Go doing a GET request and building the Querystring

Question

I am pretty new to Go and don't quite understand everything as yet. In many of the modern languages Node.js, Angular, jQuery, PHP you can do a GET request with additional query string parameters.

Doing this in Go isn't quite a simple as it seems, and I can't really figure it out as yet. I really don't want to have to concatenate a string for each of the requests I want to do.

Here is the sample script:

package main

import (
    "fmt"
    "io/ioutil"
    "net/http"
)

func main() {
    client := &http.Client{}

    req, _ := http.NewRequest("GET", "http://api.themoviedb.org/3/tv/popular", nil)
    req.Header.Add("Accept", "application/json")
    resp, err := client.Do(req)

    if err != nil {
        fmt.Println("Errored when sending request to the server")
        return
    }

    defer resp.Body.Close()
    resp_body, _ := ioutil.ReadAll(resp.Body)

    fmt.Println(resp.Status)
    fmt.Println(string(resp_body))
}

In this example you can see there is a URL, which requires a GET variable of api_key with your api key as the value. The problem being that this becomes hard coded in the form of:

req, _ := http.NewRequest("GET", "http://api.themoviedb.org/3/tv/popular?api_key=mySuperAwesomeApiKey", nil)

Is there a way to build this query string dynamically?? At the moment I will need to assemble the URL prior to this step in order to get a valid response.

I suppose nothing, but it's not really a elegant sollution, just thought there is a better way of doing things in Go. You see the action changes, the method and then you have to string everything together. — Rudi Strydom, Jun 04 '15 at 19:40
You can use [`url.Values`](https://golang.org/pkg/net/url/#Values)'s [`Encode`](https://golang.org/pkg/net/url/#Values.Encode) method. You could also use [`URL.String`](https://golang.org/pkg/net/url/#URL.String) to build up the whole URL. — Dave C, Jun 04 '15 at 20:46

jcbwlkr · Answer 1 · 2016-12-05T20:25:03.103

309

As a commenter mentioned you can get Values from net/url which has an Encode method. You could do something like this (req.URL.Query() returns the existing url.Values)

package main

import (
    "fmt"
    "log"
    "net/http"
    "os"
)

func main() {
    req, err := http.NewRequest("GET", "http://api.themoviedb.org/3/tv/popular", nil)
    if err != nil {
        log.Print(err)
        os.Exit(1)
    }

    q := req.URL.Query()
    q.Add("api_key", "key_from_environment_or_flag")
    q.Add("another_thing", "foo & bar")
    req.URL.RawQuery = q.Encode()

    fmt.Println(req.URL.String())
    // Output:
    // http://api.themoviedb.org/3/tv/popular?another_thing=foo+%26+bar&api_key=key_from_environment_or_flag
}

http://play.golang.org/p/L5XCrw9VIG

edited Dec 05 '16 at 20:25

answered Jun 05 '15 at 02:41

jcbwlkr

7,789
2
22
28

Any solution to make the query string parameters in your wanted order? The keys are sorted in `Encode()` method. – qtopierw Jan 13 '18 at 10:48
1

@artificerpi if it was critical for some reason you could reimplement what they do inside the Encode method https://golang.org/src/net/url/url.go?s=24222:24253#L845 But I would wonder why it mattered. – jcbwlkr Jan 15 '18 at 18:58
4

You don't need to use NewRequest if you're not doing anything with it. You can just use `url.Parse("https://something.com/")` instead or even create an URL object directly. – Richard Mar 25 '19 at 05:43
one thing that is confusing me in this is that it reads like the OP first makes the request, and then after it's finished, goes back and updates the URL. I'm sure that's not the case, so hoping someone can clear up for me why we're first making the API call and then setting the URL we need – Jake Boomgaarden Sep 05 '20 at 22:19
3

@JakeBoomgaarden the `http.NewRequest` function does not actually call the remote server. It just constructs a `*http.Request` variable for you to use. You can tweak the request to your liking then send it off by passing it to the `*http.Client.Do` method. – jcbwlkr Sep 08 '20 at 19:52

score 58 · Answer 2 · answered Apr 21 '18 at 12:56

Use r.URL.Query() when you appending to existing query, if you are building new set of params use the url.Values struct like so

package main

import (
    "fmt"
    "log"
    "net/http"
    "net/url"
    "os"
)

func main() {
    req, err := http.NewRequest("GET","http://api.themoviedb.org/3/tv/popular", nil)
    if err != nil {
        log.Print(err)
        os.Exit(1)
    }

    // if you appending to existing query this works fine 
    q := req.URL.Query()
    q.Add("api_key", "key_from_environment_or_flag")
    q.Add("another_thing", "foo & bar")

    // or you can create new url.Values struct and encode that like so
    q := url.Values{}
    q.Add("api_key", "key_from_environment_or_flag")
    q.Add("another_thing", "foo & bar")

    req.URL.RawQuery = q.Encode()

    fmt.Println(req.URL.String())
    // Output:
    // http://api.themoviedb.org/3/tv/popularanother_thing=foo+%26+bar&api_key=key_from_environment_or_flag
}

score 57 · Answer 3 · answered Jul 11 '19 at 09:29

57

Using NewRequest just to create an URL is an overkill. Use the net/url package:

package main

import (
    "fmt"
    "net/url"
)

func main() {
    base, err := url.Parse("http://www.example.com")
    if err != nil {
        return
    }

    // Path params
    base.Path += "this will get automatically encoded"

    // Query params
    params := url.Values{}
    params.Add("q", "this will get encoded as well")
    base.RawQuery = params.Encode() 

    fmt.Printf("Encoded URL is %q\n", base.String())
}

Playground: https://play.golang.org/p/YCTvdluws-r

answered Jul 11 '19 at 09:29

Janek Olszak

4,067
1
28
22

1

Agreed, great answer! – pow-il Nov 18 '20 at 17:02
2

This should really be the accepted answer. – Paul Nov 28 '20 at 12:45
Totally agree, creating a request is overkill for something small like this. – kurczynski Mar 15 '22 at 15:17

Go doing a GET request and building the Querystring

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