[SOLVED] webserver does not receive files
lazarus / 38-Networking and Web Programming

----------------------------------------------------------------------
egsuh · 2022-12-04 14:04

I was replying to questions on webserver related questions, but
ironically I cannot make my own webserver to receive files. I think I
had similar experience in the past, but cannot remember how I solved
it.

Server-side code is as following. It will display the local file
names.

Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i procedure
TwmWebTests.upfileRequest(Sender: TObject; ARequest: TRequest; 
AResponse: TResponse; var Handled: Boolean);var   i: integer;   s:
string;begin   s := Format('Files uploaded (count=%d):<br>',
[ARequest.Files.Count]);   for i := 0 to ARequest.Files.Count-1 do
begin       s += ARequest.Files[i].LocalFileName + '<br>';   end; 
 AResponse.Content := s;   Handled:= True;end; 
And html file that sends file to the server is :

Code: Text  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i <!Doctype
html><html><head></head><body> <form method="post"
action="http://localhost/myapp/upfile" target="_blank">File Upload 
<input type="file" name="files"><br>  <input
type="submit"></form> </body></html> 
When I select a file and submit from the browser, it displays
following. So, there are no problem in setting up webserver itself.

Quote     Files Uploaded (count=0):

Server is Windows Server 2019 Standard.

----------------------------------------------------------------------
Leledumbo · 2022-12-04 23:47

Quote from: egsuh on December 04, 2022, 03:04:16 pm<form method="post"
action="http://localhost/myapp/upfile" target="_blank">File Upload
File upload requires:
Code: Text  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i
enctype="multipart/form-data" in the form attribute, cannot be with
the default "application/x-www-form-urlencoded".

----------------------------------------------------------------------
egsuh · 2022-12-05 05:27

Wow, now it works. Following is displayed on submission.

QuoteFiles uploaded (count=1):
C:\Windows\TEMP\CGI00016.tmp
There are many web pages that explain how to write HTML for file
uploads, but I couldn't find any example that specified enctype.

----------------------------------------------------------------------
Leledumbo · 2022-12-06 06:43

Quote from: egsuh on December 05, 2022, 06:27:44 amThere are many web
pages that explain how to write HTML for file uploads, but I couldn't
find any example that specified enctype.
No idea with that, when I first learned it back about 15 years ago,
all tutorials specifically mention it. I guess today people don't
really write HTML by hand and forget about this detail? I really don't
know. But I do know many frameworks provide a function that only needs
to be passed any element, typically div, then it magically turns into
a file upload, hiding it from the programmer.

----------------------------------------------------------------------
krolikbest · 2022-12-07 19:20

Hi,
to avoid create new thread I linked my question there. So simply code:
Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i program
project1; {$mode objfpc}{$H+} uses   {$IFDEF UNIX}{$IFDEF
UseCThreads}  cthreads,  {$ENDIF}{$ENDIF}  sysutils, fphttpapp,
HTTPDefs, fpwebfile, httproute; procedure Test(aReq: TRequest; aRes:
TResponse);var  n: Integer;  f: TUploadedFile;  i: Integer;  s :
string; begin   n := AReq.Files.Count;   if n = 0 then   begin     
aRes.Code:=200;     
aRes.Contents.LoadFromFile(ExtractFilePath(ParamStr(0)) +
'fileUploader.html');   end      else   begin     f := AReq.Files[0]; 
   f.LocalFileName:=ExtractFilePath(ParamStr(0)); //apparently this
does not set a default folder     s:=f.LocalFileName;   
 ARes.Code:=200;     ARes.Content := s;   end;end; begin 
Application.Port:=8080;  HTTPRouter.RegisterRoute('/',@Test, true); 
Application.threaded := true;  Application.Initialize; 
Application.Run;end.                      It works but question is how
can I set a specified folder on server side  for uploaded files? Using
LocalFileName as I did seems to be incorrect..

Regards,

----------------------------------------------------------------------
krolikbest · 2022-12-07 19:37

Ach, ok. Solved

----------------------------------------------------------------------
Leledumbo · 2022-12-08 03:41

Left a trace for future people who have the same problem.

As in most other implementations, uploaded files are typically
downloaded into a temporary directory. In fcl-web implementation, it's
in whatever returned by GetTempDir, which typically reads TEMP
environment variable so users can override it from the command line.
There are however, multiple other ways to do the same from application
level:
TRequest.DefaultRequestUploadDir, this is a class variable, can be set
in global context not requiring any instance to be made firstWrite
your own HTTP server implementation with a TRequest descendant that
overrides GetTempUploadFileName (to be honest, this is too much)Just
pick your option.

----------------------------------------------------------------------
krolikbest · 2022-12-08 08:19

Hi,
 currently I'm writing without access to the code so a bit from memory
but I did something like:
 Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i      f :=
AReq.Files[0];     s:=f.FileName;     ARes.Code:=200;   
 ARes.Contents.LoadFromStream(f.Stream); //this line loads contents of
the uploaded file     ARes.Contents.SaveToFile(your_path_file); //this
line save this contents to the new created file     ARes.Content :=
s; Of course such solution still writes in TempDir as well and to
avoid it one need to follow your tips. Then there is no need to use my
solution.

Regards,

----------------------------------------------------------------------
egsuh · 2022-12-08 09:29

Aren’t there anything like f.savetofile(your_file_path)?

----------------------------------------------------------------------
krolikbest · 2022-12-08 11:30

Hi

 f is TUploadedFile class according to
https://wiki.freepascal.org/fpWeb_Tutorial#Reading_File_Uploads.
Didn't check if is it possible to simply use f.saveToFile.

----------------------------------------------------------------------
Leledumbo · 2022-12-08 14:49

Quote from: egsuh on December 08, 2022, 10:29:44 amAren’t there
anything like f.savetofile(your_file_path)?
Unfortunately, no. This is the interface of TUploadedFile:
Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i  
TUploadedFile = Class(TCollectionItem)  Private    FContentType:
String;    FDescription: String;    FDisposition: String;   
FFieldName: String;    FFileName: String;    FLocalFileName: String; 
 FSize: Int64;    FStream : TStream;  Protected    // Note that this
will free the file stream, to be able to close it - file is share deny
write locked!    Procedure DeleteTempUploadedFile; virtual;   
function GetStream: TStream; virtual;    Procedure FreeStream;
virtual;  Public    Destructor Destroy; override;    Property
FieldName : String Read FFieldName Write FFieldName;    Property
FileName : String Read FFileName Write FFileName;    Property Stream :
TStream Read GetStream;    Property Size : Int64 Read FSize Write
FSize;    Property ContentType : String Read FContentType Write
FContentType;    Property Disposition : String Read FDisposition Write
FDisposition;    Property LocalFileName : String Read FLocalFileName
Write FLocalFileName;    Property Description : String Read
FDescription Write FDescription;  end; You can, however, make your own
implementation that has such a method by extending this class and
assigning it to httpdefs' UploadedFileClass variable. But for me, this
interface is enough for most use cases. The Stream property is the key
to do everything possible to uploaded files.

----------------------------------------------------------------------
krolikbest · 2022-12-08 19:07

Hi,
Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i      
SetLength(f, AReq.Files.Count);      f[0] := AReq.Files[0];     
ms:=TMemoryStream.Create;      f[0].Stream.Position:=0;     
ms.CopyFrom(f[0].Stream, f[0].Stream.Size);     
ms.SaveToFile(ExtractFilePath(ParamStr(0))+'\tst.txt');      ms.Free;
           
Regards,