Question

警告：mysqli_query（）期望参数1为mysqli，给定字符串   第45行的C：\ xampp \ htdocs \ tutoringEnrollmentList.php

警告：mysqli_fetch_array（）期望参数1为mysqli_result，   在第60行的C：\ xampp \ htdocs \ tutoringEnrollmentList.php中给出的null

这里是第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);

应该是

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

Answer 2

@agam

答案是正确的，但对于您，我想建议您，您应该检查此特定查询here。对于这样的问题你可能看起来here，我发现它在我早期的php和mysql中更有用

Answer 3

尝试检查此行＃是否导致错误

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

试图在php上显示mysql数据有这2个错误

3 个答案: