【发布时间】:2016-04-07 05:50:07
【问题描述】:
我是初学者和文凭学生...我不知道错误是什么...请帮我解决错误...
<?php
$servername="localhost";
$username="root";
$password="";
$dbname="slr";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO software (soft_id, soft_name, installed_date, expiry_date, product_key)
VALUES ('2', 'Dhurga', '2016-01-01', '2016-04-30', 'stevenreega@gmail.com')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
<input type="button"value="Finish"onclick="history.go(-2);return true;">
</table>
mysqli_close($conn);
?>
【问题讨论】:
-
<input type="button"value="Finish"onclick="history.go(-2);return true;"> </table>在 php 标签内做什么?? -
缺少 PHP 标签。