How can I send a MemoryStream to a cgi application ?

hi fausomen, maybe this works:

//1x idhttp-component,
//1x Memo-component

procedure TForm1.Button1Click(Sender: TObject);
var src,cont,tmp:Tmemorystream;
    ds,boundary:string;
    //helping procedure to write string with linefeed...
       procedure wr(const txt:string);
       var ds:string;
       begin
         ds:=txt+#13#10;
         src.Write(ds[1],length(ds));
       end;
begin

src  :=Tmemorystream.Create;
cont :=Tmemorystream.Create;


boundary:='---------------------------thisStringHasToBeUNIQUE';

//your http-content-type is multipart
idhttp1.Request.ContentType:='multipart/form-data; boundary='+boundary;

//the binary-file:
wr('--'+boundary);
wr('Content-Disposition: form-data; name="userfile"; filename="C:\originalfilename.bmp"');
wr('Content-Type: application/octet-stream');
wr('');

  //integrate the bin-data...
  tmp:=tmemorystream.Create;
  tmp.LoadFromFile('c:\binary.bmp');
  tmp.seek(0,0); src.CopyFrom(tmp,tmp.size);
  tmp.free;

wr(''); //last-linefeed is absolutely necessary!

//close your multipart...
wr('--'+boundary+'--');
wr('');

 IdHTTP1.Post('http://www.host.com/response.php?var1=content1',src,cont);

 cont.seek(soFromBeginning,0);
 memo1.Lines.LoadFromStream(cont);

src.free;
cont.free;
end;


//your response.php-file could be like this:
<?
$tmpfilename=$_FILES["userfile"][tmp_name];
$orgfilename=$_FILES["userfile"][name];

system('cp '.$tmpfilename.' /tmp/'.$orgfilename);

/*
the structure of _$FILES["blalaber"] is
Array
(
    [name] => testfile.zip
    [type] => application/x-zip-compressed
    [tmp_name] => /tmp/phppZxrmP
    [error] => 0
    [size] => 29868
)
*/
?>
---------------
hope, this is what r you searching for...

luwo

Hi luwo,

I need to use a Delphi cgi application to receive image.
My client application is ready to send the image (using IdHTTP1.Post) but I dont know how to transfer the stream (from client application) to a variable MemoryStream in the cgi application.

Thanks

faustomen:
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