How to pass an URL to a fasthttp-router, in Golang?

Question

The application consists in outputting a bash command to a buffer. The catch: the arguments are urls.

Which is working for simple cases, like: "duckduckgo". I.e., http://localhost:8080/free-riding/duckduckgo works perfectly.

Output:

   #[1]DuckDuckGo (HTML)

   [2]About DuckDuckGo
   ____________________ Submit

References

   1. https://duckduckgo.com/opensearch_html_v2.xml
   2. https://duckduckgo.com/about.html

But, http://localhost:8080/free-riding/www.estadao.com.br/politica/blog-do-fausto-macedo/dia-d-julgamento-bolsonaro-inelegivel-tse-reuniao-embaixadores/ will give:

Not Found.

Thus, how can I encode an url-argument to be passed to a router, using fasthttp?

Here is the code,

package main

import (
    "fmt"
    "log"
    "os/exec"

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

func Lynx(ctx *fasthttp.RequestCtx) {
    var url string = ctx.UserValue("url").(string)
    cmd := exec.Command("lynx", "--dump", url)
    stdout, err := cmd.Output()

    if err != nil {
        log.Fatal(err)
        return
    }
    fmt.Fprintf(ctx, "%s\n", stdout)
}

func main() {
    r := router.New()
    r.GET("/free-riding/{url}", Lynx)
    log.Fatal(fasthttp.ListenAndServe(":8080", r.Handler))
}

Escape the slashes, or use a [catch-all parameter](https://github.com/fasthttp/router#catch-all-parameters) instead of `{url}`. — mkopriva, Jun 29 '23 at 13:06
I mixed both, I will post the final code, as a solution. Thank you. — BuddhiLW, Jun 29 '23 at 13:33

BuddhiLW · Accepted Answer · 2023-06-29T13:41:56.777

The final code, which worked just fine, using @mkopriva 's commentary: Escape the slashes, or use a catch-all parameter instead of {url}.

func Lynx(ctx *fasthttp.RequestCtx) {
    var urlstring string = ctx.UserValue("url").(string)
    urlparsed, err := url.Parse(urlstring)
    urlparsedstring := urlparsed.String()
    fmt.Fprintf(ctx, "URL: %s\n", urlparsedstring)

    cmd := exec.Command("lynx", "--dump", urlparsedstring)
    stdout, err := cmd.Output()

    if err != nil {
        log.Fatal(err)
        return
    }
    fmt.Fprintf(ctx, "%s\n", stdout)
}

func main() {
    r := router.New()
    r.GET("/free-riding/{url:*}", Lynx)
    log.Fatal(fasthttp.ListenAndServe(":8080", r.Handler))
}

Showing the output: https://imgur.com/a/8f3rpjv

score 0 · Answer 2 · answered Jun 29 '23 at 13:34

you should encode it using URL encoding. Here is a small change in your code to include URL encoding and decoding:

package main

import (
    "fmt"
    "log"
    "net/url"
    "os/exec"

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

func Lynx(ctx *fasthttp.RequestCtx) {
    urlParam := ctx.UserValue("url").(string)
    decodedURL, err := url.QueryUnescape(urlParam)
    if err != nil {
        log.Fatalf("Error decoding URL: %v", err)
        return
    }
    cmd := exec.Command("lynx", "--dump", decodedURL)
    stdout, err := cmd.Output()

    if err != nil {
        log.Fatalf("Error executing command: %v", err)
        return
    }
    fmt.Fprintf(ctx, "%s\n", stdout)
}

func main() {
    r := router.New()
    r.GET("/free-riding/{url}", Lynx)
    log.Fatal(fasthttp.ListenAndServe(":8080", r.Handler))
}

Unfortunately, didn't work https://imgur.com/a/PZjpfAV – BuddhiLW Jun 29 '23 at 13:39 — BuddhiLW, Jun 29 '23 at 13:39

How to pass an URL to a fasthttp-router, in Golang?

2 Answers2