bbram Posted January 8, 2011 Share Posted January 8, 2011 Hello I am getting the following errors when I am running the code below. I am not 100% for sure what the error's mean. If possible can you guide help to figure out what is wrong with my code? Notice: Undefined index: Userid in C:\wamp\www\checklogin.php on line 13 Notice: Undefined index: password in C:\wamp\www\checklogin.php on line 14 Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\checklogin.php on line 26 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\checklogin.php on line 29 <?php $host="localhost"; $username="root"; $password=""; $db_name="movedb"; $tbl_name="security"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("Cannot Select Database"); // username and password sent from form $myusername=$_POST['Userid']; $mypassword=$_POST['password']; $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); //this creates an array with the results $row= mysql_fetch_array($result); echo ("HI"); //this gets the value held in the 'user_level' column for the user //$user_level = $row['user_level']; if($count==1) { echo ("passed"); echo $myusername; echo $mypassword; /* // Register $myusername, $mypassword and redirect to file "login_success.php" session_register("myusername"); session_register("mypassword");*/ header("location:login_success.php"); } else { echo ("failed"); echo $myusername; echo $mypassword; //header("location:access_denied.php"); } /*else { echo "Wrong Username or Password"; }*/ ?> Link to comment https://forums.phpfreaks.com/topic/223784-errors-with-passing-variables/ Share on other sites More sharing options...
tomtimms Posted January 9, 2011 Share Posted January 9, 2011 Need to find out why your query isn't working, after this line $result=mysql_query($sql); use mysql_error(); and your query error will display on the page. From there we can see what the problem is.... Also try and use if(isset($_POST['Userid']){$myusername=$_POST['Userid'];} else {$myusername = "Not Set";} and echo out $myusername to see if you are even getting the passed variable in the URL. Link to comment https://forums.phpfreaks.com/topic/223784-errors-with-passing-variables/#findComment-1157201 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.