While循环中的PHP While循环

Question

试图运行一个脚本，从我的数据库中通过电子邮件发送给用户他们的个人时间表。

代码应该类似于this...select表1中的所有用户，其中用户希望每周发送电子邮件。

然后获取这些user_ids并运行查询，以获得表2中每个用户的每周计划。

如果表2中的日期小于或等于代码运行时的1周，并且它不等于今天的日期，则设置变量$email_content。

如果表2中的日期超过1周，那么代码应该知道它已经完成，并且这里想要继续发送()为该用户收集的信息，并仅将该信息发送给该用户，然后继续发送给下一个用户。

在回显“中断”的地方，它本质上是在标记每个用户的日程安排之间的分隔，但它回显了大约20次。理想情况下，这部分将成为我放置mail()函数的位置，因此我不能让它执行mail()函数20次。

我已经尝试了大量不同的代码变体，但似乎都不能正确理解。总而言之，我需要获取user1的info1，然后将info1发送到user1，然后继续获取user2的info2，然后将info2发送到user2，依此类推……

<?php

// GET USERS WHO WANT THEIR WEEKLY SCHEDULE EMAILED TO THEM (Table 1)
$sql = "SELECT * FROM XXXXXXX WHERE weekly_email = 'Yes'";
$result = mysqli_query($connection, $sql);
while ($row = mysqli_fetch_array($result)) {
    $id .= "'" . $row['user_id'] . "', ";
}
$id = trim($id, ', ');

// GET THOSE USERS' SCHEDULES (Table 2)
$sql_email = "SELECT * FROM XXXXXXXX WHERE user_id IN ($id) ORDER BY user_id, date, start_time ASC";

$result_email = mysqli_query($connection, $sql_email);

while ($row_email = mysqli_fetch_array($result_email)) {
    $truck_name_email = $row_email['truck_name'];
    $location_name_email = $row_email['location_name'];
    $date_email = $row_email['date'];
    $address_email = $row_email['address'];
    $x = strtotime("7 days");

    $start_email = $row_email['start_time'];
    $end_email = $row_email['end_time'];

    if ($date_email <= date("m/d/Y l", $x)) {

        if ($date_email !== date("m/d/Y l")) {
            if (!empty($address_email)) {
                $address_email2 = explode(",", $row_email['address'], 2);
                $email_content = $date_email . substr($date_email, 11) . " - "
                    . $location_name_email . ", " . $address_email2[0] . ", "
                    . $start_email . "-" . $end_email . "<br/>";

                echo $email_content;
            } elseif (empty($address_email)) {
                $email_content .= $date_email . " - " . $location_name_email
                    . ", " . $start_email . "-" . $end_email . "<br/>";

                echo $email_content;
            }
        }
    }
    if ($date_email >= date("m/d/Y l", $x)) {
        echo "break";
        // break;
    }
}

Answer 1

我只是说，你应该注意潜在的SQL注入，在任何用户可控制的字符串进入SQL语句之前对它们进行转义。例如，如果$row['user_id']是一个字符串，则必须执行mysqli_escape_string

看起来你在代码中遗漏了一些东西：

    $x = strtotime("7 days");

    $start_email = $row_email['start_time'];
    $end_email = $row_email['end_time'];

    if ($date_email <= date("m/d/Y l", $x)) {

        if ($date_email !== date("m/d/Y l")) { // I think that you have a $x here?
            if (!empty($address_email)) {
                $address_email2 = explode(",", $row_email['address'], 2);
                $email_content = $date_email . substr($date_email, 11) . " - "
                    . $location_name_email . ", " . $address_email2[0] . ", "
                    . $start_email . "-" . $end_email . "<br/>";