How to read Query Params in fasthttp without URL decoding

Question

I am processing a GET request in Go using fasthttp.

The query parameter test in this request is .%2A%2Ftoday%2F.%2A.

I am using POSTMAN to create the request, and the URL generated is:

http://localhost:3000/apiname/?test=.%252A%252Ftoday%252F.%252A

ctx.QueryArgs().Peek("test") gives me .*/today/.* instead of the original .%2A%2Ftoday%2F.%2A

I know I cannot partially encode/decode the request URL. Is there any way to get the original param as is?

Maybe no, what is the use-case of this? A hack is again encode. — Eklavya, Jun 03 '20 at 08:18
You can do this way `url.QueryEscape(".*/today/.*")` using `net/url` package — Eklavya, Jun 03 '20 at 08:45
@Eklavya Thanks! You can post this as an answer , I'll accept. — Alzio, Jun 03 '20 at 09:15

score -1 · Answer 1 · answered Aug 08 '21 at 10:49

Are you sure? I've just tested it and I'm getting the result you want.

This is the minimal working example:

package main

import (
    "fmt"
    "log"

    "github.com/fasthttp/router"
    "github.com/valyala/fasthttp"
)

func Test(ctx *fasthttp.RequestCtx) {
    ctx.Response.SetBodyString(string(ctx.QueryArgs().Peek("test")))
    fmt.Println(string(ctx.QueryArgs().Peek("test")))
}

func main() {

    r := router.New()
    r.GET("/test", Test)
    log.Fatal(fasthttp.ListenAndServe("127.0.0.1:8080", r.Handler))
}

This is the command line output after the GET request:

$ go run main.go                                                                                                                
.%2A%2Ftoday%2F.%2A

and this is the response in Postman:

You seem to be passing double encoded param. – lifeisshubh Aug 14 '21 at 10:25 — lifeisshubh, Aug 14 '21 at 10:25

How to read Query Params in fasthttp without URL decoding

1 Answers1