How to unzip a string in language Go

Question

I have some Go code below which is has PK zipped string. How to get the contents of each file in this zip string?

Below is an attempt using string manipulation I can get the contents of the first file. Is there another way to get the contents without using string functions?

package main

import (
    "fmt"
    "strings"
)

func main() {
 
    x := "PK\x03\x04\n\x00\x00\x00\x00\x00\x14OOQ\xddDYc\v\x00\x00\x00\v\x00\x00\x00\t\x00\x00\x00file1.txttextfileonePK\x03\x04\n\x00\x00\x00\x00\x00\x1aOOQq?\xb5]\t\x00\x00\x00\t\x00\x00\x00\t\x00\x00\x00file2.txttextfile2PK\x01\x02?\x00\n\x00\x00\x00\x00\x00\x14OOQ\xddDYc\v\x00\x00\x00\v\x00\x00\x00\t\x00$\x00\x00\x00\x00\x00\x00\x00 \x00\x00\x00\x00\x00\x00\x00file1.txt\n\x00 \x00\x00\x00\x00\x00\x01\x00\x18\x00#~\xfb\xa7\x85\xa2\xd6\x01#~\xfb\xa7\x85\xa2\xd6\x01\x00\xa8\xdfE\xe7\x8b\xd6\x01PK\x01\x02?\x00\n\x00\x00\x00\x00\x00\x1aOOQq?\xb5]\t\x00\x00\x00\t\x00\x00\x00\t\x00$\x00\x00\x00\x00\x00\x00\x00 \x00\x00\x002\x00\x00\x00file2.txt\n\x00 \x00\x00\x00\x00\x00\x01\x00\x18\x00u\xa7v\xaf\x85\xa2\xd6\x01u\xa7v\xaf\x85\xa2\xd6\x01\xa8\x9eˏ\x85\xa2\xd6\x01PK\x05\x06\x00\x00\x00\x00\x02\x00\x02\x00\xb6\x00\x00\x00b\x00\x00\x00\x00\x00"

    i := strings.Index(x , "PK")
    right := x[i+1:]
    i = strings.Index(right , "PK")
    left := right[:i]
    i = strings.LastIndex(left , ".txt")
    file1contents := left[i+4:]
    fmt.Println("file1contents : ", file1contents)
}

Output of above file1contents : textfileone

What have you tried? Include your code. What problems did you encounter? — Jonathan Hall, Oct 15 '20 at 07:01
I have only limited knowledge in go (about 5 minutes worth ) didn’t really know where to start. But again thanks to SO community kudos to Brits. — ozmike, Oct 15 '20 at 12:47
If you only have 5 minutes experience with Go, then the first thing to do is read about Go. Start with [A Tour of Go](https://tour.golang.org/). — Jonathan Hall, Oct 15 '20 at 12:49
You got an answer (lucky for you), by asking a question that doesn't conform to site guidelines, and will be removed. So I guess for a certian, narrow, weasly definition of "works", it worked for you, in this case. — Jonathan Hall, Oct 15 '20 at 12:59
Some one upvoted the answer beside me. This info is hard to find. But not a great question I will admit — ozmike, Oct 15 '20 at 13:04
There's nothing wrong with the answer. The problem is the question. — Jonathan Hall, Oct 15 '20 at 13:07
At the time I wrote the question really had no idea what else to write . It’s a unusual use case .. But I tried to make it accessible to a guru ..and the gods answered! — ozmike, Oct 15 '20 at 13:18

score 2 · Answer 1 · answered Oct 15 '20 at 01:46

The zip package in the standard library will do this for you:


func main() {
    z := []byte("PK\x03\x04\n\x00\x00\x00\x00\x00\x14OOQ\xddDYc\v\x00\x00\x00\v\x00\x00\x00\t\x00\x00\x00file1.txttextfileonePK\x03\x04\n\x00\x00\x00\x00\x00\x1aOOQq?\xb5]\t\x00\x00\x00\t\x00\x00\x00\t\x00\x00\x00file2.txttextfile2PK\x01\x02?\x00\n\x00\x00\x00\x00\x00\x14OOQ\xddDYc\v\x00\x00\x00\v\x00\x00\x00\t\x00$\x00\x00\x00\x00\x00\x00\x00 \x00\x00\x00\x00\x00\x00\x00file1.txt\n\x00 \x00\x00\x00\x00\x00\x01\x00\x18\x00#~\xfb\xa7\x85\xa2\xd6\x01#~\xfb\xa7\x85\xa2\xd6\x01\x00\xa8\xdfE\xe7\x8b\xd6\x01PK\x01\x02?\x00\n\x00\x00\x00\x00\x00\x1aOOQq?\xb5]\t\x00\x00\x00\t\x00\x00\x00\t\x00$\x00\x00\x00\x00\x00\x00\x00 \x00\x00\x002\x00\x00\x00file2.txt\n\x00 \x00\x00\x00\x00\x00\x01\x00\x18\x00u\xa7v\xaf\x85\xa2\xd6\x01u\xa7v\xaf\x85\xa2\xd6\x01\xa8\x9eˏ\x85\xa2\xd6\x01PK\x05\x06\x00\x00\x00\x00\x02\x00\x02\x00\xb6\x00\x00\x00b\x00\x00\x00\x00\x00")
    r, err := zip.NewReader(bytes.NewReader(z), int64(len(z)))
    if err != nil {
        panic(err)
    }
    for _, f := range r.File {
        fmt.Printf("Contents of %s:\n", f.Name)
        rc, err := f.Open()
        if err != nil {
            panic(err)
        }
        c, err := ioutil.ReadAll(rc)
        if err != nil {
            panic(err)
        }
        fmt.Printf("%s\n\n", c)
        rc.Close()
        fmt.Println()
    }

}

Try it in the playground

Truely awesome and helpful effort..glad there are some 'real' programmer still left in this world many thanks.. — ozmike, Oct 15 '20 at 04:35
Here is the reverse process ie going from byte array to zip if any one needs it https://stackoverflow.com/questions/12440387/zip-a-byte-array-in-go — ozmike, Nov 19 '20 at 02:56

How to unzip a string in language Go

1 Answers1