我整天试图解决问题,但没有成功。 表单不会发送消息而不会写入错误。
form.html
<div class="ok" id="ok"></div>
<div id="data">
<div id="alert" class="alert_ig"></div>
<form id="form" class=" clearfix" method="POST" action="">
<div class="col-sm-12">
<div class="form-group">
<input name="name" type="text" id="name" value="{$smarty.cookies.nameuser}" class="form-control" placeholder="Name"/>
</div>
</div>
<div class="col-sm-12">
<div class="form-group">
<input name="email" type="email" id="email" value="{$smarty.cookies.emailuser}" class="form-control" placeholder="E-mail"/>
</div>
</div>
<div class="col-sm-12">
<div class="form-group">
<textarea id="message" class="form-control" rows="5" name="text" placeholder="Message"></textarea>
</div>
</div>
<div class="col-sm-12">
<div class="form-group">
<button style="margin-top:10px" id="submit" type="submit" class="bot_g">Send</button>
</div>
</div>
</form>
<script>
$(document).ready(function() {
$('#submit').click(function(e){
var form = $(this);
var error = false;
if (!error) {
var data = form.serialize();
$.ajax({
type: 'POST',
url: '{$home}/system/modules/contacts/send.php',
dataType: 'json',
data: data,
success: function(data){
if (data.error.length > 0) {
$('#alert').html(""+data['error']+"");
$('#email').addClass("fill");
} else {
$('#ok').html('send.');
$( "#data" ).css( "display","none" );
}
},
return false;
)};
});
</script>
</div>
send.php
<?php
require_once '../../inc/core.php';
$name=Text(trim($_POST['name']));
$email=Text(trim($_POST['email']));
$subject=Text(trim($_POST['subject']));
$text=Text(trim($_POST['text']));
if(empty($name)){
$json['error'] = 'Come on, you have a name don\'t you?';
echo json_encode($json);
exit;
}
if (mb_strlen($name) < 2 || mb_strlen($name) > 250){
$json['error'] = 'Your name must consist of at least 2 characters!';
echo json_encode($json);
exit;
}
if(empty($email)){
$json['error'] = 'No Email, No Message!';
echo json_encode($json);
exit;
}
if (mb_strlen($email) < 5 || mb_strlen($email) > 64){
$json['error'] = 'Your email must consist of at least 5 characters!';
echo json_encode($json);
exit;
}
if (!filter_var($email, FILTER_VALIDATE_EMAIL)){
$json['error'] = 'Unknown characters in your e-mail!';
echo json_encode($json);
exit;
}
if(empty($subject)){
$json['error'] = 'Um...yea, you have to write something to send this form.';
echo json_encode($json);
exit;
}
if (mb_strlen($subject) < 5 || mb_strlen($subject) > 250){
$json['error'] = 'Your subject must consist of at least 5 characters!';
echo json_encode($json);
exit;
}
if (mb_strlen($text) < 2 || mb_strlen($text) > 10000){
$json['error'] = 'Thats All? Really?';
echo json_encode($json);
exit;
}
$mailer = new phpmailer();
$mailer->ContentType = "text/html";
$mailer->From = $email;
$mailer->Subject = 'New message from '.$home;
$mailer->Body ="Subject: ".$subject."<br/>
Name: ".$name."<br/>
E-mail: ".$email."<br/>
".nl2br($text);
$mailer->AddAddress($setup['emailadmin'], '');
$mailer->Send();
$json['error'] = 0;
echo json_encode($json);
?>
这个代码什么时候出现问题? 我必须使用这个send.php格式,但可能会将ajax和html表单更改为工作...
感谢
答案 0 :(得分:0)
var form = $(this);
在这一行$(this);指向提交按钮,因此var表单存储您的提交按钮而不是表单元素本身。
您可能在客户端看不到错误,但如果您尝试调试发送到服务器端的数据,您将看到错误。
尝试使用与表单元素匹配的选择器替换$(this)。
此外,如果您想在服务器返回错误(例如500错误)时通知您的用户,您可以向$ ajax函数添加错误回调以及您已有的成功回调。
请在此处查看http://api.jquery.com/jquery.ajax/文档。
还要在事件处理程序的末尾添加e.preventDefault(),以防止在您想要执行ajax提交时提交表单的默认操作。