gzip

问题 I have a bunch of files that come from some web requests, and some are gziped, i need to unpack them and print them as a string. This is the first time I try using golang, I tried some examples I found online but can't get it working. Here's the last test I was trying: package main import ( "bytes" "compress/gzip" "fmt" "io/ioutil" ) func main() { content := []byte{72,84,84,80,47,49,46,49,32,50,48,48,32,79,75,13,10,84,114,97,110,115,102,101,114,45,69,110,99,111,100,105,110,103,58,32,99,104

问题 I have work this for weeks, very hope for your help!!! please forgive my poor english. First, I think it's necessary to describe the Application Scenario: what data I want to decompress?---- the data is come from the network traffic of the internet . In these traffic, there are some data are compressed by gzip and store in the http or tcp packet, if the data size is huge and large than the maxlength of tcp payload, it will be sliced and transmiss. I can extract the compressed data from these

问题 I want encrypt this file with a default password programming in R. How can I do? zip("C:/Users/user/Downloads/myarchive.zip", files="C:/Users/user/Downloads/example.txt", flags = "-r9X") Thank you!!! 回答1: PASSWORD <- "BillMurray" zip("C:/Users/user/Downloads/myarchive.zip", files="C:/Users/user/Downloads/example.txt", flags = paste("--password", PASSWORD)) 来源： https://stackoverflow.com/questions/40112701/i-would-like-do-a-zip-compression-with-password-in-r

问题 On a remote host I have a bash script sending a simple gzipped YAML file to my Ruby Sinatra endpoint: #!/bin/bash /bin/gzip -c /tmp/test.yaml > /tmp/test.gz curl -i <hostname>:<port>/last_run_report -H "Content-Type: application/xml" -H "Content-Encoding: gzip" --data-binary @/tmp/test.gz My sample Ruby app is: require 'sinatra' require 'zlib' require 'stringio' set :port, <port> set :bind, "<ip>" post '/last_run_report' do sio = StringIO.new(request.body.to_s) gz = Zlib::GzipReader.new(sio)

问题 I have a byte stream that is a concatenation of sections, where each section is composed of a header plus a deflated byte stream. I need to split this byte stream sections but the header only contains information about the data in uncompressed form, no hint about the compressed data length so I can advance properly in the stream and parse the next section. So far the only way I found to advance past the deflated byte sequece is to parse it according to the this specification. From what I

问题 I have a byte stream that is a concatenation of sections, where each section is composed of a header plus a deflated byte stream. I need to split this byte stream sections but the header only contains information about the data in uncompressed form, no hint about the compressed data length so I can advance properly in the stream and parse the next section. So far the only way I found to advance past the deflated byte sequece is to parse it according to the this specification. From what I

问题 I have a binary file with contents created by zlib.compress on Python, is there an easy way to open and decompress it in Clojure? import zlib import json with open('data.json.zlib', 'wb') as f: f.write(zlib.compress(json.dumps(data).encode('utf-8'))) Basicallly it isn't a gzip file, it is just bytes representing deflated data. I could only find these references but not quite what I'm looking for (I think first two are most relevant): deflateclj_hatemogi_clojure/deflate.clj funcool/buddy-core

问题 JSON and Gzip is a simple way to serialize data. These are widely implemented across programming languages. Also this representation is portable across systems (is it?). My question is whether json+gzip is good enough (less then 2x cost) compared to very efficient binary serialization methods? I'm looking for space and time costs while serializing various kinds of data. 回答1: Serialising with json+gzip uses 25% more space than rawbytes+gzip for numbers and objects. For limited precision

问题 JSON and Gzip is a simple way to serialize data. These are widely implemented across programming languages. Also this representation is portable across systems (is it?). My question is whether json+gzip is good enough (less then 2x cost) compared to very efficient binary serialization methods? I'm looking for space and time costs while serializing various kinds of data. 回答1: Serialising with json+gzip uses 25% more space than rawbytes+gzip for numbers and objects. For limited precision

问题 JSON and Gzip is a simple way to serialize data. These are widely implemented across programming languages. Also this representation is portable across systems (is it?). My question is whether json+gzip is good enough (less then 2x cost) compared to very efficient binary serialization methods? I'm looking for space and time costs while serializing various kinds of data. 回答1: Serialising with json+gzip uses 25% more space than rawbytes+gzip for numbers and objects. For limited precision