wlminimal

wlminimal

How to get a link(url) from uploaded image in Amazon S3

Hi!
Now I am trying to upload image using ex_aws for Amazon AWS S3

Here is my code

  def upload_file_to_S3(bucket_name, s3_filename, file_binary) do
    S3.put_object(bucket_name, s3_filename, file_binary, [{:acl, :public_read}])
    |> ExAws.request()
  end

and I hope(?) it returns with link(uploaded image url) but it didn’t return it
This is first time using ex_aws, and trying to find docs about it but can’t find it.

How can I get an upload image url like " https://s3-us-west-1.amazonaws.com/text-marketing/03da408158584cd38a51aea77754d86c.jpg"

so I can use it in html ‘a’ tag or save it to the database.

Marked As Solved

joaquinalcerro

joaquinalcerro

You can find a example here:

defmodule Epr.Services.AwsS3 do
  def put_object(bucket, filename, binary_file, opts \\ []) do
    ExAws.S3.put_object(bucket, filename, binary_file, opts)
    |> ExAws.request()
  end

  def delete_object(bucket, filename) do
    ExAws.S3.delete_object(bucket, filename)
    |> ExAws.request()
  end

  def presigned_url(
        method,
        bucket,
        filename,
        config \\ ExAws.Config.new(:s3),
        opts \\ [expires_in: 7200]
      ) do
    {:ok, url} = ExAws.S3.presigned_url(config, method, bucket, filename, opts)
    url
  end
end

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