get file name boto3 python - amazon-s3

Question

I'm a new in flask and boto3 and want create a simple upload form to s3 amazon and i need to save file on s3 with the existing filename and return link to this file.
Two issues:
1) In the example below file always uploads with a name 'test'. If upload file name will be 'my file.pdf' i need to upload file to s3 with the same name ('file.pdf').
I believe it can be done with request but i do not know how exactly. How it can be done?
2) How to return link to the file that i just have uploaded? (I have no idea)
below is my code
#app.route('/')
def index():
return '''
<form method="post" enctype="multipart/form-data" action="upload">
<input type="file" name="file" multiple>
<input type="submit">
</form>
'''
#app.route('/upload', methods=['POST', 'GET'])
def upload():
s3 = boto3.resource('s3')
s3.Bucket('dimkzn').put_object(='test', Body=request.files['file'])
return 'file save! to S3'
if name=='main':
app.run(debug=True) code here

Answer 1

In your upload function you are missing the Key parameter names where you have ="test" which is why every file is being saved as test. You can pass through the filename from the file object within request.files
#app.route('/upload', methods=['POST', 'GET'])
def upload():
s3 = boto3.resource('s3')
s3.Bucket('dimkzn').put_object(Key=request.files['file'].filename, Body=request.files['file'])
return 'file save! to S3'
The URL path will then be https://s3-<region>.amazonaws.com/dimkzn/<filename>
If you wish to upload the file into a subfolder, folders don't actually exist within S3 but you can create folder structures by changing the key of the file. E.G
#app.route('/upload', methods=['POST', 'GET'])
def upload():
s3 = boto3.resource('s3')
s3.Bucket('dimkzn').put_object(Key=f"media/example/{request.files['file'].filename}", Body=request.files['file'])
return 'file save! to S3'
Will upload the file to:
https://s3-<region>.amazonaws.com/dimkzn/media/example/<filename>

Answer 2

I think your problem is in the line:
s3.Bucket('dimkzn').put_object(='test', Body=request.files['file'])
You're missing an argument name before the ='test'. I can't see from your code what request.files['file'] returns: is that the filename or the file content?
Correct syntax is
object = bucket.put_object(
Body=b'bytes'|file,
Key='filename')
...where Body is the content of the file, Key is the filename.
Once successful, you can construct the link yourself as follows:
https://s3-<region>.amazonaws.com/<bucketname>/<filename>
Good luck!

Answer 3

Your Missing "Key", should be like this:
s3.Bucket('dimkzn').put_object(Key='test', Body=request.files['file']