我正在开发一个必须上传 zip 文件的应用程序。zip 文件基本上是一个静态网站,因此它有很多文件和几个子目录。我已经玩了一段时间的 rubyzip gem 并且不知道如何简单地从中提取文件。关于我可以在哪里阅读一些示例的任何指示?我相信以前有人遇到过这个问题。rubyzip 的文档不是很好,所以我希望有人能给我一些指示。
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1 回答
1
给你,一个超级神奇的多线程 zip-to-S3 上传器,我根本没有测试过 - 发疯!看来我晚了三年。
class S3ZipUploader
require 'thread'
require 'thwait'
require 'find'
attr_reader *%i{ bucket s3 zip failed_uploads }
def initialize(zipfilepath, mys3creds)
# next 4 lines are important
@s3 = AWS::S3.new(access_key_id: mys3creds[Rails.env]['aws_access_key'],
secret_access_key: mys3creds[Rails.env]['aws_secret_access_key'],
region: 'us-west-2')
@bucket = @s3.buckets[ mys3creds[Rails.env]['bucket'] ]
@failed_uploads = []
@zip = Zip::File.open(zipfilepath)
end
def upload_zip_contents
rootpath = "mypath/"
desired_threads = 10
total_entries = zip.entries.count
slice_size = (total_entries / desired_threats).ceil
threads = []
zip.entries.each_slice(slice_size) do |e_arr|
threads << Thread.new do |et|
e_arr.each do |e|
result = upload_to_s3(rootpath + e.name, e.get_input_stream.read)
if !result
@failed_uploads << {name: e.name, entry: e, error: err}
end
end
end
end
ThreadsWait.all_waits(*threads)
end
def upload_file_to_s3(filedata,path, rewrite_basepath)
retries = 0
success = false
while !success && retries < 3
success = begin
obj = bucket.objects[path]
obj.write(Pathname.new(outputhtml))
obj.acl = :public_read
success = true
rescue
retries += 1
success = false
end
end
return success
end
end
uploader = S3ZipUploader.new("/path/to/myzip.zip", MYS3CREDS)
uploader.upload_zip_contents
于 2016-04-18T02:41:43.443 回答