whlie loop

[dan_t]

This is probably a simple question, but if you are looping through a while loop using an if - else statement

why would it echo the answer the amount of times there are data fields?

Like if you had three users in a database and you had an if statement to validate a user name and password. If the user name and password passed it would say good - if it failed it would say no good.

But instead it prints - good no good good, or no good good good, ect.???

[kenrbnsn]

Code?

[dan_t]

This is on of the hundred ways I have tried it:

$query = "SELECT username, password FROM userTable";

	$result = mysql_query($query);

	while($row = mysql_fetch_array($result)) {
		$logged = (($row['username'] == $username) && ($row['password'] == $password));

			if(!$logged){
				echo "Sorry";
			} else {
				echo "Good!";
			}

 

I also tried a switch statement, but have become frazzled.

[kenrbnsn]

What are you trying to accomplish with this code?

 

Ken

[dan_t]

I did have code sending it to one page or another printing the content and username on the one page, but it printed everything multiple times so now I'm just trying to fix that problem first.

 

[kenrbnsn]

If you only want records where the username & password match those in the DB, do that via the query:

<?php
$query = "SELECT username, password FROM userTable where username='$username' and password='$password'";
?>

This assume that the values of $username & $password have been properly sanitized.

 

Ken

[dan_t]

"This assume that the values of $username & $password have been properly sanitized"

What do you mean by sanitized?

[maxudaskin]

You just have to break the loop once found. Better yet, change your SQL to get the password from the database using the username.

 

Instead of

<?php
$query = "SELECT username, password FROM userTable";

	$result = mysql_query($query);

	while($row = mysql_fetch_array($result)) {
		$logged = (($row['username'] == $username) && ($row['password'] == $password));

			if(!$logged){
				echo "Sorry";
			} else {
				echo "Good!";
			}
                        }
}

   

Do this:

<?php
$sql = 'SELECT `password` FROM `userTable` WHERE username = \'' . $username . '\'';
$query = mysql_query($sql);
if(mysql_num_rows($query) < 1) { // If there are no results found with that username, tell the user
    echo 'Incorrect Username';
} else { // Use and else statement so that it doesn't give errors when checking the password (no result means no password to check)
    $result = mysql_result($query);
    if($result != $password) {
        echo 'Incorrect Password';
    } else {
        echo 'Login Success!';
    }
}