2017-08-17 184 views
0

试图结合下拉字段到身体这里发送的电子邮件是我的代码。只返回1下拉。想要将消息和消息1从下拉列表中合并到$ comment字段尝试了多种在线解决方案,但似乎找不到可行的解决方案。请帮助PHP电子邮件结合字段发送电子邮件

$admin_email = "[email protected]"; 
    $email = $_REQUEST['email']; 
    $subject = "Beverage Request"; 
    $comment = $_REQUEST['message']); 


    //send email 
    mail($admin_email, "$subject", $comment, "From:" . $email); 

    //Email response 
    echo "Thank you for contacting us!"; 
    } 

    //if "email" variable is not filled out, display the form 
    else { 
?> 

<form method= "post" action= "<?php echo $_SERVER [ 'PHP_SELF' ] ;?>" /> 
    <table> 
    <tr> 
     <td>Email: <input name="email" type="text" /><br /> 
     </td> 
    </tr> 
    <tr> 
     <td>Beverage Choice </td> 
     <td><select id="message1" Name="message1"> 
       <option value="Beer">Beer</option> 
       <option value="Wine">Wine</option> 
       <option value="Cidar">Cidar</option> 
      </select> 
     </td> 

     <td>Size of bottle </td> 
     <td><select id="message" Name="message"> 
       <option value="Sample">Sample</option> 
       <option value="12 oz">12 oz</option> 
       <option value="22 oz">22 oz</option> 
       <option value="32 oz">32 oz</option> 
      </select> 
     </td> 


    </tr> 
    <tr> 
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在我看来,因为这篇文章会回答你的问题[html-php-form-input-as-array](https://stackoverflow.com/questions/20184670/html-php-form-input-as-array ) – chilly

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你可以阅读$ _POST ['message1'],就像你做'消息'对应。顺便说一句:你应该检查$ _POST ['email']是否真的包含一个邮件地址(并且不会超过)。否则,有人可以接管你的邮件表格并发送他自己的邮件,使用你的服务器发送垃圾邮件 –

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这里出现语法错误 –

回答

0
$admin_email = "[email protected]"; 

$email = $_REQUEST['email']; 

$subject = "Beverage Request"; 

$comment = $_REQUEST['message']." ".$_REQUEST['message1']; 

您需要连接信息和MESSAGE1价值在$注释变量来存储。

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完美这正是我所期待的。非常感谢和快速回复。我不是一个PHP编码器,所以我迷路了。 :) –