Question

我是cronjobs的新手，在学校上过一些课程，但是老师对此并不了解，谢谢......我写了一个脚本，在3天后没有激活用户时自动删除用户，或者不活跃90天。

但是我想通过cronjob运行该脚本...我的主机使用Directadmin并且我已经使用了google很多但没有任何作用...... :(

我知道计时器是如何工作的，但在某个地方出错 - ， -

也许你们这里有人可以帮助我？

这是我使用的行：/ usr / bin / php -q / home / 用户名 / domains / Domain_name / public_html / 文件夹 / _ dell.php

这是_dell.php

中的php脚本

BTW php脚本有效，当我自己运行时它运行完美（y）

<?php

    include "includes/conn.php";

    // Delete users die na 3 dagen nog niet zijn geactiveerd.
    $ongeactiveerd = time() - (60 * 60 * 24 * 3); // staat gelijk aan 3 dagen.
    $ongeactiveerd1=mysqli_query($server,"SELECT * FROM users WHERE rank < 1 AND status = 0 AND laatstgezien < $ongeactiveerd");

            if(mysqli_num_rows($ongeactiveerd1) > 0) { 

            $test = array();
                while($row = mysqli_fetch_array($ongeactiveerd1)) {

                    $test[] = $row['username'];
                }


                foreach($test as $testkey => $testuser) {
                    $test[$testkey] = "'".mysqli_real_escape_string($server,$testuser)."'";
                }

                $homo = join(', ', $test);

                echo $homo;

                $query = "DELETE FROM users WHERE username IN ($homo)";
                $query1 = "DELETE FROM pm WHERE verzender IN ($homo)";
                $query2 = "DELETE FROM pm WHERE ontvanger IN ($homo)";

                echo "<br>".$query."<br><br>".$query1."<br><br>".$query2;
                //exit();
                $result = mysqli_query($server,$query) or die (mysqli_error($server));
                $result1 = mysqli_query($server,$query1) or die (mysqli_error($server));
                $result2 = mysqli_query($server,$query2) or die (mysqli_error($server));

            } 
            else {
                echo "Er zijn geen on-geactiveerde gebruikers.<br><br>";
            }


        // Delete users die 90 dagen inactief zijn.
        $inactief = time() - (60 * 60 * 24 * 90); // staat gelijk aan 90 dagen || 3 maanden.
        $status1 = 1;
        $inactief1=mysqli_query($server,"SELECT * FROM users WHERE rank < 1 AND status = 1 AND laatstgezien < $inactief");

            if(mysqli_num_rows($inactief1) > 0) { 

            $test1 = array();
                while($row = mysqli_fetch_array($inactief1)) {

                    $test1[] = $row['username'];
                }


                foreach($test1 as $test1key => $test1user) {
                    $test1[$test1key] = "'".mysqli_real_escape_string($server,$test1user)."'";
                }

                $homo1 = join(', ', $test1);

                echo $homo1;

                $queryy = "DELETE FROM users WHERE username IN ($homo1)";
                $queryy1 = "DELETE FROM pm WHERE verzender IN ($homo1)";
                $queryy2 = "DELETE FROM pm WHERE ontvanger IN ($homo1)";

                echo "<br>".$queryy."<br><br>".$queryy1."<br><br>".$queryy2."<br>";
                //exit();
                $resultt = mysqli_query($server,$queryy) or die (mysqli_error($server));
                $resultt1 = mysqli_query($server,$queryy1) or die (mysqli_error($server));
                $resultt2 = mysqli_query($server,$queryy2) or die (mysqli_error($server));

            } 
            else {
                echo "Er zijn geen inactieve gebruikers.";
            }
?>

如何在directadmin中使用cronjob运行php脚本

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