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好的，相当直接的 php 联系表。它在 ie 中提交和发送数据，并将用户带到感谢页面。但是在 google chrome 和 firefox 中，我的用户都被带到了感谢页面，但他们的表单数据没有发送到电子邮件。
这是我正在使用的代码：
一页上的表单数据

<table width="400" border="0" align="center" cellpadding="0" cellspacing="1">
<tr>
<td><form name="form1" method="post" action="send_wl.php">
<table width="100%" border="0" cellspacing="1" cellpadding="3">
<tr>
<td width="16%">First Name</td>
<td width="2%">:</td>
<td width="82%"><input name="name" type="text" id="name" size="50"></td>
</tr>
<tr>
<tr>
<td width="16%">Last Name</td>
<td width="2%">:</td>
<td width="82%"><input name="lastname" type="text" id="lastname" size="50"></td>
</tr>
<tr>
<td>Address</td>
<td>:</td>
<td><textarea name="address" cols="50" rows="4" id="address"></textarea></td>
</tr>
<tr>
<td>Date of Birth</td>
<td>:</td>
<td><input name="dob" type="date" id="dob" size="50"></td>
</tr>
<tr>
<td>Health Care Number</td>
<td>:</td>
<td><input name="phn" type="text" id="phn" size="50"></td>
</tr>
<tr>
<td>Phone</td>
<td>:</td>
<td><input name="ac" type="text" id="ac" size="3"><input name="phone" type="text" id="phone" size="7"></td>
</tr>
<tr>
<td>Email</td>
<td>:</td>
<td><input name="email" type="text" id="email" size="50"></td>
</tr>
<tr>
<td>Comments</td>
<td>:</td>
<td><textarea name="comment" cols="50" rows="4" id="comment"></textarea></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Submit"> <input type="reset" name="Submit2" value="Reset"></td>
</tr>
</table>
</form>

另一个页面上的 php 脚本“send_wl.php”

<?php 
    $to = "ouremail@gmail.com";
    $from = $_POST['email'] ; 
    $name = $_POST['name'] ; 
    $headers = "From: $from"; 
    $subject = "New Patient Data"; 

    $fields = array(); 
    $fields["name"] = "name";
    $fields["lastname"} = "lastname"; 
    $fields["address"] = "address"; 
    $fields["email"] = "email";
    $fields["ac"] = "ac"; 
    $fields["phone"] = "phone"; 
    $fields["dob"] = "dob";
    $fields["phn"] = "phn";
    $fields["comment"] = "comment"; 

    $body = "We have received the following information:\n\n"; foreach($fields as $a => $b){    $body .= sprintf("%20s: %s\n",$b,$_REQUEST[$a]); } 

    $headers2 = "From: noreply@ourwebsite.com"; 
    $subject2 = "Thank you for contacting us"; 
    $autoreply = "Thank you for contacting us. Somebody will get back to you as soon as possible, usualy within 48 hours. If you have any more questions, please consult our website at www.ourwebsite.com";


    if($from == '') {print "You have not entered an email, please go back and try again";
    } else { 
        if($name == '') {print "You have not entered a name, please go back and try again";
        } else { 
            $send = mail($to, $subject, $body, $headers); 
            }
            if($send) 
                {header( "Location: http://www.thewebsite.com/thankyou.html" );} 
            else 
                {print "We encountered an error sending your mail, please notify mailreciepient@gmail.com"; } 
        }
    }
?>

我也对网站进行了所有建议的更改，但此表单仍然无法正常工作。

我是否也应该从 $fields["lastname"] = "lastname" 中删除 = "lastname" 我知道这似乎是一个愚蠢的问题，但老实说，我在这个问题上大发雷霆。

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2 回答 2

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PHP 不依赖于浏览器，它是服务器，所以只要运行相同的服务器，输出就会相同。

您的问题实际上是您的代码。

您对 $fields 的所有引用都必须是： $fields["dob"] 因为它是一个哈希映射。这样，在获取数据时，它会正确执行...因为您的密钥在搜索时是错误的。

if($from == '') {
  print "You have not entered an email, please go back and try again";
} elseif ($name == ''){
  print "You have not entered a name, please go back and try again";
} else { 
    $send = mail($to, $subject, $body, $headers); 
    $send2 = mail($from, $subject2, $autoreply, $headers2); 
    if($send && $send2){  //<--   check both flags?
        header( "Location: http://www.thewebsite.com/thankyou.html" );
    }else{
        print "We encountered an error sending your mail, please notify mailreciepient@gmail.com";
    }
}
于 2013-10-24T18:21:36.517 回答
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您是否尝试过使用 $_POST 而不是 $_REQUEST？任何 $fields{"fieldname"} 也应该是 $fields["comment"]

于 2013-10-24T18:17:08.920 回答