我从table1将该数据提取到表单输入值中,并尝试将其插入到table2中。然后,我希望将其从表1中删除。但是,在完成INSERT之后,DELETE不起作用。没有IF,有什么方法可以做到这一点吗?怎么样? 看到我试过了:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "lock_access";
$conn = mysqli_connect($servername, $username, $password);
if(!$conn)
{ echo 'Not connected';
}
if(!mysqli_select_db($conn,$dbname))
{
echo 'DB Not selected'; }
$fn = $_POST['full_name'];
$fln = $_POST['flatno'];
$gn = $_POST['gender'];
$pcn = $_POST['pcardno'];
$sql = "INSERT INTO approved (full_name, flat_number, gender, new_card) VALUES('$fn', '$fln', '$gn', '$pcn')";
if(mysqli_query($conn, $sql))
{
DELETE FROM lock_access WHERE full_name='$fn';
}
else {
echo "Application not received. ";
}
?>