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This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data"
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|"Body" -> {"image" -> File[image]}|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

Edit 2 This evidently uses expect: "100-continue", which is not supported by some common load balancers that may be between you and your server. Thus, it may not work in many cases.

This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data",
      "Expect" -> "" (* See edit 2.5 below *)
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|
  "Body" -> {"image" -> File[image]},
  "Expect" -> "" (* See edit 2.5 below *)
|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

Edit 2, 2.5 Mathematica's HTTP client uses Expect: "100-continue", which is not supported by some common load balancers that may be between you and your server. To disable that, add the header "Expect" -> "".

This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data"
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|"Body" -> {"image" -> File[image]}|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

Edit 2 This evidently uses expect: "100-continue", which is not supported by many common load balancers and other devices that may be between you and your server. Thus, it may not work in many cases.

This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data"
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|"Body" -> {"image" -> File[image]}|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

Edit 2 This evidently uses expect: "100-continue", which is not supported by some common load balancers that may be between you and your server. Thus, it may not work in many cases.

This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data"
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|"Body" -> {"image" -> File[image]}|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

This is a hack, but it works.

UploadFile[url_, filePath_, urlParams___] := With[
  {
    bytes = Import[filePath, "Byte"],
    filename = StringJoin[FileBaseName[filePath], ".", FileExtension[filePath]]
  },
  URLExecute[
    url,
    urlParams,
    "Method" -> "POST",
    "MultipartElements" -> {
      {"file\"; filename=\"" <> filename, "application/octet-stream", bytes}
    },
    "Headers" -> {
      "Accept" -> "application/json; charset=UTF-8",
      "Content-Type" -> "multipart/form-data"
    }
  ]
]

Notice that I'm stuffing the filename parameter in the first element of "MultipartElements" along with the field name in order to get it in Content-Disposition for that element.

(* Get a new ephemeral RequestBin from http://requestb.in/, then put URL below: *)
UploadFile["http://requestb.in/1hyjbdl1", "ExampleData/rose.gif"]

result:

Content-Disposition: form-data; name="file"; filename="rose.gif"

Theoretically you could also specify "Content-Disposition" -> "form-data; filename=\"foo\"" in the request header, but the server would have to be expecting exactly one file in the form-data and parse accordingly, whereas the above approach with filenames in each part I think is more commonly expected.

Edit

In version 11, there's a no-hack way to do this:

image = FindFile["ExampleData/Ocelot.jpg"];
req = HTTPRequest[url, <|"Body" -> {"image" -> File[image]}|>];
URLRead[req, "Body"]

Docs (see Scope > "To send files...")

Edit 2 This evidently uses expect: "100-continue", which is not supported by many common load balancers and other devices that may be between you and your server. Thus, it may not work in many cases.