I'm trying to resume an upload using indy (HTTP Post), the code looks like this (using Delphi 2010, Indy 10.4736):
IdHttp.Head('http://localhost/_tests/resume/large-file.bin');
ByteRange := IdHttp.Response.ContentLength + 1;
// Attach the file to post/upload
Stream := TIdMultipartFormDataStream.Create;
with Stream.AddFile('upload_file', 'D:\large-file.bin', 'application/octet-stream') do
begin
HeaderCharset := 'utf-8';
HeaderEncoding := '8';
end; // with
with IdHTTP do
begin
IOHandler.LargeStream := True;
with Request do
begin
ContentRangeStart := ByteRange;
ContentRangeEnd := (Stream.Size - ByteRange);
ContentLength := ContentRangeEnd;
ContentRangeInstanceLength := ContentLength;
end; // with
Post('http://localhost/_tests/resume/t1.php', Stream);
end; // with
but upload resume doesn't work :(
I looked into Indy's code, it seems that this function in IdIOHandler.pas
TIdIOHandler.Write()
always deal with complete streams/files (since the parameter ASize: TIdStreamSize seems to be always 0, which according to the code means sending the full file/stream).
This prevents indy from resuming the upload.
My question is: is it possible to avoid sending the full file?
Setting content range didn't change anything. I also tweaked indy's code (modified 3 lines) to make indy obey to the content range / stream position, but it's buggy and indy always end up hanging in IdStackWindows.pas because of an infinite timeout here:
TIdSocketListWindows.FDSelect()
As I told you in your earlier question, you have to post a TStream
containing the remaining file data to upload. Don't use TIdMultipartFormDataStream.AddFile()
, as that will send the entire file. Use the TStream
overloaded version of TIdMultipartFormDataStream.AddFormField()
instead.
And TIdHTTP
is not designed to respect the ContentRange...
properties. Most of the Request
properties merely set the corresponding HTTP headers only, they do not influence the data. That is why your edits broke it.
Try this:
IdHttp.Head('http://localhost/_tests/resume/large-file.bin');
FileSize := IdHttp.Response.ContentLength;
FileStrm := TFileStream.Create('D:\large-file.bin', fmOpenRead or fmShareDenyWrite);
try
if FileSize < FileStrm.Size then
begin
FileStrm.Position := FileSize;
Stream := TIdMultipartFormDataStream.Create;
try
with Stream.AddFormField('upload_file', 'application/octet-stream', '', FileStrm, 'large-file.bin') do
begin
HeaderCharset := 'utf-8';
HeaderEncoding := '8';
end;
with IdHTTP do
begin
with Request do
begin
ContentRangeStart := FileSize + 1;
ContentRangeEnd := FileStrm.Size;
end;
Post('http://localhost/_tests/resume/t1.php', Stream);
end;
finally
Stream.Free;
end;
end;
finally
FileStrm.Free;
end;
With that said, this is a GROSS MISUSE of HTTP and multipart/form-data
. For starters, the ContentRange
values are in the wrong place. You are applying them to the entire Request
as a whole, which is wrong. They would need to be applied at the FormField
instead, but TIdMultipartFormDataStream
does not currently support that. Second, multipart/form-data
was not designed to be used like this anyway. It is fine for uploading a file from the beginning, but not for resuming a broken upload. You really should stop using TIdMultipartFormDataStream
and just pass the file data directly to TIdHTTP.Post()
like I suggested earlier, eg:
FileStrm := TFileStream.Create('D:\large-file.bin', fmOpenRead or fmShareDenyWrite);
try
IdHTTP.Post('http://localhost/_tests/upload.php?name=large-file.bin', FileStrm);
finally
FileStrm.Free;
end;
.
IdHTTP.Head('http://localhost/_tests/files/large-file.bin');
FileSize := IdHTTP.Response.ContentLength;
FileStrm := TFileStream.Create('D:\large-file.bin', fmOpenRead or fmShareDenyWrite);
try
if FileSize < FileStrm.Size then
begin
FileStrm.Position := FileSize;
IdHTTP.Post('http://localhost/_tests/resume.php?name=large-file.bin', FileStrm);
end;
finally
FileStrm.Free;
end;
I already explained earlier how to access the raw POST
data in PHP.