2017-08-17 190 views
0

試圖結合下拉字段到身體這裏發送的電子郵件是我的代碼。只返回1下拉。想要將消息和消息1從下拉列表中合併到$ comment字段嘗試了多種在線解決方案,但似乎找不到可行的解決方案。請幫助PHP電子郵件結合字段發送電子郵件

$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> 
+0

在我看來,因爲這篇文章會回答你的問題[html-php-form-input-as-array](https://stackoverflow.com/questions/20184670/html-php-form-input-as-array ) – chilly

+0

你可以閱讀$ _POST ['message1'],就像你做'消息'對應。順便說一句:你應該檢查$ _POST ['email']是否真的包含一個郵件地址(並且不會超過)。否則,有人可以接管你的郵件表格併發送他自己的郵件,使用你的服務器發送垃圾郵件 –

+0

這裏出現語法錯誤 –

回答

0
$admin_email = "[email protected]"; 

$email = $_REQUEST['email']; 

$subject = "Beverage Request"; 

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

您需要連接信息和MESSAGE1價值在$註釋變量來存儲。

+0

完美這正是我所期待的。非常感謝和快速回復。我不是一個PHP編碼器,所以我迷路了。 :) –