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Found 3 results

  1. Hello We have a testing site where users create a profile then take an aptitude test. Was working fine several months ago but we're trying to set up tests now and when the user enters their information and hits submit, we get these errors. (first it was the white screen, then I turned on the errors and received this) Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in ~/inc/connect.inc.php on line 11 Notice: Undefined variable: i_price in ~/profile-2.inc.php on line 97 Fatal error: Call to undefined function session_register() in ~/profile-2.inc.php on line 101 I'm assuming the Fatal Error is what's causing the profile not to be set up(and thus get the white screen). The code in profile-2 is: # Create new record: $now = time(); $password_name = random_password(PASS_LENGTH); $password_hash = bin2hex(mhash(MHASH_MD5, $password_name)); $i_qry2 = db_qry("INSERT INTO reports(createdate,passhash,fname,lname,age,gender,sport,loc,phone,address1,address2,city,state,zip,email,price) VALUES ($now,'$password_hash','$f_firstname','$f_lastname',$f_age,$f_gender,'$f_sport','$f_loc','$f_phone','$f_address1','$f_address2','$f_city','$f_state','$f_zip','$f_email','$i_price')") or die("Error: profile-2, SQL request error #2 ".mysql_error()); $i_sid = mysql_insert_id($sql_link); # Register SID in session: session_register('r_id'); <<<<this is line 101 $_SESSION['r_id'] = $i_sid; session_register('r_pass'); $_SESSION['r_pass'] = ''; session_register('r_ccode'); $_SESSION['r_ccode'] = stripslashes($f_ccode); Does anyone have a suggestion on what might be the problem and how I go about fixing it? Thanks in advance.
  2. Hi everyone, I'm studying ICT in Belgium and we have the task to design a website. So I'm working on a website for like parties, events and stuff. So I wrote the following code : <?php $link = mysqli_connect("localhost","root","","partyguide") or die ('Er ging iets mis: ' . mysqli_connect_error($link)); $sql = "SELECT * FROM evenementen"; if(!empty($_POST)) { $sql.="WHERE id='" .$_POST["evenement_datum"]."'"; } $result = mysqli_query($link,$sql); ?> <html> <head> <title>Evenementen</title> </head> <body> <?php if(empty($_POST)) { $link = mysqli_connect("localhost","root","","partyguide") or die ('Er ging iets mis: ' . mysqli_connect_error($link)); ?> <form name="form1" action="<?php echo($_SERVER["PHP_SELF"]);?>" method="post"> Kies een datum : <select name="evenement_datum"> <?php while($rij = mysqli_fetch_array($result)){ echo("<option value=\"".$rij['datum']."\">".$rij['datum']."</option>\n"); }?> </select> <input type="Submit" value="Toon evenementen!"> </form> <?php }else{ ?> <table width="1000" height="500" align="center" border="1" bordercolor="blue"> <?php while($rij = mysqli_fetch_array($result)){ ?> <tr> <td><?php echo $rij['datum']; ?></td> <td><?php echo $rij['plaats']; ?></td> <td><?php echo $rij['tijdstip']; ?></td> <td><?php echo $rij['naam'] ?></td> </tr> <?php } ?> </table> <?php } ?> </body> </html> So I'm trying to have an option box which displays the date's of events in my database, If i click a date, it has to display all the events in a table, but whenever I click one, following error pops up : Catchable fatal error: Object of class mysqli_result could not be converted to string in D:\www\evenementen.php on line 11 Can anyone help me with this? My teacher doesn't know how to solve this so I hope anyone of you could...
  3. I'm new to coding and prepared statements. I'm getting: "Fatal error: Call to a member function bind_param() on a non-object in /home/content/01/3251001/html/Manage-Users/acc-files/updateRecord.php on line 17" For this: <?php include "db_connect.php"; if(isset($_POST['UserID'])){ $stmt = $con -> prepare('UPDATE UserList SET Status = ?, FirstName = ?, LastName = ?, Username = ?, Email = ?, Department = ?, Manager = ?, WHERE UserID = ?'); $stmt -> bind_param('isssssss', /* Line 17 */ $_POST['Status'], $_POST['FirstName'], $_POST['LastName'], $_POST['Username'], $_POST['Email'], $_POST['Department'], $_POST['Manager'], $_POST['UserID']); $stmt->execute(); } Thank you in advance for helping.
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