find parse error implementing Kevin's scripts

**almond** · Oct 4, 2001 12:15

Kevin's article, "Managing Users with PHP Sessions and MySQL" is the clearest I've seen. I thought I had adapted it to my needs, which are basically just to log users in who already have a username and password, then maintain their sessions. I keep getting the following error when I try the test page. I've spent a couple of hours trying to find the error. Please spot it if you can. (I've used "my..." where I'm hiding my actual directories.)

Error message:

Parse error: parse error in /home/myusername/mylib/accesscontrol.php on line 63

Here's the test page :

<?php // include("accesscontrol.php"); ?>
<?php require('/home/myusername/mylib/accesscontrol.php');?>
<html>
<head>
<title> Members-Only Page </title>
</head>
<body>
<p>Welcome, <?=$username?>! You have entered a members-only area of the site. Don't you feel special?</p>
</body>
</html>

Here's my slightly modified version of Kevin's script :

<?php // accesscontrol.php
include("common.php");
include("db.php");
session_start();

if(!isset($txtName)) {
?>

<html>
<head>
<title> Please Log In for Access </title>
</head>
<body>
<h2> Login Required </h2>
<p>You must log in to access this area of the site.<p>
<form method="post" action="<?=$PHP_SELF?>">
User ID: <input type="text" name="txtName" size="12"><br>
Password: <input type="password" name="password" SIZE="12"><br>
<input type="submit" value="Log in">
</form></p>
</body>
</html>

<?php
exit;

session_register("txtName");
session_register("password");

dbConnect("my_database");
$sql = "SELECT * FROM sspass WHERE ss = '$txtName' AND pwd = '$password'";
$result = mysql_query($sql);
if (!$result) {
error("A database error occurred while checking your login details.");
}

if (mysql_num_rows($result) == 0) {
session_unregister("txtName");
session_unregister("password");
?>

<html>
<head>
<title> Access Denied </title>
</head>
<body>
<h2> Access Denied </h2>
<p>Your user ID or password is not found. To try logging in again, click
<a href="<?php echo $PHP_SELF; ?>">here</a>.</p>
</body>
</html>

<?php
exit;
}

?>

**Technosailor** · Oct 4, 2001 12:24

see if this works

PHP Code:


<?php 

exit; 



session_register("txtName"); 

session_register("password"); 



dbConnect("my_database"); 

$sql = "SELECT * FROM sspass WHERE ss = '$txtName' AND pwd = '$password'"; 

$result = mysql_query($sql); 

if (!$result) { 

error("A database error occurred while checking your login details."); 

} 



if (mysql_num_rows($result) == 0) { 

session_unregister("txtName"); 

session_unregister("password"); 

}

?>

You'd need the '}' at the end.
Sketch

**Paul S** · Oct 4, 2001 12:25

Which one is line 63?

**almond** · Oct 4, 2001 12:39

It worked.

Many, many thanks, Stretch.

**almond** · Oct 4, 2001 12:58

I've played with it some more, and it is passing the username to subsequent pages properly, but it's accepting any password. It's supposed to be checking the password against the one stored in the database. Any idea why it's accepting bad passwords?

**Technosailor** · Oct 4, 2001 13:11

this might do it:

PHP Code:


<form method="post" action="<?=$PHP_SELF;?>">

I put a semicolon at the end of your PHP statement

Sketch
PS...not STRETCH

**freddydoesphp** · Oct 4, 2001 16:05

A semi-colon will give you an error. Its either

PHP Code:


<form method="post" action="<?=$PHP_SELF?>">

or

PHP Code:


<form method="post" action="<? echo $PHP_SELF; ?>">

But not both

**Technosailor** · Oct 4, 2001 16:08

I was thinking the second statement....
Sketch

**almond** · Oct 4, 2001 19:34

Sorry about the "Stretch," Sketch.

I probably did something similar to this script I'm trying to get working.

**almond** · Oct 4, 2001 23:22

Back to the original fix: Adding that brace got the page to display properly and use the correct username, but I wonder if it isolated the subsequent HTML from the if statement so that the HTML doesn't execute.

if (mysql_num_rows($result) == 0) {
session_unregister("txtName");
session_unregister("password");
?>

Does adding a closing brace here isolate the html below from the "if" above?

<html>
<head>
<title> Access Denied </title>

Is there some way to more explicitly force the "Access Denied" page to appear if the login data are wrong?

**almond** · Oct 5, 2001 10:20

Inserting a closing brace after the first "exit;" made the script work perfectly.

Thank you, Kevin Yank, for by far the best articles on PHP and mySQL I've found anywhere.

**Paul S** · Oct 5, 2001 10:34

This,

PHP Code:


<?php

if (mysql_num_rows($result) == 0) { 

    session_unregister("txtName"); 

    session_unregister("password"); 

?> 



<html> 

<head> 

<title> Access Denied </title>

 

<?php } ?>

is exactly the same as

PHP Code:


<?php

 if (mysql_num_rows($result) == 0) { 

     session_unregister("txtName"); 

     session_unregister("password");  



     echo '<html> 

       <head> 

       <title> Access Denied </title>';

 }

?>

User Tag List

Thread: find parse error implementing Kevin's scripts

Thread Tools

Display

find parse error implementing Kevin's scripts

problem solved

Bookmarks

Bookmarks

Posting Permissions