Question

我的网络服务器上有一个简单的php脚本, 我需要用 HTTP POST 上传一个文件, 我在德尔斐正在这样做。

我和Indy的代码在这里,但有保证它不会起作用,我无法弄清楚我做错了什么。我怎么能看到我发送到服务器上的东西有这样一个工具呢?

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为此目的有TIdMultipart FormDataStream :

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;

Answer 2

从 Indy 调用 PHP 时会因用户代理失败, 然后会发生403个错误。

试试这个,它帮我修好了它:

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

你失去了两个字符--。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;

友情链接