can any one help me with correcting the code or can any one give me another code for using it in the login form.
this is my problem:-
i had make a login form using in oracle 9i form builder >>>in this form i have three text boxes one for intering user name and the second one for entering the password and the third text box is not visible and it is used for counting the tries.
In addition i have a three buttons , one is for login and the two others are not visible and they are a show main menu button and a exit button. For login button i had put a WHEN-BUTTON-PRESSED trigger in the login button and it must check if the user name and the password match what it is on the login table so it allow the user to see the show main menu button otherwise if the user name or the password are wrong and has been putted wrong for 3 times of trying then it will show the exit button.
and this is a picture of the login form in the design view.
and this is the code for theWHEN-BUTTON-PRESSED trigger on the login button.
_________
declare
alertNum number;
dummy1 tbl_login.USER_NAME%type;
dummy2 tbl_login.PASSWORD%type;
begin
select tbl_login.USER_NAME into dummy1 from tbl_login where tbl_login.USER_NAME = :LOGIN.USER_NAME;
select tbl_login.PASSWORD into dummy2 from tbl_login where tbl_login.PASSWORD = :LOGIN.PASSWORD;
if :LOGIN.TRIES<3 then
if sql% found
then
set_item_property('LOGIN.SHOW_MENU', visible, property_true);
set_item_property('LOGIN.SHOW_MENU', enabled, property_true);
else
message ('Invalid password....try again');
:LOGIN.TRIES := :LOGIN.TRIES+1;
:LOGIN.USER_NAME := null;
:LOGIN.PASSWORD := null;
end if;
else
message ('Exceeded Number of tries..press exit button');
set_item_property('LOGIN.EXIT', visible, property_true);
set_item_property('LOGIN.EXIT', enabled, property_true);
end if;
end;___________
can any one help me correcting the code of the WHEN-BUTTON-PRESSED trigger on the login form or can any one give me another code for using it in the login form.
i hope to get some help from the experts>>>