用户点击提交后提供反馈

时间:2013-04-22 02:35:31

标签: javascript html

用户点击“提交”后,如何获得弹出框或简单的“谢谢”?最简单的方法:

<div id="suggestion">
    <div class="control-group">
        <div class="controls">
            <input type="submit" value="Send" class="btn btn-info" />

这是PHP:

<?php
    // Email of sender
    $videogamename = $_POST['namevideogame']; 

    // From (Sender Name)
    $artistname = $_POST['nameofartist'];

    $header = "from: Game AMV";

    // Message
    $message = "Name of Video Game: ".$videogamename."\r\n\nName of Artist/Song: ".$artistname."\r\n\nLink: ".$_POST['link'];

    // Enter your email address
    $to ='something@gmail.com';

    $send_contact = mail($to, 'New Song Suggestion from gameamv.com', $message, $header);

    // Check, if message sent to your email 
    // display message "We've recived your information"
    if($send_contact){
        header("location: http://gameamv.com/");
    } else {
        echo "Error. The form was not submitted. Please try again.";
    }
?>

2 个答案:

答案 0 :(得分:1)

简单 - 为您的<form>元素提供可识别的id值。

然后您将按如下方式挂钩onsubmit事件:

document.getElementById('myThankfulForm').addEventListener('submit', function () {
    alert("Thankful Form is Thanking you!");
}, false);

这是一个演示jsFiddle

答案 1 :(得分:0)

如果您希望在发送电子邮件后弹出(在页面重新加载后),我会在重定向到目标网页之前在会话值中添加一条消息。然后在目标网页上,检查是否设置了会话值,是否有登录页面回显一些Javascript以提醒消息,然后将会话值设置为空。

当然我假设这是一个应用程序。