Question

我正在尝试创建一个URL缩短脚本，当用户提交他们的URL时，它应该返回一个包含缩短的URL的文本字段，但它不是，请你看看我的代码，看看你是否可以停止问题？

感谢。

<html>
<head></head>

<body>

<?php

if ($_POST['shortbtn']) {
    $url = $_POST['url'];

    if ($url) {

        require("./connect.php");

        $charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
        $len = 5;
        $numrows = 1;

        while ($numrows != 0) {
            for ($i = 0; $i <= $len; $i++) {
                $rand = rand() % strlen($charset);
                $tmp = substr($charset, $rand, 1);
                $code .= $tmp;
            }

            $query = mysql_query("SELECT * FROM items WHERE code='$code'");
            $numrows = mysql_num_rows($query);
        }

        $date = date("F d, Y");

        mysql_query("INSERT INTO items VALUES('', '$url', '$code', '$date')");

        $query = mysql_query("SELECT * FROM items WHERE code='$code'");
        $numrows= mysql_num_rows($query);

        if ($numrows == 1) {
            $site = "http://dthms.dev:8888/projects/url-short/";
            echo "Your URL: <input type='text' value='$site/$code' size='40'>";
        } else
            echo "Not Added";

        mysql_close();

    } 

    else echo "<script type='text/javascript'>window.location = './'</script>";
} else echo "";

?>

</body>
</html>

在您提交网址时，它会返回“未添加”

if ($numrows == 1) {
                $site = "http://dthms.dev:8888/projects/url-short/";
                echo "Your URL: <input type='text' value='$site/$code' size='40'>";
            } else
                echo "Not Added";

为什么我的脚本没有显示正确的输出？

0 个答案: