带有 indy 的 Http Post [英] Http Post with indy

问题描述

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，我正在 Delphi 中执行此操作.

I have a simple php script on my web server which I need to upload a file using HTTP POST, which I am doing in Delphi.

这是我与 Indy 的代码，但显然它不起作用，我无法弄清楚我做错了什么.如何查看我在服务器上发送的内容是否有这样的工具?

Here is my code with Indy but aparantely it won't work and I can't figure out what i am not doing properly. How can I view what I send on the server is there such a tool ?

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;

推荐答案

Indy 有 TIdMultipartFormDataStream 用于此目的:

Indy has TIdMultipartFormDataStream for this purpose:

procedure TForm1.SendPostData;
var
  Stream: TStringStream;
  Params: TIdMultipartFormDataStream;
begin
  Stream := TStringStream.Create('');
  try
   Params := TIdMultipartFormDataStream.Create;
   try
    Params.AddFile('File1', 'C:	est.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;

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