Http Post with indy [英] Http Post with indy

问题描述

我在网络服务器上有一个简单的php脚本，我需要使用HTTP POST上传一个文件，我正在使用Delphi。

这是我的代码与印度，但不可靠，它不会工作，我无法弄清楚我没有做正确的事情。如何查看我在服务器上发送的是否有这样的工具？

  procedure TForm1.btn1Click（Sender：TObject）; 
 var 
 fname：string; 
 MS，dump：TMemoryStream; 
 http：TIdHTTP; 
 
 const 
 CRLF =＃13＃10; 
 begin 
如果PromptForFileName（fname，''，''，''，''，false）然后
 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 -数据; 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））; 
除
 ShowMessage（'无法绑定套接字'）; 
结束
结束
结束
  

解决方案

Indy有 TIdMultipartFormDataStream 为此目的：

 程序TForm1.SendPostData; 
 var 
 Stream：TStringStream; 
参数：TIdMultipartFormDataStream; 
 begin 
 Stream：= TStringStream.Create（''）; 
 try 
参数：= TIdMultipartFormDataStream.Create; 
 try 
 Params.AddFile（'File1'，'C：\test.txt'，'application / octet-stream'）; 
 try 
 HTTP.Post（'http://posttestserver.com/post.php'，Params，Stream）; 
除了
在E：异常do 
 ShowMessage（'POST期间遇到错误'+ E.Message）; 
结束
 ShowMessage（Stream.DataString）; 
 finally 
 Params.Free; 
结束
 finally 
 Stream.Free; 
结束
结束
  




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.

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;

解决方案

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:\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;

这篇关于Http Post with indy的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！