On probleemi miten
käännän alla olevan vb- koodin delphi- koodiksi
jos joku vois jeesata olis hieno homma
Dim pvm As String
Private Sub Command1_Click()
Dim pvm As String
Dim paluu as String
Dim pvm as String
pvm = Format(Now,"yyyymmdd")
paluu = CryptString2(pvm, True)
End Sub
Public Function CryptString2(txtString As String, EnCrypt As Boolean) As String
On Error GoTo errhandler
Dim x As Integer
Dim outString As String
Dim iLen As Integer
Dim sFirstSeed As String
Dim sSecondSeed As String
Dim iSeed As Integer
If EnCrypt Then
sFirstSeed = Left(txtString, 1)
sSecondSeed = Mid(txtString, 2, 1)
iSeed = (Asc(sFirstSeed) Asc(sSecondSeed)) Mod 2
iLen = Len(txtString)
For x = 1 To iLen
outString = Chr((Asc(Mid$(txtString, x, 1)) Xor iSeed) 2) & outString
Next
outString = Chr(Asc(sFirstSeed) * 2 3) & outString
outString = outString & Chr(Asc(sSecondSeed) * 2 - 3)
Else
sFirstSeed = Chr((Asc(Left(txtString, 1)) - 3) 2)
sSecondSeed = Chr((Asc(Right(txtString, 1)) 3) 2)
iSeed = (Asc(sFirstSeed) Asc(sSecondSeed)) Mod 2
iLen = Len(txtString) - 1
For x = 2 To iLen
outString = Chr((Asc(Mid$(txtString, x, 1)) Xor iSeed) - 2) & outString
Next
End If
CryptString2 = outString
Exit Function
errhandler:
MsgBox "Error in Crypt" & vbCrLf & "Error: " & Err.Description & vbCrLf & "Number: " & Err.Number
CryptString2 = ""
End Function
käännän alla olevan vb- koodin delphi- koodiksi
jos joku vois jeesata olis hieno homma
Dim pvm As String
Private Sub Command1_Click()
Dim pvm As String
Dim paluu as String
Dim pvm as String
pvm = Format(Now,"yyyymmdd")
paluu = CryptString2(pvm, True)
End Sub
Public Function CryptString2(txtString As String, EnCrypt As Boolean) As String
On Error GoTo errhandler
Dim x As Integer
Dim outString As String
Dim iLen As Integer
Dim sFirstSeed As String
Dim sSecondSeed As String
Dim iSeed As Integer
If EnCrypt Then
sFirstSeed = Left(txtString, 1)
sSecondSeed = Mid(txtString, 2, 1)
iSeed = (Asc(sFirstSeed) Asc(sSecondSeed)) Mod 2
iLen = Len(txtString)
For x = 1 To iLen
outString = Chr((Asc(Mid$(txtString, x, 1)) Xor iSeed) 2) & outString
Next
outString = Chr(Asc(sFirstSeed) * 2 3) & outString
outString = outString & Chr(Asc(sSecondSeed) * 2 - 3)
Else
sFirstSeed = Chr((Asc(Left(txtString, 1)) - 3) 2)
sSecondSeed = Chr((Asc(Right(txtString, 1)) 3) 2)
iSeed = (Asc(sFirstSeed) Asc(sSecondSeed)) Mod 2
iLen = Len(txtString) - 1
For x = 2 To iLen
outString = Chr((Asc(Mid$(txtString, x, 1)) Xor iSeed) - 2) & outString
Next
End If
CryptString2 = outString
Exit Function
errhandler:
MsgBox "Error in Crypt" & vbCrLf & "Error: " & Err.Description & vbCrLf & "Number: " & Err.Number
CryptString2 = ""
End Function
Vastaukset 1
Vanhimmat
Tässä olis Delphi käännös. Testasin itse koodia käytännössä ja se toimii hyvin.
procedure TForm1.Button1Click(Sender: TObject);
var
Pvm: string;
Paluu: string;
begin
Pvm := FormatDateTime('yyyymmdd',Now);
Paluu := CryptString2(Pvm,True);
end;
function CryptString2(TxtString: string; EnCrypt: boolean): string;
var
X: Integer;
OutString: string;
iLen: Integer;
sFirstSeed: Char;
sSecondSeed: Char;
iSeed: Integer;
begin
try
if EnCrypt then
begin
sFirstSeed := TxtString[1];
sSecondSeed := TxtString[2];
iSeed := (Byte(sFirstSeed) Byte(sSecondSeed)) mod 2;
iLen := Length(TxtString);
for X := 1 to iLen do
OutString := Char((Byte(TxtString[X]) xor iSeed) 2) OutString;
OutString := Char(Byte(sFirstSeed) * 2 3) OutString;
OutString := OutString Char(Byte(sSecondSeed) * 2 - 3);
end
else
begin
sFirstSeed := Char((Byte(TxtString[1]) - 3) div 2);
sSecondSeed := Char((Byte(TxtString[Length(TxtString)]) 3) div 2);
iSeed := (Byte(sFirstSeed) Byte(sSecondSeed)) mod 2;
iLen := Length(TxtString) - 1;
for X := 2 to iLen do
OutString := Char((Byte(TxtString[X]) xor iSeed) - 2) OutString;
end;
Result := OutString;
except
MessageDlg('Error in crypt',mtError,[mbOK],0);
Result := '';
end;
end;