How to get the file name from a simple curl post request in aws lambda throu aws API gateway - amazon-web-services

Question

I have the following curl request,
curl -X POST "https://foo123.execute-api.us-east-1.amazonaws.com/default/test-2" \
-H "accept: application/json" \
-H "Content-Type: multipart/form-data" \
-F "file=#wiki.png;type=image/png"
Due to the existing pipeline structure, I am not able to change the present architecture of this curl request, I want to access the name of this file in the lambda function when I am using aws API gateway as the trigger.
I am aware I can simply send the file name in header but that doesn't work for the pipeline I am going to use it with, so my question is.
How can I access the name of this file inside a lambda function in aws, given that I can't change my curl request.
To further give clarity to my question in flask APIs we are able to get the name of this uploaded file by simply using something like.
args = upload_parser.parse_args()
uploaded_file = args['file']
required_file_name = uploaded_file.filename

Answer 1

Interesting problem, actually you get the values of the body as a base64 encoded image you will have to either use any of the existing libraries to parse it for the name and content from body or write your own parser, your can refer to the code snippet below for reference :
import cv2
import os
import base64
import numpy as np
import email
def http_api(event):
post_data = base64.b64decode(event['body'])
# fetching content-type
try:
content_type = event["headers"]['Content-Type']
except:
content_type = event["headers"]['content-type']
# concate Content-Type: with content_type from event
ct = "Content-Type: "+content_type+"\n"
# parsing message from bytes
msg = email.message_from_bytes(ct.encode()+post_data)
if msg.is_multipart():
multipart_content = {}
# retrieving form-data
for part in msg.get_payload():
# checking if filename exist as a part of content-disposition header
if part.get_filename():
# fetching the filename
file_name = part.get_filename()
multipart_content[part.get_param('name', header='content-disposition')] = part.get_payload(decode=True)
img_str = multipart_content["file"]
nparr = np.fromstring(img_str, np.uint8)
image = cv2.imdecode(nparr, cv2.IMREAD_COLOR)
return image