How to read urls using "With open()"?

Question

I was doing this and worked perfectly fine:

def b64_image(filename):
    with open("static/img/"+filename, 'rb') as f:
        b64 = base64.b64encode(f.read())
        return b64.decode('utf-8')

But now I uploaded the images to an S3 Bucket so I edited the code:

def b64_image(filename):
    with open("https://mybucket.s3-sa-east-1.amazonaws.com/" + filename, 'rb') as f:
        b64 = base64.b64encode(f.read())
        return b64.decode('utf-8')

And I'm getting the fllowing error:

FileNotFoundError: [Errno 2] No such file or directory:

How can I make this code work? How can I make the code to look for the actual URL image? Thanks in advance.

You need to use boto3 to access files on S3. Have you looked at it? — Roy2012, May 26 '20 at 19:55

score 0 · Accepted Answer · answered May 26 '20 at 20:02

0

In order to use open on files in S3 you will need to use a 3rd party library. In particular, this maintained and active library, smart-open does just that.

answered May 26 '20 at 20:02

gold_cy

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I haven`t used that, but it's definitely an option, thank you! – May 26 '20 at 20:13

How to read urls using "With open()"?

1 Answers1