我在javascript上非常糟糕,我正在尝试进行AJAX调用,但是我从表单中获取了我的值,它只是刷新页面,甚至无法观察AJAX调用是否成功。
这是我的HTML:
<form class="form-horizontal" id="subscribe-form" name="subscribe-form" onSubmit="return subscribe();">
<fieldset>
<p>
Subscribe
</p>
<div class="input-prepend input-append">
<input name="subscribwEmail" class="span2" id="InputEmail" type="email" placeholder="Email">
<input type="submit" class="btn btn-inverse" />
</div>
</fieldset>
</form>
JS:
function subscribe() {
var emailForm = $('#subscribwEmail').val();
$('#subscribe-form').hide(); // hide email form
$('#subscribeDiv').prepend('<img id="process" src="http://www.mydomain.com/assets/img/process.gif" />')
$.ajax({
type: "POST",
url: "../../actions/ajax-subscribe.php",
data: {
email: emailForm
},
dataType: "json",
success: function (data) {
if (data[0] == 1) { // test if response was 1 or 2
$('#process').hide(); // hide img
$('#subscribeDiv').append('<p>Thank you for subscribing!</p>');
} else {
$('#process').hide(); // hide img
$('#subscribeDiv').append('<p>There was an error subscribing the email.</p>');
}
}
})
};
Ajax的subscribe.php
<?php
include ('phpfunctions.php');
$email = $_POST['email'];
$ip = $_SERVER['REMOTE_ADDR'];
$con = mysqli_connect(DB_HOST,DB_USER,DB_PASS,DB_NAME) or die('ERROR WITH SQL CONNECTION CONTACT ADMIN');
$email = cleanString($con, $email);
$query = "INSERT INTO `subscribers` (`email`, `ip`) VALUES ('$email', '$ip');";
if ($result = mysqli_query($con, $query)) {
echo json_encode(1); // all ok inserted
} else {
echo json_encode(0); // failed
}
?>
答案 0 :(得分:6)
您不会取消点击事件,因此表单会提交。
将return false;
添加到subscribe
方法的末尾。
答案 1 :(得分:0)
您必须订阅方法返回错误
此外,您可以验证电子邮件,如果无效,您可以停止发送订阅返回false
您还可以停用提交按钮。 (formObj带有subscribe参数)
formObj.submit.disabled = true;
formObj.submit.value = 'Log In...';
但如果出现问题,你必须启用它。