试图在php上显示mysql数据有这2个错误

Question

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\tutoringEnrollmentList.php on line 45

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\tutoringEnrollmentList.php on line 60

here are lines 41-78

<?php
$dbc = mysqli_connect ('localhost', 'root', '', 'studentDB')
    or die (mysqli_connect_error());
$sql = "SELECT * FROM `tutoringservice`";
$data = mysqli_query($sql, $dbc);

echo "<table border=1>
<tr>
<td> # </td>
<td> Date </td>
<td> Last Name </td>
<td> First Name </td>
<td> Email </td>
<td> Student ID </td>
<td> Subject </td>
<td> Message </td>
<td> Tutoring Day </td>
</tr>";

while($record = mysqli_fetch_array($data)){
echo "<tr>";
echo "<td>" . $record['#'] . "</td>";
echo "<td>" . $record['Date'] . "</td>";
echo "<td>" . $record['Last Name'] . "</td>";
echo "<td>" . $record['First Name'] . "</td>";
echo "<td>" . $record['Email'] . "</td>";
echo "<td>" . $record['Student ID'] . "</td>";
echo "<td>" . $record['Subject'] . "</td>";
echo "<td>" . $record['Message'] . "</td>";
echo "<td>" . $record['Tutoring Day'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($dbc);



?>

Answer 1

You have used the incorrect order in

$data = mysqli_query($sql, $dbc);

It should be like

$data = mysqli_query($dbc, $sql);