Question

如果我单击通知按钮，则会发送消息，但不会将其作为电子邮件发送，并且返回的页面也没有再次显示该表，我已回显该消息并且它已存在，但无法解决很多问题

     <?php require_once("include/session.php");?>
    <?php require_once("include/dataconnect.php");?>
    <?php require_once("include/functions.php");?>
    <?php include("include/mheader.php");?>
   >
    <p>

        You can contact us so as to give you relevant numbers to speak to.Thank you.
    </p>
    <?php
    $submit = $_POST['Notify'];
    $message = mysql_real_escape_string(htmlentities(strip_tags($_POST['message'])));
    //echo "$message";
    //die();
    if('POST' === $_SERVER['REQUEST_METHOD']) 
    {
    if ($message)
    {
    //Get Email Address
        $emails = mysql_query("SELECT reusers.email FROM reusers INNER JOIN repplac ON reusers.username = repplac.Uname AND reusers.username = '{$_SESSION['username']}'")or die(mysql_error());
        //$emails = mysql_query("SELECT reusers.email FROM reusers INNER JOIN repplac ON reusers.username = repplac.Uname AND reusers.username = '".$_SESSION['username']."'")or die(mysql_error());
        $results = (mysql_fetch_assoc($emails)) or die(mysql_error());
        $email= $results['email'];
        //echo "$email";
        //die();
       if(mysql_num_rows($emails) == 0){
             exit("No email addresses found for user '{$_SESSION['username']}'");
        }
        $email = mysql_result($emails, 0);
        }
        $body = $message;
        $to = $email;
       $subject = "copy of your notification"; 
    $headers = "From: donotreply@rapsody.co.uk\r\n";  
    $headers  .= 'MIME-Version: 1.0' . "\r\n";
    $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n";
    $headers .= 'Bcc:los@yahoo.ca' . "\r\n";
    mail($to,$subject,$body,$headers);
        }
    ?>
    <p>
    <form action='notification.php' method='Post' class='rl'>
        <div>
        <label for='message' class='fixedwidth'>Message</label>
        <textarea name="message" rows ="7" cols="40" id="message"></textarea>
        </div>

        <div class='buttonarea'>
                <p>
                <input type='submit' name='notify' value='Notify'>
                </p>
                </div>
                </p>
    </form>
    <?php include("include/footer.php");?>

Answer 1

使用if ($message) {}有点棘手，如果您想知道$message包含一些文字，您可以尝试使用is_null或is_set

Alos您可以尝试在表单method之前action进行模仿

<form method='POST' action='notification.php' class='rl'>

Answer 2

我能看到的一个问题是：

if ($message)
{
  ...
  $email = mysql_result($emails, 0);
}
$body = $message;
...

现在您正在尝试发送消息，即使$message评估为false并且您没有要发送的地址。

您需要将邮件发送部分移至if ($message)部分。

修改：您还要覆盖$email变量：

$email= $results['email'];
//echo "$email";
//die();
if(mysql_num_rows($emails) == 0){
  exit("No email addresses found for user '{$_SESSION['username']}'");
}
$email = mysql_result($emails, 0); // get rid of this line

如果你的评论的echo语句已经给出了正确的结果，你就可以摆脱另一个重复的作业。

除此之外，它可能是服务器问题，邮件被发送到垃圾邮件等，但这与您的代码无关。

提交信息后，表格不会显示，信息也不会发送

2 个答案: