【发布时间】:2016-08-05 12:27:08
【问题描述】:
每次我提交表单时,预期的重置功能都不会清除我的表单。请问我应该如何清除我的表单我的代码(HTML、PHP、JS)如下。
HTML 代码:
<form id="main-contact-form" class="contact-form" name="contact-form" method="post" action="sendemail.php">
<div class="col-sm-5 col-sm-offset-1">
<div class="form-group">
<label>Name *</label>
<input type="text" name="name" class="form-control" required="required">
</div>
<div class="form-group">
<label>Email *</label>
<input type="email" name="email" class="form-control" required="required">
</div>
<div class="form-group">
<label>Phone</label>
<input type="number" class="form-control">
</div>
<div class="form-group">
<label>Company Name</label>
<input type="text" class="form-control">
</div>
</div>
<div class="col-sm-5">
<div class="form-group">
<label>Subject *</label>
<input type="text" name="subject" class="form-control" required="required">
</div>
<div class="form-group">
<label>Message *</label>
<textarea name="message" id="message" required="required" class="form-control" rows="8"></textarea>
</div>
<div class="form-group">
<button type="submit" name="submit" class="btn btn-primary btn-lg" required="required">Submit Message</button>
</div>
</div>
Javascript 代码如下:
jQuery(function($) {'use strict',
......
......
var form = $('#main-contact-form');
form.submit(function(event){
event.preventDefault();
// Serialize the form data.
var formData = $(form).serialize();
var form_status = $('<div class="form_status"></div>');
$.ajax({
type: 'POST',
url: $(this).attr('action'),
data: formData,
beforeSend: function(){
form.prepend( form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Email is sending...</p>').fadeIn() );
}
}).done(function(data){
$("#main-contact-form")[0].reset();
form_status.html('<p class="text-success">' + data.message + '</p>').delay(3000).fadeOut();
});
});
......
......
});
最后是我的 PHP 代码:
<?php
$status = array(
'type'=>'success',
'message'=>'Thank you for contacting us. As soon as possible we will contact you!'
);
$fail = array(
'type'=>'fail',
'message'=>'Please try again. Your mail could not be delivered!'
);
$name = @trim(stripslashes($_POST['name']));
$email = @trim(stripslashes($_POST['email']));
$subject = @trim(stripslashes($_POST['subject']));
$message = @trim(stripslashes($_POST['message']));
$email_from = $email;
$email_to = 'email@mail.com';//replace with your email
$body = 'Name: ' . $name . "\n\n" . 'Email: ' . $email . "\n\n" . 'Subject: ' . $subject . "\n\n" . 'Message: ' . $message;
$success = @mail($email_to, $subject, $body, 'From: <'.$email_from.'>');
header('Content-type: application/json');
if($success){
echo json_encode($status);
}
else{
echo json_encode($fail);
}
?>
【问题讨论】:
-
我强烈建议你停止使用
@,而不是忽略错误或可能的错误,处理它们。 -
你不是已经发布了这个/类似的吗? stackoverflow.com/q/38769155
-
我相信 form_status var 没有在“done”函数中定义......以防万一尝试在那里重新定义它......同时检查浏览器调试控制台是否有错误。
-
你的 Stealth 编辑添加下面的答案包含
$("#main-contact-form")[0].reset();是可耻的。
标签: php jquery html ajax forms