CODEWITHSUNDEEP
php
Olaf van Gelder
Olaf van Gelder

Reputation: 41

Trying to access array offset on value of type null (PHP)

problem resolved (in comments) i am making a project for school and i am getting this error Trying to access array offset on value of type null on this line of code

<div class="user-name"><?php echo $guests["firstName"] ?></div>

it is in this code and i retrieve the information out of a database

foreach((array)$result as $guests)
        {
    ?>
    <div class="posts-container">
        <div class="post-header">
            <div class="user-details">
                <div class="user-name"><?php echo $guests["firstName"] ?></div>
                <div class="user-email"></div>
            </div>
            <div class="time"></div>
        </div>
        <div class="post-message">
            <h3></h3>
            <p></p>
        </div>
    </div>
    <?php } ?>

people asked for the sql code and the type/structure of $result type/structure: object(mysqli_result)#2 (5) { ["current_field"]=> int(0) ["field_count"]=> int(9) ["lengths"]=> NULL ["num_rows"]=> int(5) ["type"]=> int(0) sql/database:

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "guestbook";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id, firstName, lastName, email, title, message, URL, showEmail, Date FROM guestbook";
$result = $conn->query($sql);

$conn->close();

Upvotes: 4

Views: 43932

Answers (3)

azibom
azibom

Reputation: 1944

That is because this line

$result = $conn->query($sql);

That is not what you want, you need to fetch the result like this

if ($result->num_rows > 0)
while($guest = $result->fetch_assoc()) {
{?>
    <div class="posts-container">
        <div class="post-header">
            <div class="user-details">
                <div class="user-name"><?php echo $guest["firstName"] ?></div>
                <div class="user-email"></div>
            </div>
            <div class="time"></div>
        </div>
        <div class="post-message">
            <h3></h3>
            <p></p>
        </div>
    </div>
<?php } ?>

Upvotes: 1

Aleksey Malashenok
Aleksey Malashenok

Reputation: 69

You can try something like this

<?php
//Array to be shown
$guests = [
        ['firstName' => 'Alex', 'email' => '[email protected]'] ,
        ['firstName' => 'Jhon', 'email' => '[email protected]']
];
?>
<div class="posts-container">
    <div class="post-header">
        <div class="user-details">
        <!-- foreach alternate syntax -->
        <?foreach ($guests as $guest):?>
            <div class="user-name"><?=$guest['firstName']?></div>
            <div class="user-email"><?=$guest['email']?></div>
        <!-- foreach alternate syntax -->
        <?endforeach;?>
    </div>
    <div class="time"></div>
</div>
<div class="post-message">
    <h3></h3>
    <p></p>
</div>

Upvotes: -1

Syscall
Syscall

Reputation: 19764

You are trying to access to a null value as an array.

In your case, $guest value is null. So, accessing to $guest['something'] will throw a "Notice: Trying to access array offset on value of type null".

Two suggestions :

  1. Check $guest value (quick) :

    foreach((array)$result as $guests)
{
    if (!is_array($guests)) {
        continue;
    }

  2. Check $result value (better) :

    If $result is a result of a SQL query, maybe try to add conditions to avoid NULL results. Or, use array_filter() before to remove empty values.

Upvotes: 1

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