How do I pass arguments to my handler

GoGorilla

Go Problem Overview


I am trying to pass my database object along to my handlers, instead of having a global object. But I don't know if this is possible, I'm using Gorilla Mux package, and I can see that it takes a closure as a second param.

// https://github.com/gorilla/mux/blob/master/mux.go#L174
// HandleFunc registers a new route with a matcher for the URL path.
// See Route.Path() and Route.HandlerFunc().
func (r *Router) HandleFunc(path string, f func(http.ResponseWriter,
	*http.Request)) *Route {
	return r.NewRoute().Path(path).HandlerFunc(f)
}

Which then defines the params i can use, ideally i would like to have a third param like this.

// In my main
router.HandleFunc("/users/{id}", showUserHandler).Methods("GET")

func showUserHandler(w http.ResponseWriter, r *http.Request, db *gorm.DB) {
	fmt.Fprintf(w, "We should fetch the user with id %s", vars["id"])
}

Is there a workaround? Or do I need a global db object? I am new to Go, so please explain a potential answer in detail.

Go Solutions


Solution 1 - Go

Welcome to Go.

It is acceptable to have global variables and specially database objects.

However, there are few ways to workaround that if you prefer not to, for example you can create a struct and define your showHandler on it.

type Users struct {
	db *gorm.DB
}

func (users *Users) showHandler(w http.ResponseWriter, r *http.Request) {
	//now you can use users.db
}
func (users *Users) addHandler(w http.ResponseWriter, r *http.Request) {
	//now you can use users.db
}

// setup
users := &Users{db: createDB()}
router.HandleFunc("/users/{id}", users.showHandler).Methods("GET")
router.HandleFunc("/users/new", users.addHandler)
//etc

Another approach is creating a wrapper function:

db := createDB()
router.HandleFunc("/users/{id}", func(w http.ResponseWriter, r *http.Request) {
	showUserHandler(w, r, db)
}).Method("GET")

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionMartinElvarView Question on Stackoverflow
Solution 1 - GoOneOfOneView Answer on Stackoverflow