jordanwhite
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Everything posted by jordanwhite
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Hey Guys Im following PHPACADEMY tutorial for integrating login. Im on the second video The problem im getting is when i enter in my login and password from my table in my database [(alex, abc) I tried copying them directly also] "I get the error that user doesn't exist" index.php <html> <form action='login.php' method='POST'> Username: <input type='text' name='username'><br> Password: <input type='password' name='password'><br> <input type='submit' value='Log In'> </form> </html> Login.php <?php $username = $_POST['username']; $password = $_POST['password']; if ($username&&$password) { $connect = mysql_connect("localhost","root","") or die ("couldnt connect"); mysql_select_db("phplogin") or die ("could not connect to db"); $query = mysql_query("SELECT * FROM users WHERE username='$username'"); $numrow = mysql_num_rows($query); if ($numrows!=0) { //code to login } else die("that user doesn't exist"); } else die("Please Enter username and password"); ?>
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wow, cant believe I forgot that. Thanks for the help though buddah
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the error just came up instantly "MySQL query failed!"
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changed the code still getting error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\searchans.php on line 15
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Hey guys, im new around here, and i have just recently started teaching myself PHP, by looks of this site i seem very experienced but im eager to learn I seem to get an error with this code. The point of this code is to look into my database and pull out all entries with the first name _____ (whatever the user inputs) Error: Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\searchans.php on line 12 Code: Filename:search.php <html> <form action='searchans.php' method='POST'> First Name: <input type='text' name='firstname'><br> <input type='submit' value='Search'> </form> </html> Code: Filename:searchans.php <?php $firstname = $_POST['firstname']; $con = mysql_connect("localhost","peter","abc123"); if (!$con) { die('Could not connect: ' . mysql_error()); } $query = mysql_query("SELECT * FROM Persons WHERE FirstName='$firstname'"); while($row = mysql_fetch_array($result)) { echo $row['FirstName'] . " " . $row['LastName'] . " " . $row['Age']; echo "<br />"; } ?>