Question

当网站打开X秒后，我想更新一个MySQL字段。我从MySQL获取了Seconds / Time，并且想在秒数结束后在MySQL中进行更新。

我尝试过

sleep($adddisplaytime);

但是网站等待完成，并且没有先运行

打开网站几秒钟后，是否可以运行我的更新？

$query1 = "UPDATE ads SET views = views+1, costs = costs+price WHERE id = '".$adid."'";

可以在PHP或MySQL中使用

Answer 1

注意：这会做您想要的，但是可能会被反复击中AJAX端点的人利用，您需要为此提供一些保护。

您将需要一个附加的PHP文件，该PHP的工作是仅更新数据库。您需要从页面加载脚本中取出该更新。

您的HTML / JS / PHP初始加载

<script>
setTimeout(function() {
    $.ajax('/your/ajax/endpoint.php', {
        data: {
            'adid': 'your id'
            /*
               If this is in your PHP file, you can echo the ID straight there.
               Not totally recommended, but that's one way An additional / 
               better way is to add it to a div with a data attribute and 
               use jQuery to select the data off of there
            */
        }
    }); // Probably lots more you can do here, but in this case, for simplicity, just sending and that's it
}, 2000); // This will do a 2 second wait
</script>

您的新附加PHP文件位于/your/ajax/endpoint.php

<?php
// THIS FILE DOES THE UPDATE
$adid = $_POST['adid'];

// As mentioned by tadman in his comment.. I would use prepared statements 
$query1 = "UPDATE ads SET views = views+1, costs = costs+price WHERE id = ?";

try {
    $dbh = new PDO($dsn, $user, $password);

    $sth = $dbh->prepare($query1);
    $sth->execute(array($adid));
} catch (PDOException $e) {
    echo 'Connection failed: ' . $e->getMessage();
}

注意：

再次，为了安全起见，您真的要考虑让您的第一个PHP脚本生成一个唯一ID（并将其存储在db中），并将其传递给页面，并让AJAX将该唯一ID与adid一起发送，并且如果您提供的唯一ID仅在数据库中，那么您会知道这是合法请求。从数据库中删除唯一ID，然后进行更新。

Answer 2

如果要在打开页面后等待几秒钟，然后运行update语句，请在页面顶部编写以下代码：-

echo "<script> setTimeout(function(){}, 2000) ; </script>" ;

$query1 = mysqli_query($con, "UPDATE ads SET views = views+1, costs = costs+price WHERE id = '".$adid."'");

秒后更新PHP MySQL

2 个答案: