原谅我,但我认为我的技能水平太深了:数据库连接设置在包括:
这是代码:
<html>
<body>
<?php
include "inc.php";
//print_r ($_POST);
//if "email" is filled out, send email
if (isset($_POST['email']))
{
//send email
While($row=mysql_fetch_array($querymail))
{
$to= $row['email'];
$subject = $_REQUEST['subject'] ;
$message = $_REQUEST['message'] ;
$email= 'stats@my email.com';
mail($to, $subject,
$message, "From:" . $email);
echo "email sent to ".$to;
}
}
else
//if "email" is not filled out, display the form
{
$Pnum=implode(',',$_POST['checkemail']);
$querymail=mysql_query("select sp.email_address as email from stats_player sp where sp.player_num IN ($Pnum)");
While($row=mysql_fetch_array($querymail))
{
echo "<table>";
echo "<tr><td>".$row['email']."</tr.</td>";
}
echo "<form method='post' action=''>
Subject: <input name='subject' type='text'><br>
Message:<br>
<textarea name='message' rows='15' cols='40'>
</textarea><br>
<input type='submit'>
</form>";
}
?>
</body>
</html>
当我允许时,print_r ($_POST);
我会按预期显示所有 player_nums。
该页面以正确的电子邮件列表打开(此表将消失,仅用于验证数据),我得到了表单。
当我点击表单上的提交按钮时,我得到了错误。
Warning: implode() [function.implode]: Invalid arguments passed in /home..../results/admin/emailall.php on line 31
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home..../results/admin/emailall.php on line 33
我认为,因为我有这个表格重新提交给自己,所以它正在清除我在初始查询中的内容。我以为我已经用 isset 绕过了它,但我可能(可能是)错了。
(注意我在粘贴时删除了一些空行,因此错误行将不对应)