Help my php says this

Question

<?php
$conn = mysqli_connect('localhost','root','','bursary');

$table_content = '
<table class="table-responsive col-md-12 table-users">
<thead>
<tr>
<th class="col-md-1">Name</th>
<th class="col-md-2">ID Number</th>
<th class="col-md-3">Registration Number</th>
<th class="col-md-4">Institution Name</th>
<th class="col-md-5">Bank Name</th>
<th class="col-md-6">Account Name</th>
<th class="col-md-7">Account Number</th>
</tr>
</thead>
';
$sql = $conn->query ("SELECT firstName, sirName, user_id, Regno, institutionName, bankName, accountName, accountNo FROM profile");

while ($user = mysqli_fetch_array($sql)) {
$name = strtoupper($user['firstName']." ".$user['sirName']);
$userID = $user['user_id'];
$btn_AD_id = "AD".$userID;
$btn_DEL_id = "DEL".$userID;
$table_content .= '
<tr>
<td class="col-md-1">'.$user['$name'].'</td>
<td class="col-md-2">'.$user['user_id'].'</td>
<td class="col-md-3">'.$user['Regno'].'</td>
<td class="col-md-4">'.$user['institutionName'].'</td>
<td class="col-md-5">'.$user['bankName'].'</td>
<td class="col-md-6">'.$user['accountName'].'</td>
<td class="col-md-7">'.$user['accountNo'].'</td>
</tr>
';
}
$table_content .='</table>';

?>

Answer 1

$user['$name'] You have $name variable inside single quotes ``.

Change them to double quotes or remove them.
$user[$name]

You can't use variables inside single quotes. It's displayed as string.

Answer 2

Maybe you meant to display the $name variable itself?
You are now using it as array index. That doesn’t exist.

Answer 3

The column doesn’t exist in your database.

Answer 4

It gives this error :
Notice: Undefined index: $name in C:\xampp\htdocs\bursery\my_application.php on line 27

Answer 5

hy toni thnc it worked...but now it says:
undefined index: (displays data from the database) on line 27