VincentBlackdot
How to post data in a url
how do i post data to a url with its parameters eg
www.sendmessage.com/client/xml/send?client&pass&message&phone
i want to pick the fields client, pass, message and phone from the db and pass them to a function like
def send_msg(entry) do
{:ok, url} ->
options = [
hackney: [basic_auth: {url.client, url.pass}],
ssl: [verify: :verify_none]]
{:ok, post_params} = prepare_request_params(entry, url)
case HTTPoison.post(String.trim(url.url), post_params, headers, options) do
{:ok, %HTTPoison.Response{status_code: 200, body: body}} ->
IO.inspect(process_response_body(body))
end
end
def prepare_request_params(entry, url) do
{:ok,Poison.encode!(%{
"username" => "#{url.client}",
"password" => "#{url.pass}",
"recipient" => ["#{entry.phone}"],
"message" => "#{entry.message}"
})}
end
First Post!
dmitrykleymenov
The short answer is you can’t. POST request is created for hiding its body. And it’s hard to understand how will you use it, since code is invalid. I’m pretty sure you need something different than using url parameters in post request. Explain what do you need to achieve in a bigger scale.
If you build your url inside your app, you don’t need to interpolate username and pass to it, transfer them separately.
Most Liked
gregvaughn
Aside: this is a very strange construction. Don’t use the ! version of Poison.encode if you want an :ok tuple as a return value. Just take away the ! and you’ll get an :ok tuple. (Also, Poison is an old library not much used for new code these days).
Last Post!
dmitrykleymenov
I would check the requirements of API. Are you sure that they need url parameters in url for post request with doubling it inside POST body? May be it’s GET, or they don’t need params in url?
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