我已经按照发送联系表单的简单教程,但似乎无法正常工作。请有人帮忙。
以下表格:
<table width="400" border="0" cellspacing="1" cellpadding="0" align="center">
<tbody>
<tr>
<td>
<form action="send.php" method="post" name="form1">
<table width="100%" border="0" cellspacing="1" cellpadding="3">
<tbody>
<tr>
<td width="16%">Name</td>
<td width="2%">:</td>
<td width="82%"><input id="Name" type="text" name="Name" size="50" /></td>
</tr>
<tr>
<td>Email</td>
<td>:</td>
<td><input id="customer_mail" type="text" name="customer_mail" size="50" /></td>
</tr>
<tr>
<td>Subject</td>
<td>:</td>
<td><input id="Subject" type="text" name="Subject" size="50" /></td>
</tr>
<tr>
<td>Detail</td>
<td>:</td>
<td><textarea id="detail" cols="50" name="detail" rows="4"></textarea></td>
</tr>
<tr>
<td></td>
<td></td>
<td><input type="submit" name="Submit" value="Submit" /> <input type="reset" name="Submit2" value="Reset" /></td>
</tr>
</tbody>
</table>
</form>
</td>
</tr>
</tbody>
</table>
这是我的PHP:
<?php
$to ='kirsty.harris1985@gmail.com';
$header="from: $name <$mail_from>";
$mail_from="$customer_mail";
$Subject="$Subject";
$detail="$detail";
$send_contact=mail($to,$header,$Subject,$detail);
if($send_contact){
echo "We've recived your contact information";
} else {
echo "ERROR";
}
?>
这是错误:
服务器错误 检索http://nqmedia.co.uk/send_contact.php时网站遇到错误。它可能已关闭以进行维护或配置不正确。 以下是一些建议: 稍后重新载入此网页。 HTTP错误500(内部服务器错误):服务器尝试完成请求时遇到意外情况。
此外,该网站还有www.nqmedia.co.uk供人们查看。
答案 0 :(得分:0)
将您的PHP更改为以下内容:
<?php
$to ='kirsty.harris1985@gmail.com';
$name = $_POST['name'];
$customer_mail = $_POST['customer_mail'];
$Subject = $_POST['Subject'];
$detail = $_POST['detail'];
$header="From: $name <$customer_mail>";
$send_contact=mail($to,$Subject,$detail,$header);
if($send_contact){
echo "We've recived your contact information";
} else {
echo "ERROR";
}
?>
没有正确定义变量,您需要使用$_POST
来检索表单中的值。
$mail_from
(现在$customer_mail
)在定义之前就已使用过。
mail()
函数中使用的参数使用顺序错误。查看http://php.net/manual/en/function.mail.php
检查表单上的action
属性。它说“send.php”,但错误显示“send_contact.php”
答案 1 :(得分:0)
鉴于您的输入已相应命名 试试这个:
$name = $_POST['Name'];
$mail_from = $_POST['customer_mail'];
$header = "from: $name <$mail_from>";
$Subject = $_POST['Subject'];
$detail = $_POST['detail'];
答案 2 :(得分:0)
你永远不会抓住所有的帖子值。
将此代码设为send.php:
<?php
$name = $_POST['Name'];
$mail_from = $_POST['customer_mail'];
$subject = $_POST['Name'];
$body = $_POST['detail'];
$to ='kirsty.harris1985@gmail.com';
$header="from: $name <$mail_from>";
$send_contact=mail($to,$Subject,$body,$header);
if($send_contact){
echo "We've received your contact information";
}
else {
echo "ERROR";
}
?>
另外,你拼错了#34;收到&#34;你的回音,所以我修正了拼写错误。