php - 如何让 php 在循环外回显值？

Question

我希望能够将此人的姓名放入电子邮件中。但是我不知道如何在循环外保留值。另外，我想知道如果结果不止一个，这是否可行。

这是 php:

// WHO HAS BEEN A MEMBER FOR ONE YEAR
function joinDateFilter(){
    $query = mysql_query("SELECT * FROM user WHERE date_joined = DATE_SUB(curdate(), INTERVAL 1 YEAR)");
    $mail_to = "";
    while ($row = mysql_fetch_array($query)){
        echo $row['name']." - ".$row['email']."\n";
        $mail_to = $row['email'].", ";
        $name = $row['name'];
    }
    if (!empty($mail_to)){
        sendEmail($mail_to);
    }
}

// SEND EMAIL
function sendEmail($mail_to) {
    $from    = "webmaster@mydomain.com";
    $message = "Happy Anniversary!";
    $headers = 'From: '.$from."\r\n" .
    'Reply-To:'.$_POST['email']."\r\n" .
    "Content-Type: text/html; charset=iso-8859-1\n".
    'X-Mailer: PHP/' . phpversion();
    mail($mail_to, "Congratulations ".$name."!", $message, $headers);
}

在上面的例子中，邮件从不读取他们的名字，因为这个值似乎在 while 循环之外被遗忘了。有人可以帮助我吗？

Answer 1

目前，您的代码会在 while 循环的每次迭代中覆盖存储在 $mail_to 中的电子邮件 ID。因此，它将只有最后一个用户的电子邮件 ID。并且您需要每个用户的名称(存储在 $name 中并在函数中使用)。只要找到用户，您就需要发送电子邮件。试试这个:

// WHO HAS BEEN A MEMBER FOR ONE YEAR
function joinDateFilter(){
    $query = mysql_query("SELECT * FROM user WHERE date_joined = DATE_SUB(curdate(), INTERVAL 1 YEAR)");
    while ($row = mysql_fetch_array($query)){
        echo $row['name']." - ".$row['email']."\n";
        $mail_to = $row['email'];
        $name = $row['name'];
        if (!empty($mail_to))
            sendEmail($mail_to, $name);
    }
}

// SEND EMAIL
function sendEmail($mail_to, $name) {
    $from    = "webmaster@mydomain.com";
    $message = "Happy Anniversary!";
    $headers = 'From: '.$from."\r\n" .
    'Reply-To:'.$_POST['email']."\r\n" .
    "Content-Type: text/html; charset=iso-8859-1\n".
    'X-Mailer: PHP/' . phpversion();
    mail($mail_to, "Congratulations ".$name."!", $message, $headers);
}