❓ Help Undefined variable $row in <b>

[Alieson09]

<br /><b>Warning</b>: Undefined variable $row in <b>C:\xampp\htdocs\test3\update_profile.php</b> on line <b>22</b><br /><br /><b>Warning</b>: Trying to access array offset on value of type null in <b>C:\xampp\htdocs\test3\update_profile.php</b> on line <b>22</b><br />

 

[🤖]

These warnings suggest that there is an issue with the variable $row being undefined or null in the update_profile.php file. This could be because the variable was not properly initialized or because it is out of scope.

To resolve this issue, you should first check if the variable $row is being properly initialized and if it is in the correct scope. You can do this by checking the code and making sure that the variable is defined before it is used.

If the variable is not being properly initialized, you should initialize it with a default value before using it. You can do this by assigning it a default value of null or an empty array.

If the variable is in the wrong scope, you should move it to the correct scope or pass it as a parameter to any functions that use it.

Overall, it is important to make sure that all variables are properly initialized and in the correct scope to avoid errors like these.

 

[Alieson09]

These warnings suggest that there is an issue with the variable $row being undefined or null in the update_profile.php file. This could be because the variable was not properly initialized or because it is out of scope.

To resolve this issue, you should first check if the variable $row is being properly initialized and if it is in the correct scope. You can do this by checking the code and making sure that the variable is defined before it is used.

If the variable is not being properly initialized, you should initialize it with a default value before using it. You can do this by assigning it a default value of null or an empty array.

If the variable is in the wrong scope, you should move it to the correct scope or pass it as a parameter to any functions that use it.

Overall, it is important to make sure that all variables are properly initialized and in the correct scope to avoid errors like these.

<?php
session_start();

if (!isset($_SESSION['username'])) {
    header('Location: login.php');
}
$conn = mysqli_connect('localhost', 'root', '', 'user_login');
$username = $_SESSION['username'];
$query = "SELECT * FROM users WHERE username='$username'";
$result = mysqli_query($conn, $query);
if (isset($_POST['update_profile'])) {
    $email = $_POST['email'];

    $query = "UPDATE users SET email='$email' WHERE username='$username'";
    mysqli_query($conn, $query);
    echo "Profile updated successfully";
}
?>
<h3>Update Profile</h3>
<form method="POST" action="">
    <label>Email:</label>
    <input type="email" name="email" value="<?php echo $row['email']; ?>" required>

    <input type="submit" name="update_profile" value="Update Profile">
</form>

<a href="logout.php">Logout</a>

 

[🤖]

<?php
session_start();

if (!isset($_SESSION['username'])) {
    header('Location: login.php');
}
$conn = mysqli_connect('localhost', 'root', '', 'user_login');
$username = $_SESSION['username'];
$query = "SELECT * FROM users WHERE username='$username'";
$result = mysqli_query($conn, $query);
if (isset($_POST['update_profile'])) {
    $email = $_POST['email'];

    $query = "UPDATE users SET email='$email' WHERE username='$username'";
    mysqli_query($conn, $query);
    echo "Profile updated successfully";
}
?>
<h3>Update Profile</h3>
<form method="POST" action="">
    <label>Email:</label>
    <input type="email" name="email" value="<?php echo $row['email']; ?>" required>

    <input type="submit" name="update_profile" value="Update Profile">
</form>

<a href="logout.php">Logout</a>

The issue with the undefined variable $row is that it has not been fetched from the database using the mysqli_fetch_assoc() function.

You can fix this issue by adding the following code after the line $query = "SELECT * FROM users WHERE username='$username'";:

$row = mysqli_fetch_assoc($result);

This code will fetch the row containing the user's information from the database and store it in the $row variable. You can then use this variable to display the user's current email in the form by changing the value attribute in the email input field to value="<?php echo $row['email']; ?>".

The updated code should look like this:

<?php
session_start();

if (!isset($_SESSION['username'])) {
    header('Location: login.php');
}
$conn = mysqli_connect('localhost', 'root', '', 'user_login');
$username = $_SESSION['username'];
$query = "SELECT * FROM users WHERE username='$username'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
if (isset($_POST['update_profile'])) {
    $email = $_POST['email'];

    $query = "UPDATE users SET email='$email' WHERE username='$username'";
    mysqli_query($conn, $query);
    echo "Profile updated successfully";
}
?>
<h3>Update Profile</h3>
<form method="POST" action="">
    <label>Email:</label>
    <input type="email" name="email" value="<?php echo $row['email']; ?>" required>

    <input type="submit" name="update_profile" value="Update Profile">
</form>

<a href="logout.php">Logout</a>

 

[maeorn]

Baka wla binabato yung $row sa array?

 

[Alieson09]

Baka wla binabato yung $row sa array?

ok na po solved na po