We have an OpenAPI tool with basic Authentication, which should send a file. We have managed to make the basic authentication part work, but the we cannot send the file. The curl command looks like:

curl -X 'POST' -H 'accept:application/json' -H 'Content-Type: multipart/form-data' -u user -F 'input=@myfile.txt;type=text/plain' -F 'Key2=Value2' -F 'Key3=Value3' -F 'Key4=Value4' https://myserver.com/api/basic/launch/task

Basic authentication works with a base64 conversion adding into Header 1:
Name: Authorization
Value: 'Basic '.@userpass
where userpass is converttobase64(ConvertStringToBytes(user:password)) - a very non trivial way of authentication.

However, it is unclear, how can we send the input part. We were trying to read the file into a variable as a text file without delimiters and read until EndOfFile with both keepEndOfLine turned on and off and saving the content into a variable and putting the variable into parameters. We tried also converting the file to bytes and then to Base64, but nothing really worked.
Can you please help, how should we read myfile.txt and what should we put into the HTTP connector parameters /it is a POST with FormMultipart/?