我的Web服务器上有一个简单的php脚本,我需要使用HTTP POST上传文件,这是我在Delphi中所做的。
这是我与Indy的代码,但相应地它将无法正常工作,而且我无法弄清楚自己的工作不正常。我如何查看我在服务器上发送的内容是否有这样的工具?
procedure TForm1.btn1Click(Sender: TObject);
var
fname : string;
MS,dump : TMemoryStream;
http : TIdHTTP;
const
CRLF = #13#10;
begin
if PromptForFileName(fname,'','','','',false) then
begin
MS := TMemoryStream.Create();
MS.LoadFromFile(fname);
dump := TMemoryStream.Create();
http := TIdHTTP.Create();
http.Request.ContentType:='multipart/form-data;boundary =-----------------------------7cf87224d2020a';
fname := CRLF + '-----------------------------7cf87224d2020a' + CRLF + 'Content-Disposition: form-data; name=\"uploadedfile\";filename=\"test.png"' + CRLF;
dump.Write(fname[1],Length(fname));
dump.Write(MS.Memory^,MS.Size);
fname := CRLF + '-----------------------------7cf87224d2020a--' + CRLF;
dump.Write(fname[1],Length(fname));
ShowMessage(IntToStr(dump.Size));
MS.Clear;
try
http.Request.Method := 'POST';
http.Post('http://posttestserver.com/post.php',dump,MS);
ShowMessage(PAnsiChar(MS.Memory));
ShowMessage(IntToStr(http.ResponseCode));
except
ShowMessage('Could not bind socket');
end;
end;
end;
最佳答案
Indy为此具有TIdMultipartFormDataStream
:
procedure TForm1.SendPostData;
var
Stream: TStringStream;
Params: TIdMultipartFormDataStream;
begin
Stream := TStringStream.Create('');
try
Params := TIdMultipartFormDataStream.Create;
try
Params.AddFile('File1', 'C:\test.txt','application/octet-stream');
try
HTTP.Post('http://posttestserver.com/post.php', Params, Stream);
except
on E: Exception do
ShowMessage('Error encountered during POST: ' + E.Message);
end;
ShowMessage(Stream.DataString);
finally
Params.Free;
end;
finally
Stream.Free;
end;
end;
关于delphi - 带有Indy的Http Post,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/10765661/