我有一个登录表单,当出现错误时,即"电子邮件地址未注册",该表单当前会重新加载到新页面。我需要在重新加载到新页面的同时将错误显示在同一页面上。我遵循了一个关于如何使用Ajax的教程,但它似乎不起作用,页面仍然会重新加载。ajax代码是:
$("#login_button").click(function(){
$.post($("#login_form").attr("action"), $("#login_form :input").serializeArray(), function(info){$("#login_errors").html(info);});
clearInput();
});
$("login_form").submit(function(){
return false;
});
function clearInput(){
$("#login_form: input").each(function() {
$(this).val('');
});
}
我的登录表单代码是:
<form id= "login_form" action="login.php" method="post">
<span id="login_errors"></span>
<label>Email Address</label>
<input type="text" name="email" id="email" required="required"/>
<br />
<label>Password</label>
<input type="password" name="password" id="password" required="required"/>
<br />
<div class="checkbox">
<input id="remember" type="checkbox" />
<label for="remember">Keep me signed in</label>
</div>
<div class="action_btns">
<div class="one_half last"><input type="submit" class="btn btn-blue" id="login_button" value="Login"></div>
<div class="one_half last"><a href="#" id="register_form"
class="btn">Sign up</a></div>
</div>
</form>
我将ajax脚本链接到:
<script type="text/javascript" src="login_Ajax.js"></script>
您应该防止提交按钮触发默认操作(提交)。
像这样更新你的代码:
$("#login_button").click(function(){
$.post($("#login_form").attr("action"), $("#login_form :input").serializeArray(), function(info){$("#login_errors").html(info);});
clearInput();
// Prevent the default action from occurring.
return false;
});
当您开始使用AJAX时,您必须做的第一件事是确切地知道如何通过类等元素将信息显示到div或HTML5的任何标记中。请看下一个代码,因为我会为您评论它。
//This is our DOM on JQUERY or JAVASCRIPT named as main.js
$(function() {
//First we have to listen on a event click (button) and extract the values of the inputs
$('body').on('click','.myButtonClass', function() {
$email = $('#emailID').val();
$password = $('#passwordID').val();
//We save in a object our values
$x = {
action:'register',
email:$email,
password:$password
};
//We execute ajax calling a php file for POST method
$.post('file.php',$x,function(callback) {
$('.answer').html(callback);
});
});
});
/*END OF DOM AND MAIN.JS*/
//BEGIN OUR PHP FILE WHERE WE'LL RECIEVE ALL THE DATA FROM THE AJAX FILE
<?php
//we save the action sended trough ajax
$action = $_POST['action'];
//we do a switch for determinate what value have the action variable
switch ( $action ) :
case 'register':
//we save with addslashes for security in the data base
$email = addslashes($_POST['email']);
$password = addslashes($_POST['password']);
//If your objetive is validate if the email and password were send in the form, you can do something like this
$validate = strlen($email)*strlen($password);
//This say that if email * password are > 0, the data were send. Otherwise, this operation gonna be equal zero
//and thats means that both or one of the data were not recieve in this file trough ajax.
if ( $validate > 0 ) :
echo 'You login!';
else :
echo 'You are not write all the inputs.';
endif;
break;
endswitch;
?>
/*END OF PHP FILE*/
//BEGIN OUR FORM IN HTML
<form>
<input type="email" id="emailID" placeholder="Write here your email">
<input type="password" id="passwordID" placeholder="Write here your password">
<input type="button" class="myButtonClass">
</form>
<!--This will be where we display the answer with ajax, in the element who have the class answer-->
<div class="answerswer"></div>