Http Post与indy

Question

我的网络服务器上有一个简单的PHP脚本，我需要使用HTTP POST上传文件，我在Delphi中这样做。

这是我与Indy的代码，但是它不会起作用，我无法弄清楚我做得不好。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;

Answer 1

Indy为此目的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;

Answer 2

从Indy调用PHP可能会因User-Agent而失败，然后会出现403错误。

尝试这种方式，它为我修复了它：

var Answer: string;
begin
  GetHTML:= TIdHTTP.create(Nil);
  try
    GetHTML.Request.UserAgent:= 'Mozilla/3.0';
    Answer:= GetHTML.Get('http://www.testserver.com/test.php?id=1');
  finally
    GetHTML.Free;
  end;
end;

Answer 3

你输了2个字符＆＃39; - ＆＃39;。最好这样做：

http.Request.ContentType:='multipart/form-data;boundary='+myBoundery;
fname := CRLF + '--' + myBoundery + CRLF + 'Content-Disposition: form-data; name=\"uploadedfile\";filename=\"test.png"' + CRLF;