Send emoji with indy delphi7 - delphi

i want to send emoji with indy 9.00.10 on delphi 7. i use tnt VCL Controls .
i found this url http://apps.timwhitlock.info/emoji/tables/unicode for unicode and bytes code.
how to convert this codes to delphi Constants for Send with indy.
i use this delphi code for send message to telegram bot:
procedure TBotThread.SendMessage(ChatID:String; Text : WideString;
parse_mode:string;disable_notification:boolean);
Var
Stream: TStringStream;
Params: TIdMultipartFormDataStream;
//Text : WideString;
msg : WideString;
Src : string;
LHandler: TIdSSLIOHandlerSocket;
begin
try
try
if FShowBotLink then
Text := Text + LineBreak + FBotUser;
msg := '/sendmessage';
Stream := TStringStream.Create('');
Params := TIdMultipartFormDataStream.Create;
Params.AddFormField('chat_id',ChatID);
if parse_mode <> '' then
Params.AddFormField('parse_mode',parse_mode);
if disable_notification then
Params.AddFormField('disable_notification','true')
else
Params.AddFormField('disable_notification','false');
Params.AddFormField('disable_web_page_preview','true');
Params.AddFormField('text',UTF8Encode(Text));
LHandler := TIdSSLIOHandlerSocket.Create(nil);
FidHttpSend.ReadTimeout := 30000;
FidHttpSend.IOHandler:=LHandler;
LHandler.SSLOptions.Method := sslvTLSv1;
LHandler.SSLOptions.Mode := sslmUnassigned;
FidHttpSend.HandleRedirects := true;
FidHttpSend.Post(BaseUrl + API + msg, Params, Stream);
finally
Params.Free;
Stream.Free;
ENd;
except
on E: EIdHTTPProtocolException do
begin
if E.ReplyErrorCode = 403 then
begin
WriteToLog('Bot was blocked by the user');
end;
end;
end;
end;
bytes sample for emojies:
AERIAL_TRAMWAY = '\xf0\x9f\x9a\xa1';
AIRPLANE = '\xe2\x9c\x88';
ALARM_CLOCK = '\xe2\x8f\xb0';
ALIEN_MONSTER = '\xf0\x9f\x91\xbe';
sorry for bad english!!!

The Telegram Bot API supports several forms of input:
We support GET and POST HTTP methods. We support four ways of passing parameters in Bot API requests:
URL query string
application/x-www-form-urlencoded
application/json (except for uploading files)
multipart/form-data (use to upload files)
You are using the last option.
Indy 9 does not support Delphi 2009+ or Unicode. All uses of string are assumed to be AnsiString, which is the case in Delphi 7. Any AnsiString you add to TIdMultipartFormDataStream or TStrings, even a UTF-8 encoded one, will be transmitted as-is by TIdHTTP. However, there is no option to specify to the server that the string data is actually using UTF-8 as a charset. But, according to the docs:
All queries must be made using UTF-8.
So not specifying an explicit charset might not be problem.
If you still have problems with multipart/form-data, then consider using application/x-www-form-urlencoded (use TIdHTTP.Post(TStrings)) or application/json (use TIdHTTP.Post(TStream)) instead:
procedure TBotThread.SendMessage(ChatID: String; Text: WideString; parse_mode: string; disable_notification: boolean);
var
Params: TStringList;
LHandler: TIdSSLIOHandlerSocket;
begin
if FShowBotLink then
Text := Text + LineBreak + FBotUser;
Params := TStringList.Create;
try
Params.Add('chat_id=' + UTF8Encode(ChatID));
if parse_mode <> '' then
Params.Add('parse_mode=' + UTF8Encode(parse_mode));
if disable_notification then
Params.Add('disable_notification=true')
else
Params.Add('disable_notification=false');
Params.Add('disable_web_page_preview=true');
Params.Add('text=' + UTF8Encode(Text));
LHandler := TIdSSLIOHandlerSocket.Create(nil);
try
LHandler.SSLOptions.Method := sslvTLSv1;
LHandler.SSLOptions.Mode := sslmClient;
FidHttpSend.HandleRedirects := true;
FidHttpSend.ReadTimeout := 30000;
FidHttpSend.IOHandler := LHandler;
try
try
FidHttpSend.Post(BaseUrl + API + '/sendmessage', Params, TStream(nil));
except
on E: EIdHTTPProtocolException do
begin
if E.ReplyErrorCode = 403 then
begin
WriteToLog('Bot was blocked by the user');
end;
end;
end;
finally
FidHttpSend.IOHandler := nil;
end;
finally
LHandler.Free;
end;
finally
Params.Free;
end;
end;
procedure TBotThread.SendMessage(ChatID: String; Text: WideString; parse_mode: string; disable_notification: boolean);
var
Params: TStringStream;
LHandler: TIdSSLIOHandlerSocket;
function JsonEncode(const wStr: WideString): string;
var
I: Integer;
Ch: WideChar;
begin
// JSON uses UTF-16 text, so no need to encode to UTF-8...
Result := '';
for I := 1 to Length(wStr) do
begin
Ch := wStr[i];
case Ch of
#8: Result := Result + '\b';
#9: Result := Result + '\t';
#10: Result := Result + '\n';
#12: Result := Result + '\f';
#13: Result := Result + '\r';
'"': Result := Result + '\"';
'\': Result := Result + '\\';
'/': Result := Result + '\/';
else
if (Ord(Ch) >= 32) and (Ord(Ch) <= 126) then
Result := Result + AnsiChar(Ord(wStr[i]))
else
Result := Result + '\u' + IntToHex(Ord(wStr[i]), 4);
end;
end;
end;
begin
if FShowBotLink then
Text := Text + LineBreak + FBotUser;
Params := TStringStream.Create('');
try
Params.WriteString('{');
Params.WriteString('chat_id: "' + JsonEncode(ChatID) + '",');
if parse_mode <> '' then
Params.WriteString('parse_mode: "' + JsonEncode(parse_mode) + '",')
if disable_notification then
Params.WriteString('disable_notification: True,')
else
Params.WriteString('disable_notification: False,');
Params.WriteString('disable_web_page_preview: True,');
Params.WriteString('text: "' + JsonEncode(Text) + '"');
Params.WriteString('}');
Params.Position := 0;
LHandler := TIdSSLIOHandlerSocket.Create(nil);
try
LHandler.SSLOptions.Method := sslvTLSv1;
LHandler.SSLOptions.Mode := sslmClient;
FidHttpSend.HandleRedirects := true;
FidHttpSend.ReadTimeout := 30000;
FidHttpSend.IOHandler := LHandler;
try
try
FidHttpSend.Request.ContentType := 'application/json';
FidHttpSend.Post(BaseUrl + API + '/sendmessage', Params, TStream(nil));
except
on E: EIdHTTPProtocolException do
begin
if E.ReplyErrorCode = 403 then
begin
WriteToLog('Bot was blocked by the user');
end;
end;
end;
finally
FidHttpSend.IOHandler := nil;
end;
finally
LHandler.Free;
end;
finally
Params.Free;
end;
end;
That being said, your function's Text parameter is a WideString, which uses UTF-16, so you should be able to send any Unicode text, including emojis. If you are trying to generate text in your code, just make sure you UTF-16 encode any non-ASCII characters correctly. For example, codepoint U+1F601 GRINNING FACE WITH SMILING EYES is wide chars $D83D $DE01 in UTF-16:
var
Text: WideString;
Text := 'hi ' + #$D83D#$DE01; // 'hi 😁'
SendMessage('#channel', Text, 'Markup', False);
Alternatively, you can use HTML in your text messages, so you can encode non-ASCII characters using numerical HTML entities. According to the docs:
All numerical HTML entities are supported.
Codepoint U+1F601 is numeric entity $#128513; in HTML:
var
Text: WideString;
Text := 'hi $#128513;'; // 'hi 😁'
SendMessage('#channel', Text, 'HTML', False);

Related

Remove all CRLF from a Base64 text file

How can I remove all CRLF from a from a Base64 text file to make its content only on one line?
The following code uses a function, NoLineFeed, and a combination of TStringStream and AnsiString but still some CRLF are present (near the end of the file) after the content of the file have been processed by NoLineFeed.
function NoLineFeed was excerpted from a StackOverflow post by Arnaud Bouchez: Make String into only 1 line
var
StringVal: AnsiString;
XmlFile: TStringStream;
begin
XmlFile := TStringStream.Create;
try
XmlFile.LoadFromFile('file.txt');
StringVal := NoLineFeed(XmlFile.DataString);
if Length(StringVal) > 0 then
XmlFile.Write(StringVal[1], Length(StringVal));
XmlFile.SaveToFile('converted_file.txt');
finally
XmlFile.Free;
end;
end;
{ Arnaud Bouchez }
function NoLineFeed(const s: string): string;
var i: integer;
begin
result := s;
for i := length(result) downto 1 do
if ord(result[i])<32 then
if (i>1) and (ord(result[i-1])<=32) then
delete(result,i,1) else
result[i] := ' ';
end;
An alternative approach using a TStringList:
var
lst: TStringList;
begin
lst := TStringList.Create;
try
lst.LoadFromFile('file.txt');
lst.LineBreak := ' ';
lst.SaveToFile('converted_file.txt');
finally
lst.Free;
end;
end;
Linebreaks and cariage return doesn't come always together, linebreaks sometimes come alone without cariage return and this may be the left linebreaks you have , just remove #13 characters and #10 characters separately will solve the problem ,try this code:
var
StringVal: AnsiString;
XmlFile: TStringStream;
begin
XmlFile := TStringStream.Create;
try
XmlFile.LoadFromFile('file.txt');
StringVal := StringReplace(XmlFile.DataString, #13, ' ', [rfReplaceAll]);
StringVal := StringReplace(StringVal, #10, '', [rfReplaceAll]);
if Length(StringVal) > 0 then
XmlFile.Write(StringVal[1], Length(StringVal));
XmlFile.SaveToFile('converted_file.txt');
finally
XmlFile.Free;
end;
end;

Delphi - idHTTP upload file using TIdMultipartFormDataStream

I'm trying to upload a file using TIdHTTP. The problem is the access token gets changed when the request is sent to the server.
The access token that I'm using is fJNhDM6TlcpeVmD8h3jFuPJS71sxwZB8bZBXajTRB5TNAcRa6PNXfv4J7mPxIvMdMhjy7oKdTLbsRYthpBCCqGVkj4vlojJ4BRBkLAVIBJ1DZAnMZD
The API returns
HTTP/1.1 400 Bad Request
OAuth "invalid_token" "Malformed access token fJNhDM6TlcpeVmD8h3jFu=\r\nPJS71sxwZB8bZBXajTRB5TNAcRa6PNXfv4J7mPxIvMdMhjy7oKdTLbsRYthpBCCqGVkj4v=\r\nlojJ4BRBkLAVIBJ1DZAnMZD"
There is =\r\n added to my token twice.
My code is:
function TFoo.Post(const AToken, guID, AMessage, AImageFileName: string): Boolean;
var
lParam : TIdMultipartFormDataStream;
begin
Result := False;
if not FileExists(AImageFileName) then begin
LastError := 'File not found ' + AImageFileName;
Exit;
end;
ProxyCheck;
lParam := TIdMultipartFormDataStream.Create;
try
lParam.AddFormField('message', AMessage);
lParam.AddFormField('access_token', AToken);
lParam.AddFile('source', AImageFileName);
idHTTP.Request.ContentType := 'application/x-www-form-urlencoded';
try
idHTTP.Post( UrlAPI + guID + '/photos', lParam);
Result := True;
except;
LastError := idHTTP.ResponseText + sLineBreak + idHTTP.Response.WWWAuthenticate.Text;
end;
finally
lParam.Free;
end;
end;
What am I missing here ?
By default, AddFormField() sets the TIdFormDataField.ContentTransfer property to MIME's quoted-printable format. That is where the extra =\r\n is coming from. It is a "soft" line break being inserted by quoted-printable every 76 characters. Any server that supports quoted-printable would remove "soft" line breaks during decoding. But maybe your server does not.
If you want to disable the quoted-printable behavior, you can set the ContentTransfer property to either:
a blank string:
lParam.AddFormField('access_token', AToken).ContentTransfer := '';
'7bit' (since it does not contain any non-ASCII characters):
lParam.AddFormField('access_token', AToken).ContentTransfer := '7bit';
'8bit' or binary:
lParam.AddFormField('access_token', AToken).ContentTransfer := '8bit';
lParam.AddFormField('access_token', AToken).ContentTransfer := 'binary';
In this case, I would suggest #1.
On a side note, do not set the HTTP content type when posting a TIdMultipartFormDataStream. Not only are you using the wrong media type to begin with (it should be multipart/form-data instead), but the TIdMultipartFormDataStream version of Post() will simply overwrite it anyway.
function TFoo.Post(const AToken, guID, AMessage, AImageFileName: string): Boolean;
var
lParam : TIdMultipartFormDataStream;
begin
Result := False;
if not FileExists(AImageFileName) then begin
LastError := 'File not found ' + AImageFileName;
Exit;
end;
ProxyCheck;
lParam := TIdMultipartFormDataStream.Create;
try
lParam.AddFormField('message', AMessage);
lParam.AddFormField('access_token', AToken).ContentTransfer := '';
lParam.AddFile('source', AImageFileName);
try
idHTTP.Post(UrlAPI + guID + '/photos', lParam);
Result := True;
except;
LastError := idHTTP.ResponseText + sLineBreak + idHTTP.Response.WWWAuthenticate.Text;
end;
finally
lParam.Free;
end;
end;

Delphi 2007 & Updated Indy 10

I am sending alright files (doc, pdf, xls) with english filenames but when I am sending files with greek filenames I am getting on server side ????????? characters for filename & the error message Socket Error 10053, software caused connection abort. Is there a solution for this kind of problem.
Code:
procedure TForm1.LoadFileButtonClick(Sender: TObject);
begin
OpenDialog1.Filter := 'All Files (*.*)';
OpenDialog1.FilterIndex := 1;
if OpenDialog1.Execute then
begin
Edit1.Text := ExtractFileName(OpenDialog1.FileName);
Edit3.Text := OpenDialog1.FileName;
Fstream := TFileStream.Create(OpenDialog1.FileName, fmopenread);
Edit2.Text := inttostr(Fstream.Size);
Fstream.Position := 0;
FreeandNil(FStream);
//Fstream.Free;
end;
end;
procedure TForm1.SendFileButtonClick(Sender: TObject);
var
IncommingText: string;
begin
if (opendialog1.filename<>'') and (CheckBox1.Checked = True) then begin
IdTCPClient1.iohandler.writeln(edit1.text + '#' + edit2.text + ';' + edit3.text + ',');
Sleep(2000);
try
IdTCPClient1.IOHandler.largestream:=true;
Fstream := TFileStream.Create(OpenDialog1.FileName, fmopenread);
IdTCPClient1.IOHandler.Write(Fstream, 0 ,true);
finally
Fstream.Position := 0;
FreeandNil(FStream);
//Fstream.Free;
memo1.Lines.Add('File Sent');
IncommingText := IdTCPClient1.iohandler.readln;
if IncommingText = 'DONE!' then begin
Memo1.Lines.Add('File ' +Edit1.Text +' ' +Edit2.Text +' was received successfully by the Server');
//APPLICATION.ProcessMessages;
end else begin Memo1.Lines.Add('File ' +Edit1.Text +' was not received by the Server'); end;
end; //try - finally
end else begin
showmessage('Please choose a file Or Try to connect to the Server');
end;
end;
Indy's default text encoding is ASCII (because the majority of Internet protocols are still largely ASCII based, unless they define extra extensions to support Unicode). That is why you are getting ? for non-ASCII characters. To send non-ASCII characters, you need to tell Indy which text encoding to use that is compatible with the characters you are exchanging. UTF-8 is usually the best choice for that. There are three ways you can do that:
set the global GIdDefaultTextEncoding variable in the IdGlobal unit. It is set to encASCII by default, you can set it to encUTF8 instead:
procedure TForm1.FormCreate(Sender: TObject);
begin
GIdDefaultTextEncoding := encUTF8;
end;
set the TIdIOHandler.DefStringEncoding property to TIdTextEncoding.UTF8 (or IndyTextEncoding_UTF8 if you are using Indy 10.6+):
procedure TForm1.IdTCPClient1Connected(Sender: TObject);
begin
IdTCPClient1.IOHandler.DefStringEncoding := TIdTextEncoding.UTF8;
// or:
// IdTCPClient1.IOHandler.DefStringEncoding := IndyTextEncoding_UTF8;
end;
pass TIdTextEncoding.UTF8 (or IndyTextEncoding_UTF8) directly to the AByteEncoding parameter of WriteLn():
IdTCPClient1.IOHandler.WriteLn(..., TIdTextEncoding.UTF8);
// or:
// IdTCPClient1.IOHandler.WriteLn(..., IndyTextEncoding_UTF8);
Keep in mind that you are using an Ansi version of Delphi, where string maps to AnsiString, and thus Indy has to perform an additional Ansi-to-Unicode conversion of AnsiString data before it can then apply the specified text encoding to produce the bytes it transmits. Typically, Indy uses the OS's default Ansi encoding to handle that initial conversion (so if your AnsiString data is Greek encoded, and your OS is set to Greek, you will be fine), however you can use the TIdIOHandler.DefAnsiEncoding property, or the ASrcEncoding parameter of WriteLn(), if you need to specify that your AnsiString data is using a different encoding.
As for your socket error, without seeing a call stack leading up to the error, or at least which line of your code is raising it, that is difficult to troubleshoot. My guess is that it is related to you calling ReadLn() inside of the finally block regardless of whether WriteLn() or Write() actually succeeded. That code needs to be moved out of the finally block, it does not belong there.
Try something more like this instead:
procedure TForm1.LoadFileButtonClick(Sender: TObject);
begin
OpenDialog1.Filter := 'All Files (*.*)';
OpenDialog1.FilterIndex := 1;
if OpenDialog1.Execute then
begin
Edit1.Text := ExtractFileName(OpenDialog1.FileName);
Edit3.Text := OpenDialog1.FileName;
// Indy has its own FileSizeByName() function...
Edit2.Text := IntToStr(FileSizeByName(OpenDialog1.FileName));
end;
end;
procedure TForm1.SendFileButtonClick(Sender: TObject);
var
IncommingText: string;
Strm: TFileStream;
begin
if not CheckBox1.Checked then
begin
ShowMessage('Please connect to the Server');
Exit;
end;
if OpenDialog1.FileName = '' then
begin
ShowMessage('Please choose a file');
Exit;
end;
Strm := TFileStream.Create(OpenDialog1.FileName, fmOpenRead);
try
IdTCPClient1.IOHandler.WriteLn(Edit1.Text + '#' + Edit2.Text + ';' + Edit3.Text + ',', TIdTextEncoding.UTF8);
IdTCPClient1.IOHandler.LargeStream := True;
IdTCPClient1.IOHandler.Write(Strm, 0 , True);
finally
Strm.Free;
end;
Memo1.Lines.Add('File Sent');
IncommingText := IdTCPClient1.IOHandler.ReadLn;
if IncommingText = 'DONE!' then begin
Memo1.Lines.Add('File ' + Edit1.Text + ' ' + Edit2.Text + ' was received successfully by the Server');
//APPLICATION.ProcessMessages;
end else
begin
Memo1.Lines.Add('File ' + Edit1.Text + ' was not received by the Server');
end;
end;
Lastly, just an FYI, you are setting the AWriteByteCount parameter of Write() to True, so it is going to transmit the stream size (as an Int64 because of LargeStream=True) before then sending the TStream data, so putting the file size in the WriteLn() data is redundant.

Indy multipart/form-data example [duplicate]

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;
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;
Calling a PHP from Indy can fail because of the User-Agent, then you get 403 error.
Try this way, it fixed it for me:
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;
You lost 2 characters '--'. It is better to do so:
http.Request.ContentType:='multipart/form-data;boundary='+myBoundery;
fname := CRLF + '--' + myBoundery + CRLF + 'Content-Disposition: form-data; name=\"uploadedfile\";filename=\"test.png"' + CRLF;

HTTPS post - what I'm doing wrong?

I'm making requests to the webaddress to get XML files throught the HTTPS connection. But this connection works like 50%. In most cases it fails. Usual error is "socket error #10060". Or "Error connecting with SSL. EOF was observed that violates the protocol". What I'm doing wrong?
function SendRequest(parameters: string): IXMLDocument;
var
sPostData: TStringList;
sHttpSocket: TIdHTTP;
sshSocketHandler: TIdSSLIOHandlerSocketOpenSSL;
resStream: TStringStream;
xDoc: IXMLDocument;
begin
sPostData := TStringList.Create;
try
sPostData.Add('add some parameter to post' + '&');
sPostData.Add('add some parameter to post' + '&');
sPostData.Add('add some parameter to post' + '&');
sPostData.Add(parameters);
sHttpSocket := TIdHTTP.Create;
sshSocketHandler := TIdSSLIOHandlerSocketOpenSSL.Create;
sHttpSocket.IOHandler := sshSocketHandler;
sHttpSocket.Request.ContentType := 'application/x-www-form-urlencoded';
sHttpSocket.Request.Method := 'POST';
resStream := TStringStream.Create;
sHttpSocket.Post(Self.sUrl, sPostData, resStream);
xDoc := CreateXMLDoc;
xDoc.LoadFromStream(resStream);
Result := xDoc;
resStream.Free;
sHttpSocket.Free;
sshSocketHandler.Free;
sPostData.Free;
except on E: Exception do
begin
TCommon.ErrorLog('errorLog.txt', DateTimeToStr(Now) + ' ' + E.Message);
end
end;
end;
Maybe I can do this in another way, that works like 100%, when internet connection is available?
Regards,
evilone
An "EOF" error suggests you are connnecting to a server that is not actually using SSL to begin with, or the SSL data may be corrupted.
Besides that, why are you including explicit '&' characters between your post data parameters? Don't do that, Indy will just encode them and send its own '&' characters. Also, consider using TMemoryStream instead of TStringStream to ensure IXMLDocumect.LoadFromStream() is loading the server's original raw XML data as-is, and not an altered version that the RTL/VCL produces due to Unicode handling (TStringStream is TEncoding-enabled).
Edit: Given the URL you provided, an example of calling verifyUser() would look like this:
const
ERPLYAccountCode = '...';
function verifyUser(const user, pass: string; const sessionLength: Integer = 3600): IXMLDocument;
var
sPostData: TStringList;
sHttpSocket: TIdHTTP;
sshSocketHandler: TIdSSLIOHandlerSocketOpenSSL;
resStream: TMemoryStream;
xDoc: IXMLDocument;
begin
Result := nil;
try
resStream := TMemoryStream.Create;
try
sPostData := TStringList.Create;
try
sPostData.Add('clientCode=' + ERPLYAccountCode);
sPostData.Add('request=verifyUser');
sPostData.Add('version=1.0');
sPostData.Add('responseType=XML');
sPostData.Add('responseMode=normal');
sPostData.Add('username=' + user);
sPostData.Add('password=' + pass);
sPostData.Add('sessionLength=' + IntToStr(sessionLength));
sHttpSocket := TIdHTTP.Create;
try
sshSocketHandler := TIdSSLIOHandlerSocketOpenSSL.Create(sHttpSocket);
sHttpSocket.IOHandler := sshSocketHandler;
sHttpSocket.Request.ContentType := 'application/x-www-form-urlencoded';
sHttpSocket.Post('https://www.erply.net/api/', sPostData, resStream);
finally
sHttpSocket.Free;
end;
finally
sPostData.Free;
end;
resStream.Position := 0;
xDoc := CreateXMLDoc;
xDoc.LoadFromStream(resStream);
Result := xDoc;
finally
resStream.Free;
end;
except
on E: Exception do
begin
TCommon.ErrorLog('errorLog.txt', DateTimeToStr(Now) + ' ' + E.Message);
end;
end;
end;

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