Jump to content
Latest Mentoring Zoom Meetings uploaded - June 25 2025

zhafran

Member
 View Profile  See their activity

  • Posts

    10

  • Joined

  • Last visited

zhafran's Achievements

Newbie

Newbie (1/14)

0

Reputation

  1. zhafran
    Thanks for your explanation, Ben One more thing to clear and this is because I always see the same variable in your code. So I get confuse whether it is same variable or the other one. Consider this line $_SESSION[$type][] = $value; and foreach($_SESSION[$alert] as $value) Both variable is same which is $value and both is an assignment to the same session (you mention earlier post). So this is just like confuse me a little bit
  2. zhafran
    Really? What I saw in the video you only set the variable like this : private $alertTypes; Or maybe I just missed seeing Ok enough for that part. Still not clear but better than before. I have one more question. This one is about getAlert() Consider this line : if (isset($_SESSION[$alert])) Is it the session come from this line : $_SESSION[$type][] = $value; If not, I need your explanation. Sorry to trouble you, Ben.
  3. zhafran
    Thanks for your reply Still staring at the codes to understand the real process. Hopefully the thing will clicks soon Ok when we say : $type = $this->alertTypes[0]; What I understood right now is, we want to replace the $type with $this->alertTypes and put the node 0 to it. Like this : $type[0] Am I right?
  4. zhafran
    Hi there, I'm working on Ben Falk's PHP Login Using OOP & MVC and I need detailed explanation about this function : function setAlert($value, $type = null) { if ($type == '') { $type = $this->alertTypes[0]; } $_SESSION[$type][] = $value; } What really the use of the blank array and how it works really? Hopefully Ben or anybody can explain it deeply because I just a beginner here. Thanks!
  5. zhafran
    Nah! The error all fixed now and I learned something new today Thanks a lot Ben! You're really the legend here
  6. zhafran
    Thanks Ben! The error gone already now One more thing, when I try to add new record, I got this error on records.php Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\test\crud-mysqli\records.php:42) in C:\xampp\htdocs\test\crud-mysqli\records.php on line 64 Is it because the doctype that I use that cause this error?
  7. zhafran
    Thanks Andrea to spend your time replying my topic. Although you not answering my question but it's great to see some reply from expert in noob's thread like me
  8. zhafran
    Thanks Ben, I tried many times to fix this 2 issue here ('header' and 'last name cannot be NULL') but really don't have an idea what's going wrong. Really appreciate your help!
  9. zhafran
    Anybody home? Or do I need edit my question to be more specefic to get answer here? Thanks
  10. zhafran
    Hello all, Currently I'm working with Ben Falk's tutorial on PHP CRUD Videos and I got my head spinning with some errors I'm stuck at video part 5 and below is my code (which I did based on the videos) for your reference (to help me ): <?php include ('connect-db.php'); function renderForm($first = NULL, $last = NULL, $error = NULL, $id = NULL) { ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title><?php if ($id != '') { echo "Edit Record"; } else { echo "New Record"; } ?></title> <link href="css/style.css" rel="stylesheet" type="text/css" /> <script type="text/javascript" src="js/jquery.js"></script> <link rel="shortcut icon" href="images/favicon.ico" /> </head> <body> <h1><?php if ($id != '') { echo "Edit Record"; } else { echo "New Record"; } ?></h1> <?php if($error != '') { echo "<div style='padding=4px; border=1px solid red; color:red'>" . $error . "</div"; } ?> <form action="" method="post"> <div> <?php if (!empty($id)) { ?> <input type="hidden" name="id" value="<?php echo $id; ?>" /> <p>ID: <?php echo $id; ?></p> <?php } ?> <strong>First Name: *</strong> <input type="text" name="firstname" value="<?php echo $first; ?>" /> <strong>Last Name: *</strong> <input type="text" name="lastname" value="<?php echo $last; ?>" /> <p>* Required</p> <input type="submit" name="submit" value="Submit" /> </div> </form> </body> </html> <?php } ?> <?php if (isset($_GET['id'])) { // editing existing records renderForm(NULL, NULL, NULL, $_GET['id']); } else { // create new record if (isset($_POST['submit'])) { $firstname = htmlentities($_POST['firstname'], ENT_QUOTES); $lastname = htmlentities($_POST['lastname'], ENT_QUOTES); if ($firstname == NULL || $lastname = NULL) { $error = 'ERROR: Please fill in all the required field'; renderForm($firstname, $lastname, $error); } else { if ($stmt = $mysqli->prepare("INSERT players (firstname, lastname) VALUES (?,?)")) { $stmt->bind_param("ss", $firstname, $lastname); $stmt->execute(); $stmt->close(); } else { echo "Error: Could not prepare SQL statement."; } header("Location: view.php"); } } else { renderForm(); } } $mysqli->close(); ?> And this is the error message I've got : Warning: mysqli_stmt::execute() [mysqli-stmt.execute]: (23000/1048): Column 'lastname' cannot be null in C:\xampp\htdocs\test\crud-mysqli\records.php on line 59 Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\test\crud-mysqli\records.php:42) in C:\xampp\htdocs\test\crud-mysqli\records.php on line 64 Hopefully someone can help me on this and your help will highly appreciated! Many thanks!
×
×
  • Create New...