Question

我正在尝试在数据库中插入一些东西，但是当我使用该脚本时，它会导致脚本错误的结束。

无论我订购什么订单，都无法解决它。

代码：

<?PHP
include('inc/database/db_connect.php'); //includes db settings  

function sernum()
{
    $template = 'X9XX-XX99-99XX-99X9-XXXX';
    $k = strlen($template);
    $sernum = '';
    for ($i=0; $i<$k; $i++) {
        switch($template[$i]) {
            case 'X': $sernum .= chr(rand(65, 90)); break;
            case '9': $sernum .= rand(0,9); break;
            case '-': $sernum .= '-'; break;
    }
    return $sernum;
        mysqli_query($db_connect,"INSERT INTO AlphaKeys (`AlphaKey`) VALUES ('$sernum');
        mysqli_close($db_connect);
       }
     }
}
echo '<prev>';

for ($i=0; $i < 20; $i++) echo sernum(), '<br/>';

echo '</pre>'; 
?>

和db_connect.php

<?PHP
$connect_error = "Sorry, we're experiencing downtime, please try again.";
$db_connect = mysqli_connect("localhost","********","B&5I)0U%a-B}","********") or die($connect_error);
?>

我想弄清楚，我不能......

Answer 1

纠正这一行

mysqli_query($db_connect,"INSERT INTO AlphaKeys (`AlphaKey`) VALUES ('$sernum');

到

mysqli_query($db_connect,"INSERT INTO AlphaKeys (`AlphaKey`) VALUES ('$sernum')");

意外的文件结尾

1 个答案: