Golang : Gunzip file
This tutorial demonstrates how to g-unzip a gz
file easily in Golang. It is a continuation from previous tutorial on how to unzip a zip archive.
gunzip.go
package main
import (
"compress/gzip"
"flag"
"fmt"
"io"
"os"
"strings"
)
func main() {
flag.Parse() // get the arguments from command line
filename := flag.Arg(0)
if filename == "" {
fmt.Println("Usage : gunzip sourcefile.gz")
os.Exit(1)
}
gzipfile, err := os.Open(filename)
if err != nil {
fmt.Println(err)
os.Exit(1)
}
reader, err := gzip.NewReader(gzipfile)
if err != nil {
fmt.Println(err)
os.Exit(1)
}
defer reader.Close()
newfilename := strings.TrimSuffix(filename, ".gz")
writer, err := os.Create(newfilename)
if err != nil {
fmt.Println(err)
os.Exit(1)
}
defer writer.Close()
if _, err = io.Copy(writer, reader); err != nil {
fmt.Println(err)
os.Exit(1)
}
}
Outputs (example) :
./gunzip
Usage : gunzip sourcefile.gz
This program will not produce any message upon successful decompression of gzipped file.
By Adam Ng
IF you gain some knowledge or the information here solved your programming problem. Please consider donating to the less fortunate or some charities that you like. Apart from donation, planting trees, volunteering or reducing your carbon footprint will be great too.
Advertisement
Tutorials
+24.4k Golang : How to validate URL the right way
+10.6k Golang : Simple File Server
+4.7k Linux/MacOSX : How to symlink a file?
+5.3k Golang : Qt update UI elements with core.QCoreApplication_ProcessEvents
+15.5k Golang : ROT47 (Caesar cipher by 47 characters) example
+7.9k Golang : Handle Palindrome string with case sensitivity and unicode
+15.5k Golang : Convert date format and separator yyyy-mm-dd to dd-mm-yyyy
+23.6k Find and replace a character in a string in Go
+15.9k Golang : Get file permission
+8.1k How to show different content from website server when AdBlock is detected?
+9.8k Golang : Function wrapper that takes arguments and return result example