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Form submitting when clicked

I am having an issue with this login system, when ever I click the log in button, or the sign up button it re-directs me to a white page with writing on it, That being said it is interfering with my log in action.
Here is the code that I think is causing the issue,
<form method="POST" action="" accept-charset="UTF-8">
on line 16 of the HTML code, I tried to take that code out and it stopped the re-directing but the text boxes went out of place, and the white background/background-box was not there either,
Link, HERE

You want to use preventDefault() if this is a purely Javascript: you should be able to pass the button press event into the listener when you create it:
$('.login').on('click', function(e) {
e.preventDefault();
// Will be executed on press
}
<form method="POST" class="login" accept-charset="UTF-8">
If there's no JS involved in this scenario, then you want to get rid of the action parameter entirely – leaving it as the empty string will still cause it to redirect in some cases.

As Jonathan Lonowski explained above, when the log in / sign up button is clicked, the form will post the data to the page mentioned in the action= attribute. Since this attribute is empty in your form tags, it will re-load the same page, posting the data to itself.
The data will arrive in key=value variable pairs. The variable value will be the contents of the field, the variable name will be the value of the name="" attribute on the element.
For e.g., for this field:
<input id="fname" name="first" value="Bobby" />
The data will be received like this:
$fn = $_POST['first']; //value is Bobby, or whatever user enters
On your page containing the form, add a section at the top like this:
<?php
if (isset($_POST['fname']) == true){
$fn = $_POST['fname'];
echo "Received First Name: " . $fn;
die();
}else{
?>
//Your current page, in its entirety, goes here
<?php
} //close the PHP if statement
?>
That is how you deal with a normal HTML <form> construct.
However, if you wish to use AJAX to communicate with a PHP file without changing the page, then:
(1) There is no need to use a <form> construct, just use a DIV with an input button: <input type="button" id="sub_btn" value="Submit" />
(2) Trap the button press using standard js/jQuery:
$('sub_btn').click(function(){
var fn = $('#first').val();
//etc.
$.ajax(function(){
type: 'post',
url: 'my_php_processing_file.php',
data: 'fname=' +fn+ '&lname=' etc
});
});
In your PHP processor file, the data will be received thus:
<?php
$fn = $_POST['fname'];
$ln = $_POST['lname'];
//Do your MySQL lookup here...
echo 'Received ' .$fn. ' ' .$ln;
(3) IF you do use the form construct, you can still do everything as above, but you will need to suppress the default form action of navigating to the page specified in the action= attribute (an attribute setting of action="" will post data to and reload the same page you are on).
To suppress navigating to the page specified in action= (involves page refresh, even if just action=""), use event.preventDefault(), as follows:
$('#buttonID').click(function(event){
event.preventDefault();
//remainder of button click code goes here
});

Getting Javascript/jQuery and PHP To work together

UPDATED:
Okay, Thanks to OneSneakyMofo's Help below, I have managed to use ajax to call a submit.php form and have it return for example an echo statement. My problem is that none of my $post values are being carried over, for example if my start my php script with if (isset($_POST['pizzacrustformid'])) { the javascript will return blank, also when I do a var_dump($_POST);, Nothing is being saved into it which means the data is not being carried over, the php script is just being called. Please let me know if there is something I need to do in order to get the POST information to get carried over from the form as it would with a
< Submit > Button traditionally.
I Have Updated my code on Github to reflect my progress. https://github.com/dhierholzer/Basiconlineordering Thanks Again!
ORIGINAL POST:
I am new to using jquery and having forms be submitted without loading a newpage /refreshing the page.
In my Code I have multiple forms on one page that display one at a time via fade in and out effects by hitting the next button.
My problem is now that I do this, I cannot seem to get a PHP script to activate when hitting the next button to save those form options into sessions.
So here is an example:
<!--First Pizza Form, Pick Pizza Crust Type-->
<div id="pizzacrust">
<form method="post" name="pizzacrustform" id="pizzacrustformid">
<div id="main">
<div class="example">
<div>
<input id="freshpizza" type="radio" name="pizzacrust" value="1" checked="checked"><label style="color:black" for="freshpizza"><span><span></span></span>Fresh Dough</label>
</div>
<div>
<input id="originalpizza" type="radio" name="pizzacrust" value="2"><label style="color:black" for="originalpizza"><span><span></span></span>Original</label>
</div>
<div>
<input id="panpizza" type="radio" name="pizzacrust" value="3"><label style="color:black" for="panpizza"><span><span></span></span>Deep Dish Pan</label>
</div>
</div>
</div>
</form>
</div>
<div><button href="#" id="btn">Show Pizza Size</button></div>
So this Is my First Form, One thing to pay attention to is that instead of a < Submit > button, I am using a normal button and using javascript to do the submitting part.
Here is that Javascript:
<!--Controls All Button Fades-->
$('#btn').click(function(e){
$('#pizzacrust, #btn').fadeOut('slow', function(){
$('#pizzasize, #btn2').fadeIn('slow');
$('#pizzacrustformid').submit();
});
});
and Then:
$(document).ready(function () {
$('#pizzacrustformid').on('submit', function(e) {
e.preventDefault();
});
});
Now Traditionally being a php programmer, I just had a button in my form and then my php activated by having something like:
if (isset($_POST['submitted'])) { //MY Code To save values into sessions}
I cant seem To Get a function like that working when the form is submitted via a javascript function as I have it.
Here is my full code in my GitHub which may make it easier to see more so how these forms are working together right now.
https://github.com/dhierholzer/Basiconlineordering
Please Let me know any solutions that might be possible
Thanks again.

Edit:
OP, it looks like you are wanting to do AJAX, but you don't have anywhere to submit your AJAX to. Firstly, you will need to create a file that accepts the form.
Let's call it submit.php.
With that in place, you can start working on the AJAX call. To begin, you will need to separate your code from index.php.
Take this out of index.php and put it in submit.php:
if (isset($_POST['pizzacrustformid'])) {
// use a foreach loop to read and display array elements
echo '<p>hello!<p>';
}
In your Javascript, you will need to do something like the following:
$('#btn').click(function(e){
$.ajax({
method: "POST",
url: "some.php",
data: $('#pizzacrustformid').serializeArray()
})
.done(function(data) {
alert(data); //should be "Hello world"
$('#pizzacrust, #btn').fadeOut('slow', function(){
$('#pizzasize, #btn2').fadeIn('slow');
});
})
.fail(function() {
alert( "error" );
})
.always(function() {
alert( "complete" );
});
});
What is happening here is is on submit, your form data will pass over to the submit.php page, and it will generate the PHP code. That code will hit the done function (if it's successful), call an alert, then fade out to the next section.
That should get you on the right path. I would create another branch and strip out all of the forms and work on getting this done before continuing.
Also, I would set this all up in one single form and show the first section, do some validation, and then move on to the next section before finally submitting eveyrthing you need.
Hope this helps.

I recommend you do requests via ajax, here a tutorial and examples:
http://www.w3schools.com/jquery/jquery_ajax_get_post.asp
delete all jquery functions about submit
create a file called blu.php with the php code
add the jquery code in index.php
with this you only do once request at the end. I hope this helps you.
<?php echo 'tus datos son: ';
echo ' '.$_POST["data1"];
echo ' '.$_POST["data2"];
echo ' '.$_POST["data3"]; ?>
<script>
$(document).ready(function(){
$("#btn5").click(function(){
var pizzacrust= $('input[name="pizzacrust"]:checked').val();
var pizzasize= $('input[name="pizzasize"]:checked').val();
var pizzatoppings= $('input[name="pizzatoppings"]:checked').val();
$.post("blu.php",
{
data1: pizzacrust,
data2: pizzasize,
data3: pizzatoppings
},
function(data,status){
alert("Data: " + data);
});
});
});
</script>

I think you need to using click() func call ajax, dont use on() submit. Submit action makes current page will refresh. I will review your code later, but you should to try this solution above.

Temporarily changing HTML structure on the same page in PHP when submit button is clicked?

I have PHP page that have submit button to another URL.
I want to reload the current page after the submit button clicked, and add div to the HTML.
My page url is: /foo.php, and in the HTML I have:
<button onclick="$.get('/bar', function() { ... })">Submit</button>
As you can see the form sends request to /bar page.
I want to reload the /foo.php (the current page), and change the HTML to:
<button onclick="$.get('/bar', function() { ... })">Submit</button>
<div>Thank you!</div>
My problem is how can I know that the user click on the button and the refresh was because the click, and not because just navigating.
Another thing, if it possible, I want that the new div will disappear if the user refresh the page again.

Why don't you just append the div in the success callback of the get function? You wouldn't have to reload the page.
<div id="btn_area">
<button onclick="$.get('/bar', function() { $('#btn_area').append($('<div>').html('Thank You');)})">Submit</button>
</div>
By the way, i hardly recommend to separate the javascript from the html and not put it directli in the DOM.
Another Method would be, to fire an additional form with a hidden parameter to the same side. After that, you check on the serverside the hidden parameter and display the div.
A third method is, to set a cookie in the Callback, reload the side, check the cookie, display the div and remove the cookie again.
In my opinion, the first mentioned option (add the div directly in the callback without reloading) would be by far the 'prettiest', but of course i don't know what else is going on on your site

Alternatively, you could simulate a flash session (one time use session) if you opt to do this in PHP. Consider this example:
foo.php
<?php session_start(); ?>
<form method="POST" action="bar.php">
<button type="submit" name="thank_you">Submit</button>
</form>
<?php if(isset($_SESSION['thank_you'])): ?>
<?php unset($_SESSION['thank_you']); ?>
<h1>Thank You!</h1>
<?php endif; ?>
bar.php
<?php
session_start();
if(isset($_POST['thank_you'])) {
$_SESSION['thank_you'] = true;
// processes
header('Location: foo.php');
}
?>

Demo
You can handle that in js side. Just make your request, and in callback, you can manipulate dom. You can see below;
<button>Submit</button>
$("button").on("click", function() {
var $button = $(this);
$.get("/echo/html", function() {
$button.after("<div>Thank you!</div>");
});
});

Link to a page and trigger ajax via the same button

I'm trying to make a div button which triggers ajax and links to a page which the ajax is posting data to. I have the ajax working on click, but can't seem to get the link to work at all with it, all the ways I've tried to get the link working do nothing.
I have a div button:
<div class="cman shadow1" id="search_button" value="carSearch"/>
Start Search
</div>
Which triggers:
function carSearch()
{
$.ajax({
type: "POST",
url: 'searchpost.php',
data:
{
mpg : $('.mpg').val()
},
success: function(data)
{
alert("success! "+$('.mpg').val());
}
});
}
Edit:
Feel I should mention that the page I wish to link to is atm:
<?php
session_start();
$conn = mysql_connect('localhost', 'root', '');
mysql_select_db('cdb', $conn);
if(isset($_POST['mpg']))
{
$mpg = (int)($_POST["mpg"]);
header("Location: home.php");
exit;
}
?>
I wish to use this to gather all of the search fields the user enters to then load the search results page. The above code for it is just for testing right now.
I mention this as user #Doge suggested doing document.location.href = "searchpost.php" in the success handler of the ajax, this did cause the page to change but the page didn't run the isset and the redirect, so the js variable wasn't posted, or it was but this instance of the page didn't get it.
The isset and redirect do work as using the XHR Network tab on my page the page it is to redirect to appears.

If you must do this via js then create a hidden form, populate it and submit it:
html:
<form method="post" id="hiddenform" action="searchpost.php">
<input type="hidden" name="mpg" id="mpg"/>
</form>
js:
function carSearch()
{
$('#mpg').val(yourjsvariable);
$('#hiddenform').submit();
}

jQuery Auto-submit form inside of a <div> loads in new page

I'm developing a project of "prettifying" of a part of an existing web application. There is a need of putting the existing code: search criteria form in one div, and search results in another div (forming a kind of tabs, but that's another story).
Using jQuery I was able to manage that, but right now I am struggling with the results page, which by itself is yet another form that auto-submits to another file (using document.form.submit()), which is the final search results view. This auto-submit causes that the final view quits the destination div and loads as a new page (not new window).
So, the flow is like that:
First file, let's call it "criteria.html" loads the search criteria form (inside of a div) + another div (empty) destined to be filled with search results.:
<div id="criteria">... form with search criteria here...</div>
<div id="results"></div>
On submit, using jQuery's "hide()" method, I hide the first div (surrounding the search criteria form), and make Ajax call to the second file, let's call it "results.php":
<script>
$("#criteria").hide();
$.ajax({
...,
url: "results.php",
success: function(data){
$("#results").html(data);
},
...
});
</script>
results.php searches according to given criteria, and displays an "intermediary form" (which returns as a data result of the ajax query above) with a lot of hidden fields, and at the end executes:
<script>document.form.submit();</script>
which submits to another file, let's call it "resultsView.php"
This line causes that a result page shows outside the div "results", as a new page.
As there is a lot of logic in those files (more than 700 lines each), the idea of rewriting this monster just gives me creeps.
And now the question: is this a normal behavior (opening the result outside div)?
I tried removing the document.form.submit() code and everything works fine (well, without showing the results from "resultsView.php"). It's this line that causes the viewport to reset. I also tried with empty pages (to eliminate the possibility of the interaction with contents of the pages) - still the same result.
I hope there is not too much text and the problem is clearly stated. Every suggestion of how to fix this will be greatly appreciated.

If I understand your question correctly, you need to process the final submit using ajax instead of <script>document.form.submit();</script> so that you can handle the results on-page. Traditional form submits will always reload/open the action page. If you want to avoid that you'll have to control the form submit response via ajax and handle the results accordingly... like you are doing with the first submit.
The only alternative I can think of is to make div id="results" an iframe instead, so that it contains the subsequent form submit. Of course, that unleashes further restrictions that may cause other troubles.

I am not sure if I understood your question, but maybe u can do something like this.
This is my JQuery script: [I just wait for the submission search. When it happens, I use the $.Post method to call a function that should return the Html results (You can include somenthing to hide any field you want using JQuery or css)].
<script type="text/javascript" src="http://code.jquery.com/jquery-1.3.2.js"></script>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.1.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("form#searchForm").submit(function() {
var theCity = $("select#chooseCity").val();
var theName = $("input#searchInput").val();
$.post("callProvideSearchResults.php", {theCity: theCity, theName: theName}, function(data) {
$("div#searchResults").html(data);
});
return false
});
});
</script>
This is my Body: {it consists of the choice of a city, the a form to provide the name of the person you are lookng for and the place to provide the results.
<body>
<FORM id="searchForm">
<h2>Select the city: </h2>
<select id="chooseCity">
<?php
$theCitiesOptionsHTML = "cityOptions.html";
require($thePathDataFiles.$theCitiesOptionsHTML); / A large list of cities
?>
</select>
<h2> What is the name of the person </h2>
<P> <INPUT id="searchInput" TYPE="TEXT" SIZE=50></P>
<P><INPUT TYPE="submit" VALUE="search"></P>
</FORM>
<div id="searchResults">
<!-- Here: Search results -->
</div>
</body>
// Function callProvideSearchResults.php // Just call the function that makes all the job and echo the $Html page
<?php
include "provideSearchResults.php";
$theName=$_POST['theName'];
$theCity=$_POST['theCity'];
echo provideSearchResults($theName, $theCity);
?>
// provideSearchResults.php // Database connection and search
<?php
function provideSearchResults($theName, $theCity) {
include "databaseConnection.php";
//database Queries
// Generate $theHtml using strings or ob_start, for instance
return $theHtml;
}
?>