Re: Newbie passwordcheck-program problem

Hi Ronald, > Hello there fellow programmer, > > I just started learning PHP (I got version 4 installed with Apache on a > Win98 machine). At the moment I'm trying to write a little program that > verifies a user login. > I know the program is not professional but for me it will do. Only I got > stuck with the following problem. I have written a class (I'm not very sure > if this is the right way to do this) in which I specified a function that > does the password check. > Now I like the program to check the password and return me a true or a false > which I can read from somewhere else in my program. > I wrote it like this: First of all, you don't need a class for this. Unless you expect the program to get a lot more complicated, you could just use a function. > > <html> > <head> > <title>Password Check</title> > > <?PHP class password > { > function checkpwd() > { ?> > > <FORM ACTION="<?php print("$PHP_SELF"); ?>" > METHOD="POST"> > User name is:<BR> > <INPUT TYPE="text" NAME="UserName"> <P> > > <?php echo "Give Password:<BR>" ?> > > <INPUT TYPE="PASSWORD" NAME="passwd"> > <INPUT TYPE="HIDDEN" NAME="GeefInfo" VALUE=1> <p> > <INPUT TYPE="submit"> > > <?PHP if (IsSet($this->GeefInfo)&&IsSet($this->UserName)) > { > $this->UserName=strtoupper($this->UserName); > $this->passwd=strtoupper($this->passwd); $this->GeefInfo and $this->UserName will never be set. You need to change this to something like: if (isset($HTTP_POST_VARS["GeefInfo"]) && isset($HTTP_POST_VARS["UserName"])) > > if (IsSet($this->passwd)) Similarly here. if (isset($HTTP_POST_VARS["passwd"])) > { > if ($this->passwd=="MYPASS" && > $this->UserName=="RONALD") > { > $this-> testvar=1; > exit(1); > } > else > { > $this-> testvar=0; > exit(0); I think you might be expecting exit() to do something different to what it does here. The function exit() will terminate script execution. I suspect what you wanted was return. return 1; else return 0; > } > > } > > } > } > }?> > > </head> > <body> > > <?php $exec_pwd=new password; > $vexec_pwd->checkpwd(); > print $exec_pwd; // Where is my output ????? You won't get any output. $exec_pwd contains an object (an instance of your "password" class). If you want to output the return value from checkpwd(), do: $exec_pwd = new password; print $exec_pwd->checkpwd(); I think you may be a bit confused about how variables, classes and objects work in PHP. Have a closer read through the manual. Moreover, if you're just starting out with PHP then steer clear of objects entirely - you don't need them! > ?> > > </body> > </html> > > If you can and like to help me I will be very pleased. > Thanks in advance !!! > > Ronald Hope this helps. Cheers Simon Garner