How can I rewrite this CURL multipart/form-data request without using -F?

Question

How can I rewrite the following CURL command, so that it doesn\'t use the -F option, but still generates the exact same HTTP request? i.e. so that it passes th

Answer 1

This is what I'm using, I think it's clean and doesn't need temporary files nor gobbles up RAM in case you want to upload whole files (so no reading files into memory).

# Set these two.
file='path/to/yourfile.ext'
url='http://endpoint.example.com/foo/bar'

delim="-----MultipartDelimeter$$$RANDOM$RANDOM$RANDOM"
nl=$'\r\n'
mime="$(file -b --mime-type "$file")"

# This is the "body" of the request.
data() {
    # Also make sure to set the fields you need.
    printf %s "--$delim${nl}Content-Disposition: form-data; name=\"userfile\"${nl}Content-Type: $mime$nl$nl"
    cat "$file"
    printf %s "$nl--$delim--$nl"
}

# You can later grep this, or something.
response="$(data | curl -# "$url" -H "content-type: multipart/form-data; boundary=$delim" --data-binary @-)"