Editing a zip file in memory - go

Question

I am trying to edit a zip file in memory in Go and return the zipped file through a HTTP response
The goal is to add a few files to a path in the zip file example
I add a log.txt file in my path/to/file route in the zipped folder
All this should be done without saving the file or editing the original file.

Answer 1

I have implemented a simple version of real-time stream compression, which can correctly compress a single file. If you want it to run efficiently, you need a lot of optimization.
This is only for reference. If you need more information, you should set more useful HTTP header information before compression so that the client can correctly process the response data.
package main
import (
"archive/zip"
"io"
"net/http"
"os"
"github.com/gin-gonic/gin"
)
func main() {
engine := gin.Default()
engine.GET("/log.zip", func(c *gin.Context) {
f, err := os.Open("./log.txt")
if err != nil {
c.String(http.StatusInternalServerError, err.Error())
return
}
defer f.Close()
info, err := f.Stat()
if err != nil {
c.String(http.StatusInternalServerError, err.Error())
return
}
z := zip.NewWriter(c.Writer)
head, err := zip.FileInfoHeader(info)
if err != nil {
c.String(http.StatusInternalServerError, err.Error())
return
}
defer z.Close()
w, err := z.CreateHeader(head)
if err != nil {
c.String(http.StatusInternalServerError, err.Error())
return
}
_, err = io.Copy(w, f)
if err != nil {
c.String(http.StatusInternalServerError, err.Error())
return
}
})
engine.Run("127.0.0.1:8080")
}

Answer 2

So after hours of tireless work i figured out my approach was bad or maybe not possible with the level of my knowledge so here is a not so optimal solution but it works and fill ur file is not large it should be okay for you.
So you have a file template.zip and u want to add extra files, my initial approach was to copy the whole file into memory and edit it from their but i was having complications.
My next approach was to recreate the file in memory, file by file and to do that i need to know every file in the directory i used the code below to get all my files into a list
root := "template"
err = filepath.Walk(root, func(path string, info os.FileInfo, err error) error {
if info.IsDir() {
return nil
}append(files,path)}
now i have all my files and i can create a buffer to hold all this files
buf := new(bytes.Buffer)
// Create a new zip archive.
zipWriter := zip.NewWriter(buf)
now with the zip archive i can write all my old files to it while at the same time copying the contents
for _, file := range files {
zipFile, err := zipWriter.Create(file)
if err != nil {
fmt.Println(err)
}
content, err := ioutil.ReadFile(file)
if err != nil {
log.Fatal(err)
}
// Convert []byte to string and print to screen
// text := string(content)
_, err = zipFile.Write(content)
if err != nil {
fmt.Println(err)
}
}
At this point, we have our file in buf.bytes()
The remaining cold adds the new files and sends the response back to the client
for _, appCode := range appPageCodeText {
f, err := zipWriter.Create(filepath.fileextension)
if err != nil {
log.Fatal(err)
}
_, err = f.Write([]byte(appCode.Content))
}
err = zipWriter.Close()
if err != nil {
fmt.Println(err)
}
w.Header().Set("Content-Disposition", "attachment; filename="+"template.zip")
w.Header().Set("Content-Type", "application/zip")
w.Write(buf.Bytes()) //'Copy' the file to the client