Re: Fwd: [PHP] newbie: problem with $_Post[]

Hello Richard, that is exactly what I was looking for. thanks alot Paul, I didn't know its possible to use error_reporting(), that's a good hint thanks. btw if you guys know any simple php application that I can study and learn faster than going through books please let me know. I tried Wordpress but its too complicated for a beginner like me. "Paul M Foster" <paulf@quillandmouse.com> wrote in message news:20090728043523.GS14822@quillandmouse.com... On Mon, Jul 27, 2009 at 09:01:16PM -0700, Richard S. Crawford wrote:

<form action="pass.php" method=POST> username : <input type=text name=user ><br /> password : <input type=password name=pass> <br /> <input type=submit value="go"><p> </form> <?php $user=$_POST['user']; $pass=$_POST['pass']; if(($user=="myname")&&($pass="mypass")) echo "access granted"; else echo "access denied"; ?> getting "Notice: Undefined index: user" and "Notice: Undefined index: pass". changing form action to another page will solve the problem but i want to be able to use $_POST array on the same page, how can i do it? thanks in advance /Arash

Arash, It's hard to respond when it's unclear exactly what you want to do. I'm guessing that you want to simply have the same page show both the form, and also process the form, so that you can cut down on the number of pages you have to create. The way I do this is by first checking to see if the $_POST array has been set; if it has not been set, then I know that the form wasn't filled out, and so the script needs to print out the form. If the array has been set, on the other hand, then the script can process the form data. Here's a real quick example of what I mean: <?php if (!isset($_POST)) { ?> // The $_POST array is not set, so display the form <form action="pass.php" method=POST> username : <input type="text" name="user" ><br /> password : <input type="password" name="pass"> <br /> <input type="submit" value="go"><p> </form> <?php } else { // The $_POST array *is* set, so process the data $user=$_POST['user']; $pass=$_POST['pass']; if(($user=="myname")&&($pass="mypass")) { echo "access granted"; } else { echo "access denied"; } ?> Hope that helps.

Richard's right. This is the best way to do this. Your original code had two problems. First, you're getting the error message because you're allowing E_NOTICE level error messages. You can turn these off with the error_reporting() function. Second, as written, your page paints the form and then directly tests the status of the variables before the user can respond; the whole page of PHP code executes and then waits for user response. It's only when the user responds and the page is revisited that the variables can legitimately be tested. That's why Richard's method works. The variables are only tested when the page is re-entered after the user responds. Paul -- Paul M. Foster