有这两个错误试图在PHP上显示MySQL数据

Question

警告：mysqli_query((期望参数1为mysqli，字符串给定 在C： Xampp htdocs tutoringEnrollmentlist.php中，第45行

警告：mysqli_fetch_array((期望参数1为mysqli_result， c： xampp htdocs tutoringEnrollmentlist.php在第60行60

中给出的null

这是第41-78行

<?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

您在

中使用了错误的顺序

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

应该像

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

Answer 2

尝试检查此行＃正在为您造成错误

echo "<td>" . $record['#'] . "</td>";