CODEWITHSUNDEEP
go
Cam.phiefr
Cam.phiefr

Reputation: 97

net/http : is it possible to have http.HandleFunc custom params?

I'm trying to find a way to create special routes to that would allow post request to work with a params ( /url/:param ).

I've tried something like this :

http.HandleFunc("/myroute/:myparam",myfunction)

And i was hopping to be able to get the params as of the http-request in side "myfunction" with req.Form but it keeps on failing on me and i get 404'ed.

I recognize it is something strange to ask as i could pass my params in the body of the function, but for ease of use/display i've been asked to be able to pregenerate a list of static "routes" on some preset so that it can be distributed as different urls + optional params instead of one url for everybody + params..

Upvotes: 0

Views: 3638

Answers (2)

Vallerious
Vallerious

Reputation: 640

You can define a handle function on / and then have custom parsing of the path, like splitting on / character.

http.HandleFunc("/", dispatcher)

In this way, your handler will accept everything and you can route the requests manually to the respective function. For example, PUT calls would expect some parameter after the resource (/orders/2)

logs.Info("The URL that you are calling is: " + req.URL.Path)

Output: The URL that you are calling is: /orders/2

You can also do: http.HandleFunc("/orders/", dispatcher) to narrow down the scope of your handler. In this way, you can leave the validation of the resource portion of the path to Go http package.

Upvotes: 1

KAdot
KAdot

Reputation: 2087

It's not possible with the default Mux, but you can use alternative routers like httprouter, Gorilla Mux or chi.

Here is an example of httprouter usage:

package main

import (
    "fmt"
    "github.com/julienschmidt/httprouter"
    "net/http"
)

func Hello(w http.ResponseWriter, r *http.Request, ps httprouter.Params) {
    fmt.Fprintf(w, "hello, %s!\n", ps.ByName("name"))
}

func main() {
    router := httprouter.New()
    router.GET("/hello/:name", Hello)
    http.ListenAndServe(":8080", router)
}

Upvotes: 6

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