Fatal error: Function name must be a string in C:\xampp\htdocs\test.php on line 6

Question

<?php

$conn = mysqli_connect('localhost','root','','bursary');
// Check connection

if (!$conn())
{
echo "Failed to connect to MySQL: ";
}

$result = mysqli_query($conn,"SELECT shortlist_id, user_id, application_status, amount_allocation FROM shortlist");

echo "<table border='0'>
<tr>
<th>Shortlist Id</th>
<th>Applicant Id</th>
<th>Application Status</th>
<th>Ammount Allocated</th>
</tr>";

while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . strtoupper($row['shortlist_id']) . "</td>";
echo "<td>" . strtoupper($row['user_id']) . "</td>";
echo "<td>" . strtoupper($row['application_status']) . "</td>";
echo "<td>" . strtoupper($row['amount_allocation']) . "</td>";
echo "</tr>";

}
echo "</table>";

mysqli_close($conn);
?>

Answer 1

Is the mysqli extension installed on your server?

do a phpinfo() or:

if (!extension_loaded("mysqli"))
{
echo "mysqli not loaded"
}
else
{
echo "mysqli is loaded"
}

Answer 2

Additionally to what @Xan said, I guess you need to terminate execution upon finding no connection was available, as I see it you only echo "Failed to connect...", but the code execution is continued after that.

I guess you should do
if(!$conn)
...

rather than

if(!$conn())
...

The second one (as I see in your code) looked more of a function rather than a variable : )

Answer 3

thanks ipang it worked after i used this: if (!$conn)