Validate new password against old password

I just want to say that I hate all websites that make me do that.
I know this isn't an answer, but it REALLY REALLY pisses me off.
On one site I ran out of passwords to use, I had to come up with some new ones and now I really can't keep track of my passwords.

look for password policy

where do you plan to use this ?
In delphi ?

Then you would need a centralised location to store the used passwords
And a gui for a admin to reset the password, delete users, delete the history etc

Looks like you need a password database.
For instance in oracle you can set a profile on a user
With all the restrictions you want :
http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/statements_6010.htm#i2084338

There is a whole list of options of what you can do with the oracle database here
http://download.oracle.com/docs/cd/B19306_01/network.102/b14268/toc.htm

Maybe you can extract some ideas from that

Basically I just want to verify that the new password isn't almost the same as the previous one.
I have no intention to store all passwords that were used (ever). For that I fully agree with rfwoolf.

Sum the ordinal values of the entered text and decide on your own window of acceptability.

where will you store the current password ?
in the cookie ?

Geert Gruwez:
The password will be encrypted and stored in a database.
markusventer:
Sounds good. How would that look like?

I believe this is perfect for my requiredments.
Thank you all for you info.

stored in a database ?
then the database will encrypt it for you ... so you don't need to do this yourself

well, this is valid with the accepted answer :

old password: ABCDEFGHIJ
new pass: abcdefghij
next new pass: AbCdEfGhIj

looks like a very shallow constraint

you could just as well use soundexint

If Abs(SoundExInt(OldPassword, 10) - SoundExInt(NewPassword, 10)) > 2 then
  ... password accepted
 

Not Entirely Geert, there is a big difference in the ordinal values of upper and lowercase letters.

Your statement is based on the ordinal value  of upper and lowercase letters being the same.

Apply a clever algorithm and this approach is quite easy and failsafe.

this would be a discussion about reinventing the wheel ... password policies

You might want to check the ORD-values of:
Stef1234 and Karel5678
Although completely different the ORD-values of these two strings are very close to each other. So I have to agree with Geert, the approach is a bit simple.

ugh ... cough ... cough

DelphiWizard ... i don't want to say i told you so ... but i did tell you so ... :)

Ok, enough,
it looks like you want some very simple means of password policy enforcement without too much jibberish

1: would you be satisfied if the password is stored (slightly encrypted) in the registry
2: a check to see if the new password is different (you gave some arguments in your initial Q header)

Geert>
I'm not sure you understand the OP 100%... for example why on earth would you be storing the password in the registry? From what I understand this is a database application, and for whatever reason, the application sometimes needs you to 'reset' your password (this is very common on the web for example) but they don't want you to make your password similar to the existing one - because that's what lazy users do.
All the OP seems to need is a function like this:
function IsPasswordSimilar(CurrentPassword, NewPassword) : boolean;

within the function you do a check like :
'are there 3 of the same consecutive characters in the new and old password, for example:
Mercedes123 and Hammer456 [mer = mer]

here is a class which you can use for password policies

you'll need to read around TForm1 code ... :)

unit Unit3;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, ExtCtrls, StdCtrls, ComCtrls;

type
  TPasswordPolicy = class(TObject)
  public
    function Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean; virtual;
  end;

  TPasswordPolicyType = class of TPasswordPolicy;

  TPasswordPolicies = class(TList)
  public
    constructor Create; virtual;
    procedure AddPolicy(Policy: TPasswordPolicyType);
    function Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean;
  end;

  TPasswordPolicy3Letters = class(TPasswordPolicy)
  public
    function Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean; override;
  end;

  TForm1 = class(TForm)
    Edit1: TEdit;
    Edit2: TEdit;
    Button1: TButton;
    procedure Button1Click(Sender: TObject);
  private
    fPasswordPolicy: TPasswordPolicies;
  public
    constructor Create(AOwner: TComponent); override;
  end;

var
  Form1: TForm1;

implementation

{$R *.dfm}

{ TPasswordPolicy }

function TPasswordPolicy.Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean;
begin
  ErrorMsg := '';
  Result := not SameText(OldPassword, NewPassword);
  if not Result then
    ErrorMsg := 'New password should not be the same.  Password is not case sensitive.';
end;

{ TPasswordPolicies }

constructor TPasswordPolicies.Create;
begin
  inherited Create;
  // Add default policy
  AddPolicy(TPasswordPolicy);
end;

procedure TPasswordPolicies.AddPolicy(Policy: TPasswordPolicyType);
var PolicyItem: TPasswordPolicy;
begin
  PolicyItem := Policy.Create;
  inherited Add(PolicyItem);
end;

function TPasswordPolicies.Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean;
var I: integer;
begin
  ErrorMsg := '';
  Result := True;
  for I := 0 to Count - 1 do
    if not TPasswordPolicy(Items[I]).Check(OldPassword, NewPassword, ErrorMsg) then
    begin
      Result := False;
      Break;
    end;
end;

{ TPasswordPolicy3Letters }

function TPasswordPolicy3Letters.Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean;
var I: integer;
  Temp: string;
begin
  Result := True;
  for I := 1 to Length(OldPassword) - 3 do
    if Pos(Copy(OldPassword, I, 3), NewPassword) > 0 then
    begin
      Result := False;
      ErrorMsg := '3 letters are the same and in the same order as in the old password.';
    end;
end;

constructor TForm1.Create(AOwner: TComponent);
begin
  inherited Create(AOwner);
  fPasswordPolicy := TPasswordPolicies.Create;
  fPasswordPolicy.AddPolicy(TPasswordPolicy3Letters);
end;

procedure TForm1.Button1Click(Sender: TObject);
var Emsg: string;
begin
  EMsg := '';
  if not fPasswordPolicy.Check(Edit1.Text, Edit2.Text, Emsg) then
    ShowMessage(EMsg);
end;


end.

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Have a look at the folling. Build this after my discovery of the ORD-simplicity :-)

function PasswordsDifferEnough(Const OldPassword, NewPassword : String) : Boolean;
var i, NumberOfSameChar : Integer;
begin
  Result := True;
  NumberOfSameChar := 0;
  // Check 3 positions are the same
  for i := 1 to Length(NewPassword) -3 do
  begin
    if (AnsiPos(Copy(NewPassword, i, 3), OldPassword) > 0) then
    begin
      Result := False;
      Exit;
    end;
  end;
  // Check number of same characters.
  for i := 1 to Length(NewPassword) -3 do
  begin
    if (AnsiPos(Copy(NewPassword, i, 1), OldPassword) > 0) then
      NumberOfSameChar := NumberOfSameChar + 1;
    if (NumberOfSameChar > (Length(NewPassword) div 3)) then
    begin
      Result := False;
      Exit;
    end;
  end;
end;

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Yes Geert, we are on the same track...

you could add the 50Prcnt class too :

{ TPasswordPolicy50Prcnt }

function TPasswordPolicy50Prcnt.Check(OldPassword, NewPassword: string; var ErrorMsg: string): boolean;
var
  I, n: integer;
begin
  Result := True;
  ErrorMsg := '';
  n := 0;
  for I := 1 to length(NewPassword) do
    if Pos(Uppercase(Copy(NewPassword, I, 1)), UpperCase(OldPassword)) > 0 then
      Inc(n);
  if n >= (length(NewPassword) div 2) then
  begin
    Result := False;
    ErrorMsg := '50% of the characters were resused in the new password';
  end;
end;


constructor TForm1.Create(AOwner: TComponent);
begin
  inherited Create(AOwner);
  fPasswordPolicy := TPasswordPolicies.Create;
  fPasswordPolicy.AddPolicy(TPasswordPolicy3Letters);
  fPasswordPolicy.AddPolicy(TPasswordPolicy50Prcnt);
end;

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ah, i see you have found all you need