php联系表单脚本问题?

时间:2013-08-20 10:30:42

标签: php javascript jquery html

我希望你能帮忙!

继承人的问题,我有这个php联系表单脚本(如下所示),我之前使用过,所以我知道工作,但是,由于我已经将html表单隐藏在一个新的jQuery驱动的div中:{{ 1}} 上下切换,我似乎无法发送表格,发送时的表格应该淡出然后给用户一个确认消息(如脚本中所示)但它没有做任何事情,我不知道假设有人会知道这里出了什么问题,所以我可以让这个表格再次运作?谢谢!

(<div class="toggle hidemail" id="drop-button" href=""> Email us here </div> <div class="hidden hiddenform">)

在head标签中:

    <?php

$to = 'example@gmail.com';
$subject = 'Company Name';

$name = $_POST['name'];
$email = $_POST['email'];
$message = $_POST['message'];

$body = <<<EMAIL

Hello my name is $name.

$message

From $name
my email address is $email


EMAIL;

if($name == '' || $email == '' || $message == "") {
    echo "<h5>Sorry, I think you missed  bit....</h5>";
    $error = true;
} else {
    $error = false;
    mail("example@gmail.com", $name, $email, $message);
    echo "<h5>Thank you for getting in touch. I'll get back to you ASAP.</h5>";
}


if($error == true):
?>

<article id="contact-form">

    <article id="load_area">

        <form id="my-form">

            <input type="text" name="name" placeholder="Name" />
            <input type="text" name="email" placeholder="Email" />
            <textarea name="message" placeholder="Message" /></textarea>
            <input type="button" class="submit_button" value="Send" name="send_button" />

        </form>

    </article>

</article>

<?php

endif;

?>

这是HTML:

$(document).ready(function(){
        $(".submit_button").live("click", function(){
            $("#load_area").fadeOut('slow', function(){
            $("#load_area").fadeIn();
            $.post("process.php", $("#my-form").serialize(), function(data){
                $("#load_area").html(data);
            })
            });
        })
    })

谢谢!

0 个答案:

没有答案