我有一个简报注册表单,我正在爵士乐上。下面我有这个jQuery脚本:fadeOut,removeClass在addClass后不工作
$(document).ready(function(){
$('#newsletter-signup').submit(function(){
//setup variables
var form = $(this),
formData = form.serialize(),
formUrl = form.attr('action'),
formMethod = form.attr('method'),
responseMsg = $('#signup-response');
//show response message - waiting
responseMsg.hide()
.addClass('response-waiting')
.text('Please Wait...')
.fadeIn(200);
//send data to server for validation
$.ajax({
url: formUrl,
type: formMethod,
data: formData,
success:function(data){
//setup variables
var responseData = jQuery.parseJSON(data),
klass = '';
//response conditional
switch(responseData.status){
case 'error':
klass = 'response-error';
break;
case 'success':
klass = 'response-success';
break;
}
//show reponse message
responseMsg.fadeOut(200,function(){
$(this).removeClass('response-waiting')
.addClass(klass)
.text(responseData.message)
.fadeIn(200,function(){
//set timeout to hide response message
setTimeout(function(){
responseMsg.fadeOut(200,function(){
$(this).removeClass(klass);
});
},3000);
});
});
}
});
//prevent form from submitting
return false;
});
});
,倘若你有兴趣,这里是我的PHP的片段(除去数据库凭据):
<?php
if(isset($_GET['action'])&& $_GET['action'] == "signup"){
mysql_connect(***REMOVEDforprivacy*******);
mysql_select_db(***REMOVEDforprivacy*******);
//sanitize data
$email = mysql_real_escape_string($_POST['signup-email']);
$to ='***REMOVEDforprivacy*******';
$subject = '***REMOVEDforprivacy*******';
$body = "The email address ". $email. " has been added to the email database.";
$headers = 'From: ***REMOVEDforprivacy*******' . "\r\n" .
'Reply-To: ***REMOVEDforprivacy*******' . "\r\n" .
'X-Mailer: PHP/' . phpversion();
//validate email address - check if input was empty
if(empty($email)){
$status = "error";
$message = "You did not enter an email address!";
}
else if(!filter_var($email, FILTER_VALIDATE_EMAIL)){ //validate email address - check if is a valid email address
$status = "error";
$message = "You have entered an invalid email address!";
}
else {
$existingSignup = mysql_query("SELECT * FROM signups WHERE email_address='$email'");
if(mysql_num_rows($existingSignup) < 1){
$insertSignup = mysql_query("INSERT INTO signups (email_address) VALUES ('$email')");
if($insertSignup){
$status = "success";
$message = "You have been signed up!";
mail($to, $subject, $body, $headers);
}
else {
$status = "error";
$message = "Oops, There has been a technical error!";
}
}
else {
$status = "error";
$message = "Looks like you have already registered this email address with us. Thank you for your support!";
}
}
//return json response
$data = array(
'status' => $status,
'message' => $message
);
echo json_encode($data);
exit;
}
?>
一切工作顺利了这里:
//show reponse message
responseMsg.fadeOut(200,function(){
$(this).removeClass('response-waiting')
.addClass(klass)
.text(responseData.message)
.fadeIn(200,function(){
//set timeout to hide response message
setTimeout(function(){
responseMsg.fadeOut(200,function(){
$(this).removeClass(klass);
});
},3000);
});
});
我不明白这里有什么问题。语法似乎没问题,但是,当我尝试删除大部分内容并仅使用responseMsg.fadeOut(200);
时,responseMsg不会褪色(因此您可以忘记删除类并添加新类)。我已经用Firebug检查过了,POST响应确实显示正确的错误并且成功消息正在返回......(并且实际上检查我的数据库,条目被添加并且发送了一个警告邮件)......所以除了花哨的schnazzy JQuery特效之外,所有工作都正常。我简单地以为我从页面上的另一个元素有一个JQuery冲突,所以我删除了它,并没有改变。我尝试使用$ .noconflict(),$ .noconflict(true),并用JQuery替换$的所有实例(但是,然后,页面上的其他项目无论如何都没有冲突)。我尝试删除调用以添加“响应等待”类,并在数据的JSON解析之后添加响应消息类,但没有奏效。我几乎从this awesome tutorial复制和粘贴这个脚本,所以我不知道为什么它不起作用。任何人都有线索?
UPDATE
这里FWIW是JSFiddle,但它绝对不会没有PHP DB东西工作....不知道我怎么能解决的是,球员....
更新搞定了,愚蠢的错误将我的PHP代码放在我的html标题之上......只要我被允许,就会尽快回复。
这将是不错的一个例子的jsfiddle详细说明您的问题。 – scottm 2012-07-23 20:29:45