Question

<html>
<body>
<?php

include 'index.php';
include '../db.php';
$query_email = "SELECT email FROM email_addresses"; 
$result_query_email = mysqli_query($query_email)
or die(mysqli_error());

while ($row = mysqli_fetch_array()($result_query_email)) {

//$email_user = $row[10];
$email_user = $row['email'];

$data = date("Y.m.d");
$subject = "Good news";
$message = ("Hello!");

$sentemail = mail($email_user, $subject, $message, 
"From: abc <xyz.com>\nX-Mailer: PHP/" . phpversion());
}
if($sentemail) 
{

    print("Email has been sent successfully.<br>");
}
else
{
  print("There was a problem while sending email.<br>");
}

?>

</body>
</html>

此代码用于从数据库中选择电子邮件地址并向其发送电子邮件。我的代码有什么问题，我无法收到任何电子邮件，帮助将不胜感激。谢谢！

Answer 1

设置error_reporting和display_errors以确定

中缺少)

while (mysql_fetch_array($result_query_email) {

Answer 2

这一行

while (mysql_fetch_array($result_query_email)

您需要将mysql_fetch_array()的返回值分配给变量才能在while()循环中使用它以及缺少结束)

例如

while ($emaildata = mysql_fetch_array($result_query_email)) {
{
    $email_user = $emaildata['email'];
 //.....rest of code

也

Please, don't use mysql_* functions in new code。它们不再维护，弃用过程已经开始，请参阅red box。请改为了解prepared statements，并使用PDO或MySQLi; this article会帮助您确定哪个。如果您选择PDO here is a good tutorial。

修改

此外，您似乎尝试将变量用作关联数组，您需要使用mysql_fetch_assoc()代替mysql_fetch_array()来执行此操作

Answer 3

如果您没有在源代码中提供，请在您的服务器PHP.ini中设置正确的SMTP服务器地址。

输出php数组

3 个答案: