PHP undefined variable error when getting the MySQL records

Question

I'm writing something like a "e-diary" (for schools, education project) in PHP. I have a problem with database records.

Notice: Undefined variable: name in C:\xampp\htdocs\projects\edziennik\panel.php on line 44

Notice: Undefined variable: surname in C:\xampp\htdocs\projects\edziennik\panel.php on line 44

My code:

if (!empty($_POST['name']) or !empty($_POST['surname'])){
    require('inc/database.php');
    $query = $db->prepare("SELECT * FROM `studentindex`");
    $query->execute();
    $result = $query->fetchAll();
    echo '<table class="table table-striped table-hover ">
            <thead>
                <tr>
                  <th>#</th>
                  <th>Imię</th>
                  <th>Nazwisko</th>
                  <th>Klasa</th>
                </tr>
            </thead><tbody>';
            $row = array();
    foreach ($result as $row){
      if ($row['name'] == $name or $row['surname'] == $surname){
        echo '    <tr>
                    <td>'.$row['id'].'</td>
                    <td>'.$row['name'].'</td>
                    <td>'.$row['surname'].'</td>
                    <td>'.$row['class'].'</td>
                  </tr>';
      }

Exact problem is in foreach, where the if giving that error, in other PHP versions everything was OK (with MySQL). I'm using PDO actually, in database.php is only connection that works fine.

Edit

I have error Notice: Undefined index: id in C:\xampp\htdocs\projects\edziennik\panel.php on line 48 now.

These lines:

echo '<tr>
    <td>'.$row['id'].'</td> // line 48
    <td>'.$row['name'].'</td>
    <td>'.$row['surname'].'</td>
    <td>'.$row['class'].'</td>
</tr>';

$row['name'] == $name or $row['surname'] == $surname //have you assigned value for the variables $name, $surname? — Vigneswaran S, Nov 13 '15 at 16:11
Possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — Derek, Nov 13 '15 at 16:11

c4pricorn · Accepted Answer · 2015-11-13T16:15:38.730

0

if (!empty($_POST['name']) or !empty($_POST['surname'])){

You have to add variables and assign values:

$name = $_POST['name'];
$surname = $_POST['surname'];

edited Nov 13 '15 at 16:15

answered Nov 13 '15 at 16:09

c4pricorn

3,471
1
11
12

Or, more lightweight, less confusing: replace the single occurrences of `$name` and `$surname` directly by `$_POST['name']` and `$_POST['surname']`... – arkascha Nov 13 '15 at 16:15
I have `Notice: Undefined index: id in C:\xampp\htdocs\projects\edziennik\panel.php on line 48` error now. – Kacper Nov 13 '15 at 16:16
@fox what fields are in table "studentindex"? The same you have to set in $row['...'] – c4pricorn Nov 13 '15 at 16:16
@c4pricorn in this table is person's ids (id, name, surname, class) – Kacper Nov 13 '15 at 16:19
SELECT * FROM `studentindex` - delete ` chars – c4pricorn Nov 13 '15 at 16:21
@c4pricorn the problem was this in mysql it says ID... I'm very beginner in php. – Kacper Nov 13 '15 at 16:22

PHP undefined variable error when getting the MySQL records

Edit

1 Answers1