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  1. I have need to send files to a Delphi service. I am able to send the files, but have not figured out how to extract the multipart items when received. procedure TForm15.IdHTTPServer1CommandGet(AContext: TIdContext; ARequestInfo: TIdHTTPRequestInfo; AResponseInfo: TIdHTTPResponseInfo); begin ARequestInfo.PostStream.Seek(0, soFromBeginning); Memo1.Lines.LoadFromStream(ARequestInfo.PostStream); end; Here is what I get when file is posted with a 2 other values added to the stream. What I need is a way to access each item. How do I ask for "type" and get back "OPRS" How do I ask for "ID" and get back "1234" How do I ask for "filename" and get back "test.pdf" How do I ask for actual file contents and save to memory stream to then write to disk ----------040224182640701 Content-Disposition: form-data; name="type" Content-Type: text/plain Content-Transfer-Encoding: quoted-printable OPRS ----------040224182640701 Content-Disposition: form-data; name="ID" Content-Type: text/plain Content-Transfer-Encoding: quoted-printable 1234 ----------040224182640701 Content-Disposition: form-data; name="file"; filename="test.PDF" Content-Type: application/pdf Content-Transfer-Encoding: binary %PDF-1.6 %âãÏÓ 290 0 obj << /Border [ 0 0 0 ] /Contents (þÿ