Question

我有两个MySQL表，一个有订阅者详细信息，另一个跟踪发送给他们的电子邮件，哪个广告系列，哪个版本等（分别是crawler和sent_mail）。我建了一个脚本可以通过爬虫表并基于SendGrid API，Web表单和一些javascript发送电子邮件。但我想弄清楚如何告诉脚本从抓取工作表中提取没有sent_mail表中列出的当前广告系列的电子邮件。我知道我必须使用连接，但我不确定如何在查询中构建逻辑。思考？我的尝试如下：

foreach ('SELECT crawler.email FROM crawler LEFT JOIN sent_mail ON crawler.email = sent_mail.email WHERE sent_mail.campaigns <> "$campaign"') as $address) as $address) {
    $email = new SendGrid\Email();
    $email
        ->addTo($address)
        ->setFrom($address)
        ->setSubject($title)
        ->setText($text)
        ->setHtml($text)
    ;
    $sendgrid->send($email);
    $db-> $query = "
          INSERT INTO sent_mail (email, campaigns, campaign_version)
          VALUES ($address, $campaign, $campaign_version)"; echo 'great success';
}

Answer 1

我认为您正在寻找以下代码.. 我认为以下代码可能会有所帮助

    $get_records = mysql_query('SELECT crawler.email as `email_val` FROM crawler LEFT JOIN sent_mail ON crawler.email = sent_mail.email WHERE sent_mail.campaigns <> "$campaign"');

$i = 1;
while($data = mysql_fetch_assoc($get_records)){

    $to = $data['email_val'];
    $subject = 'Your Subject';

    // Always set content-type when sending HTML email
    $headers = "MIME-Version: 1.0" . "\r\n";
    $headers .= "Content-type:text/html;charset=UTF-8" . "\r\n";

    // More headers
    $headers .= 'From: Your Name <you@you.com>' . "\r\n";


    $sendmail = mail($to,$subject,$message,$headers);
    if($sendmail){
        $insert = mysql_query("INSERT INTO sent_mail (email, campaigns, campaign_version)
          VALUES ('your value', 'your value', 'your value')");
        if($insert){
            echo  $i.'Value inserted & Mail sent to'.$data['email'];
        }else{
            echo  $i.'Value not inserted but Mail sent to'.$data['email'];
        }
    }else{
        echo  $i.'Mail sending fail to'.$data['email'];
    }
$i++;
}

注意：在运行上述代码之前，请确保已建立数据库连接。

使用PHP / MySQL自动化电子邮件活动，不确定如何构建查询

1 个答案: