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/ .
Related
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
I have the following code to connect my rails app to my FTP. This works great. However, I want to use open-uri to open the csv file so I can parse it. Any ideas how to do this? I think it's an easy thing to do but I'm missing something.
require 'net/ftp'
ftp = Net::FTP.new
ftp.connect("xxx.xxx.xx.xxx",21)
ftp.login("xxxxx","xxxx")
ftp.chdir("/")
ftp.passive = true
puts ftp.list("TEST.csv")
You'll need to use #gettextfile.
A) Get the file to a local temporary file and read its content
# Creating a tmp file can be done differently as well.
# It may also be omitted, in which case `gettextfile`
# will create a file in the current directory.
Dir::Tmpname.create(['TEST', ['.csv']) do |file_name|
ftp.gettextfile('TEST.csv', file_name)
content = File.read(file_name)
end
B) Pass a block to gettextfile and get the content one line at a time
content = ''
ftp.gettextfile('TEST.csv') do |line|
content << line
end
I'm new to ruby. Actually I'm trying to create an empty file "myfile.txt" in each of the following directories:
../../../../../TESTS/Test_A/myTest_A/
../../../../../TESTS/Test_B/myTest_B/
../../../../../TESTS/Test_C/myTest_C/
../../../../../TESTS/Test_D/myTest_D/
As you can see, the name of the Top directory is "TEST" and than after this, every directory have a different name but starts with "Test_", and than each "Test_*" directory contains only one directory and there my file should go in. I'm trying something like this:
require 'pathname'
def create_myFile
pn = Pathname.new('../../../../../TESTS/Test_*/**')
myFile = File.new("#{pn}/myFile.txt", "w+")
end
create_myFile
It doesn't work. Any suggestions?
Pathname does not accept wildcards. Dir#[] does:
Dir['../../../../../TESTS/Test_*/**'].each do |dir|
File.new File.join(dir, 'myFile.txt'), 'w+'
end
Something I'm not getting about the version process...
I have a zip file with a file inside, and I want to upload the file as a "version" of the zip:
Uploader:
version :specificFile do
process :extract_file
end
def extract_file
file = nil
Zip::ZipFile.open(current_path) do |zip_file|
file = zip_file.select{|f| f.name.match(/specificFile/)}.first
zip_file.extract(file, "tmp/" + file.name.gsub("/", "-")){ true }
end
File.open("tmp/" + file.name.gsub("/", "-"))
end
Usage:
=link_to "Specific File", instance.uploader.specificFile.url
Only this just nets me two copies of the zip. Clearly, there's something I'm missing about how version / process works, and I haven't been able to find documentation that actually explains the magic.
So how do I do this, and what am I missing?
This provided the "why", although it took a bit to understand:
How do you create a new file in a CarrierWave process?
To rephrase, when you go to create a version, carrierwave makes a copy of the file and then passes the process the file path. When the process exits, carrierwave will upload the contents of that path - not the file the process returns, which is what I thought was going on.
Working code:
version :specificFile do
process :extract_file
def full_filename (for_file = model.logo.file)
"SpecificFile.ext"
end
end
def extract_plist
file = nil
Zip::ZipFile.open(current_path) do |zip_file|
file = zip_file.select{|f| f.name.match(/specificFile/)}.first
zip_file.extract(file, "tmp/" + file.name.gsub("/", "-")){ true }
end
File.delete(current_path)
FileUtils.cp("tmp/" + file.name.gsub("/", "-"), current_path)
end
So, to make what I want to happen, happen, I:
Tell carrierwave to use a particular filename. I'm using a hardcoded value but you should be able to use whatever you want.
Overwrite the contents of current_path with the contents you want under the version name. In my case, I can't just overwrite the zip while I'm "in it" (I think), so I make a copy of the file I care about and overwrite the zip via File and FileUtils.
PS - It would be nice to avoid the duplication of the zip, but it doesn't look like you can tell carrierwave to skip the duplication.
I need to create a local xml file from a rails application and then copy it to a location on another server.
I have tried using the File.new option to create a new file but it gives me an error saying the file does not exist. After looking closer at the documentation it says that File.new opens a file that already exists.
I can't see any way to create a local file using Ruby, what am I missing?
Assuming you have built up your XML into a string, xml_string, you can do:
xml_file = open(filename, 'w')
xml_file.write xml_string
xml_file.close
Or using the block syntax to achieve this in one line:
File.open(local_filename, 'w') { |f| f.write(xml_string) }