【发布时间】:2016-05-25 10:58:05
【问题描述】:
我有一个 PHP 联系表单,我正在尝试对其实施 Ajax 以使其更加用户友好,因为联系表单是使用 javascript/jQuery 的弹出式表单,因此每次提交表单时都会刷新页面,您必须返回表单才能查看验证消息。
你能检查一下我在这里遗漏了什么吗?我已经用了一段时间了,但它仍然不起作用。如果您希望我添加 PHP 脚本,请告诉我。
Ajax 代码:
jQuery(document).ready(function ($) {
$("#ajax-contact-form").submit(function () {
var str = $(this).serialize();
$.ajax({
type: "POST",
url: "demoform.php",
data: str,
success: function (msg) {
if (msg == 'OK') {
result = '<div class="notification_ok">Your message has been sent. Thank you!</div>';
$("#fields").hide();
} else {
result = msg;
}
$("#note").html(result);
}
});
return false;
});
});
HTML 表单
<center>
<p>Please complete the form below <br />to request a <br/> <span id="free_demo">FREE DEMO</span></p>
<div id="note"></div>
<div id="fields">
<form id="ajax-contact-form" enctype="multipart/form-data">
<div><input type="text" placeholder="Name" name="fullname" id="vpbfullname" value="<?php echo strip_tags($_POST["fullname"]); ?>" class="vpb_input_fields"> <?php echo $submission_status1; ?></div>
<div><input type="text" placeholder="E-Mail" name="email" id="email" value="<?php echo strip_tags($_POST["email"]); ?>" class="vpb_input_fields"><?php echo $submission_status2; ?></div>
<div><input type="text" placeholder="Company" name="company" id="company" value="<?php echo strip_tags($_POST["company"]); ?>" class="vpb_input_fields"><?php echo $submission_status3; ?></div>
<div><input type="text" placeholder="Telephone" name="phone" id="telephone" value="<?php echo strip_tags($_POST["phone"]); ?>" class="vpb_input_fields"><?php echo $submission_status4; ?></div>
<div class="vpb_captcha_wrappers"><input type="text" placeholder="Security Code" id="vpb_captcha_code" name="vpb_captcha_code" class="vpb_input_fields"> </div>
<br/><br/><br/>
<a href="javascript:void(0);" id="refresh" onClick="vpb_refresh_aptcha();">Refresh Security Code</a></font>
<br/><br/>
<div class="vpb_captcha_wrapper"><img src="vasplusCaptcha.php?rand=<?php echo rand(); ?>" id='captchaimg' ></div><br clear="all"><br/>
<br clear="all">
<div><?php echo $submission_status; ?></div><!-- Display success or error messages -->
<div> </div>
<input type="hidden" id="submitted" name="submitted" value="1">
<input type="submit" class="vpb_general_button" value="Submit"></form> </div>
<br clear="all">
</center>
任何帮助都会很棒..谢谢
【问题讨论】:
标签: php jquery ajax forms validation