Translating from curl to uri: where do I put the -F filename parameter?

[Larry Fast]

I'm trying to convert the following curl POST command into an Ansible uri: Task.

curl -F upload_file=@my_datafile.xml http://my_destination

At the http end I'm trying to respond to the following page:

<input type="file" name="upload_file" value="" /></fieldset><p><input type="submit" name="submit" value="Submit" /></p>

getting the file contents into the Post seems easy enough but where do I stuff "upload_file" parameter?  Will I need to construct the whole POST body via a template?

A dump of the curl comms shows:

0000: POST /my_destination HTTP/1.1

0037: User-Agent: curl/7.40.0

0050: Host: 10.128.17.41

0064: Accept: */*

0071: Content-Length: 191705

0089: Expect: 100-continue

009f: Content-Type: multipart/form-data; boundary=--------------------

00df: ----5357ae70294501c5

00f5:

<= Recv header, 23 bytes (0x17)

0000: HTTP/1.1 100 Continue

=> Send data, 177 bytes (0xb1)

0000: --------------------------5357ae70294501c5

002c: Content-Disposition: form-data; name="upload_file"; filename="es

006c: 1-msg-hello-bye-20150816151607.xml"

0090: Content-Type: application/xml

00af:

=> Send data, 16384 bytes (0x4000)

0000: <?xml version="1.0" encoding="UTF-8"?>.<system>..<sys>...<timest

0040: amp>2015-08-16 15:16:07</timestamp>...<uuid>32333036-3935-584D-5

0080: 133-303330313836</uuid>...<hostname>es1-msg-hello-bye</hostname>.

00c0: ..<man_ip_address>33.21.15.23</man_ip_address>