom action in Rails Admin. As the file which I'm sending is ZIP format but after getting downloaded its a nested zip which on opening creates another zip file and just goes on.
So basically this is my custom action where I'm calling my Service which will create a zip file through gem rubyzip and I'm sending the file_path of the created zip file with is on my local machine(e.g file_path as "#{Rails.root}/abc.zip")
else
# all your code that does the work
import_params = params.require(:documents)
.permit(:email)
if import_params[:email].present?
#result = Download::Forms.call(mail_id: import_params[:email])
if #result[:status]
flash.now[:notice] = 'File Downloaded'
send_file(
#result[:file_path],
filename: #result[:file_name],
type: "application/zip"
)
else
flash.now[:error] = "Failed to Download because #{#result[:error_message]}"
end
else
flash.now[:notice] = 'Please Enter Email Address'
end
#result
end
As I also send the #result in the end so that I can also try link_to method to download the file but all in vain. I'm using Rails 4.2.6
Some more information regarding the file types is that I tried a txt file and a pdf file also to check whether they are downloading properly but the same result they got downloaded but they were not proper as txt file was containing nothing and pdf can't be opened.
The files are getting downloaded with my custom action name and not from the file name which is the right naming if I download a file from any simple action.
Any help will be really appreciated.
I want to create a temporary .zip in rails. For creating zip file I am using rubyzip gem.
Currently I am doing this:
zfname = Tempfile.new(['somename','.zip'], Rails.root.to_s + '/tmp/')
Zip::ZipFile.open(zfname.path, Zip::ZipFile::CREATE) do |zipfile|
zipfile.add(file, basepath + file)
end
This generates following error:
Zip::ZipError: Zip end of central directory signature not found
Is it possible to use Tempfile for zip? If yes, what is wrong here?
In my rails app when I needed to send zip files to user, I just stored them in a buffer and used 'send data' method in controller.
I was trying to use 'Tempfile' initially but it had an added task of removing zip files after sending it to user which is a pain.
Is this something you are looking for?
Zip::OutputStream.write_buffer do |stream|
file_paths.each_with_index do |file_path, index|
# rename the pdf
stream.put_next_entry("#{name}-#{index + 1}.pdf")
# add pdf to zip
stream.write IO.read(file_path)
end
end
So it doesn't actually look like you need or want a Tempfile. You really want a random path in the Rails.root/tmp directory. Try something like this:
zfpath = Rails.root.join('tmp', "somename-#{SecureRandom.hex(8)}.zip"
Zip::ZipFile.open(zfpath, Zip::ZipFile::CREATE) do |zipfile|
zipfile.add(file, basepath + file)
end
Update:
While it's far more complex, you can find a discussion of how to do this with a Tempfile here - http://thinkingeek.com/2013/11/15/create-temporary-zip-file-send-response-rails/ .
I have this code, which writes a zip file to disk, reads it back, uploads to s3, then deletes the file:
compressed_file = some_temp_path
Zip::ZipOutputStream.open(compressed_file) do |zos|
some_file_list.each do |file|
zos.put_next_entry(file.some_title)
zos.print IO.read(file.path)
end
end # Write zip file
s3 = Aws::S3.new(S3_KEY, S3_SECRET)
bucket = Aws::S3::Bucket.create(s3, S3_BUCKET)
bucket.put("#{BUCKET_PATH}/archive.zip", IO.read(compressed_file), {}, 'authenticated-read')
File.delete(compressed_file)
This code works already but what I want is to not create the zip file anymore, to save a few steps. I was wondering if there is a way to export the zipfile data directly to s3 without having to first create a tmpfile, read it back, then delete it?
I think I just found the answer to my question.
It's Zip::ZipOutputStream.write_buffer. I'll check this out and update this answer when I get it working.
Update
It does work. My code is like this now:
compressed_filestream = Zip::ZipOutputStream.write_buffer do |zos|
some_file_list.each do |file|
zos.put_next_entry(file.some_title)
zos.print IO.read(file.path)
end
end # Outputs zipfile as StringIO
s3 = Aws::S3.new(S3_KEY, S3_SECRET)
bucket = Aws::S3::Bucket.create(s3, S3_BUCKET)
compressed_filestream.rewind
bucket.put("#{BUCKET_PATH}/archive.zip", compressed_filestream.read, {}, 'authenticated-read')
The write_buffer returns a StringIO and needs to rewind the stream first before reading it. Now I don't need to create and delete the tmpfile.
I'm just wondering now if write_buffer would be more memory extensive or heavier than open? Or is it the other way around?