Multiform File Upload with Rest
I had seen a related question but no clear answers for my situation. My fmx/android app is connected through the RestClient and I am trying to upload "any" file to a rest server that the docs specify with these parameters.
*POST media/upload*
*This call allows a client to upload a new file. Contrary to all other calls, this is not a JSON call. Instead, it requires data to be posted with the content-type set to multipart/form-data and defines the following fields:*
Path : Path to store new file at.
File : The multipart file data.
Request
POST /media/upload
Path=Verdi/Aida/01 - Preludio.mp3
File=...
Here is one code iteration I tried but get "invalid parameter error" back from server
procedure TForm1.UpButtonClick(Sender: TObject);
var
SendFile: TMemoryStream;
//FileStrm : TFileStream; //not sure which stream method is best.
begin
RESTRequest1.Resource := 'media/upload';
RestRequest1.Method := rmPOST;
RESTRequest1.Params.Clear;
try
SendFile := TMemoryStream.Create;
SendFile.LoadFromFile(LItemPath);
RESTRequest1.Params.AddHeader('Content-Type', 'multipart/form-data');
RESTRequest1.Params.AddItem('PATH=','upload.mp4;','File=','+SendFile+;' , ctMULTIPART_FORM_DATA);
RESTRequest1.Execute;
finally
SendFile.Free;
end;
This is all new to me, Thanks For any help...
Jay
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0 |
|
Jay
2/3/2015 6:53:44 AM
embarcadero.delphi.firemonkey
4901 articles. 
4 followers.
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Jay wrote:
> RESTRequest1.Params.AddItem('PATH=','upload.mp4;','File=','+SendFile+;'
, ctMULTIPART_FORM_DATA);
SendFile is a TMemoryStream. You can't concatenate an object and a string
together.
Try something more like this instead (untested):
{code}
procedure TForm1.UpButtonClick(Sender: TObject);
var
SendFile: TMemoryStream;
begin
SendFile := TMemoryStream.Create;
try
SendFile.LoadFromFile(LItemPath);
SendFile.Position := 0;
RESTRequest1.Resource := 'media/upload';
RESTRequest1.Method := rmPOST;
RESTRequest1.Params.Clear;
RESTRequest1.Params.AddItem('PATH', 'upload.mp4', pkGETorPOST, [], ctMULTIPART_FORM_DATA);
RESTRequest1.AddBody(SendFile, ctMULTIPART_FORM_DATA);
RESTRequest1.Execute;
finally
SendFile.Free;
end;
end;
{code}
--
Remy Lebeau (TeamB)
|
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0 |
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Remy
2/3/2015 8:26:23 PM
> {quote:title=Remy Lebeau (TeamB) wrote:}{quote}
> Jay wrote:
>
> > RESTRequest1.Params.AddItem('PATH=','upload.mp4;','File=','+SendFile+;'
> , ctMULTIPART_FORM_DATA);
>
> SendFile is a TMemoryStream. You can't concatenate an object and a string
> together.
>
> Try something more like this instead (untested):
>
> {code}
> procedure TForm1.UpButtonClick(Sender: TObject);
> var
> SendFile: TMemoryStream;
> begin
> SendFile := TMemoryStream.Create;
> try
> SendFile.LoadFromFile(LItemPath);
> SendFile.Position := 0;
>
> RESTRequest1.Resource := 'media/upload';
> RESTRequest1.Method := rmPOST;
> RESTRequest1.Params.Clear;
> RESTRequest1.Params.AddItem('PATH', 'upload.mp4', pkGETorPOST, [], ctMULTIPART_FORM_DATA);
> RESTRequest1.AddBody(SendFile, ctMULTIPART_FORM_DATA);
>
> RESTRequest1.Execute;
> finally
> SendFile.Free;
> end;
> end;
> {code}
>
> --
> Remy Lebeau (TeamB)
Thanks Remy,
Unfortunately it didnt like that either. actual server error I'm getting is "invalid file parameter". But I dont know if thats a catch all message covering all param errors and cant get in touch with server programmer.
I ran process with debug and message client shoots out is "exception class EIdHTTPProtocolException http/1.0 400 bad request"
I have google pilot and or Rest Debugger and would test it that way, but not sure how to formulate a request in it with actual file involved.
Jay
|
|
0 |
|
Jay
2/4/2015 1:37:43 PM
Jay wrote:
> Unfortunately it didnt like that either. actual server error I'm
> getting is "invalid file parameter". But I dont know if thats a catch
> all message covering all param errors and cant get in touch with
> server programmer.
> I ran process with debug and message client shoots out is "exception
> class EIdHTTPProtocolException http/1.0 400 bad request"
> I have google pilot and or Rest Debugger and would test it that way,
> but not sure how to formulate a request in it with actual file
> involved.
I have never used the REST components, so I cannot help you further with
them. All I can tell you is that you have to use AddBody() in order to send
a TStream, and you need the ctMULTIPART_FORM_DATA content type. But I don't
know the correct syntax to specify additional parameters. Since the REST
components use Indy internally, you are probably better off just using Indy's
TIdHTTP and TIdMultipartFormDataStream components directly, eg:
{code}
procedure TForm1.UpButtonClick(Sender: TObject);
var
MultiPartStrm: TIdMutipartFormDataStream;
begin
MultiPartStrm := TIdMultipartFormDataStream.Create;
try
MultiPartStrm.AddFormField('PATH', 'upload.mp4');
MultiPartStrm.AddFile('File', LItemPath);
IdHTTP1.Post('http://hostname/media/upload', MultiPartStrm);
finally
MultiPartStrm.Free;
end;
end;
{code}
--
Remy Lebeau (TeamB)
|
|
0 |
|
Remy
2/4/2015 6:55:37 PM
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