How to submit form without moving to next page in PHP - php

Question

The Problem
I am trying to submit a form in php but due to the nature of what i want i need the page to not go onto the next one i just want it to submit the data and refresh the current page or whatever, at current it submits the data and goes onto page 2 which i dont want i just need it to submit the data and stay on the current page, if thats possible!
The Code
//page 1 code
<center>
<h1>What Is Jacob Dailey Doing?</h1>
<form method="post" action="jacob_dailey.php">
<select name="baby_status">
<option value="playing">Playing</option>
<option value="awake">Awake</option>
<option value="sleeping">Sleeping</option>
</select>
<br />
<input type="submit" value="Submit"/>
</form>
</center>
//page 2 code
<?php
if (isset($_POST['baby_status'])) {
$baby = $_POST['baby_status'];
setcookie("baby_status", $baby, time() + 31556926, '/'); // Data will Store For 1 Year
header('Location: ' . $_SERVER['PHP_SELF']);
}
$status = $_COOKIE['baby_status'];
echo '<center> <h1>Baby Jacob Dailey Is Currently ' . ucwords($status) . '</h1>';
if ($status == "playing") {
echo '<img src="http://cdn.sheknows.com/articles/2013/02/baby-playing-with-blocks.jpg"/>';
}
elseif ($status == "awake") {
echo '<img src="http://www.westheimphoto.com/lightbox/gallery/TaiwanStockPhotos/TWNhw1221.jpg"/>';
}
elseif ($status == "sleeping") {
echo '<img src="http://www.babycare.onlymyhealth.com/imported/images/neonatal/2012/July/19_Jul_2012/6-Months-Old-ssl.jpg"/>';
}
echo '</center>';
?>
Page 2 code shouldnt be as important but i just need it so when i click submit on page 1 it updates the information on page 2 but doesnt take me to page 2.
Cheers!

Answer 1

Your form can submit onto itself. Just in the action="xyz" either leave it (the whole action=... attribute) out entirely or else name the page that also contains the form there between quotes.
Then when you load the page you check the $_POST or $_GET array (depending on the method) to see if the submit button was pushed or if someone just navigated to the page. (You'll want to give you submit button a name="foo".)

Answer 2

action="jacob_dailey.php" in your form takes you to that page, you either paste your php code to main page and replace action with just "" or you will search AJAX and learn how to it with that

Answer 3

You can use jQuery.ajax(). Example here:
http://www.formget.com/form-submission-using-ajax-php-and-javascript/
This example uses a database, but you can use a php file to return values and read them from the response in javascript. Do not put any action to the form but enable a click event handler on the submit button to enable the function.
Also my example here: http://dev.ossipesonen.fi/alkoholilaskuri/
A very simple form where you insert values, pass them onto PHP with $_POST and then calculates the right amounts and sums, and you print them in the response.

Answer 4

Solution: Update Status Without Page Reload Using XHR and Filesystem Storage
If you want someone on another computer to see the update, then you'll need to store that information on the server. You could store the information in a database, but for this small bit of information I'm using the filesystem.
page1.php
<?php
// get baby status if available
if ( is_readable('baby_status.php') ) {
include 'baby_status.php';
}
$status = ( $status )? $status: '??';
// prepare to update select list
list($pl_check, $pl_check, $pl_check) = array('', '', '');
switch ( $status ) {
case 'playing': $pl_check = ' selected '; break;
case 'awake': $aw_check = ' selected '; break;
case 'sleeping': $sl_check = ' selected '; break;
}
?>
<center>
<h1>What Is Jacob Dailey Doing?</h1>
<form id="baby_form" method="post" action="update_baby.php">
<select id="baby_status" name="baby_status">
<option value="playing" <?php echo $pl_check ?>>Playing</option>
<option value="awake" <?php echo $aw_check ?>>Awake</option>
<option value="sleeping"<?php echo $sl_check ?>>Sleeping</option>
</select><br />
<input type="submit" value="Submit"/>
</form>
See Baby Status
</center>
<script>
// XHR/PHP/Filesystem method
function update_baby () {
var baby_status = document.getElementById('baby_status');
var status=encodeURIComponent(baby_status.options[baby_status.selectedIndex].value)
var parameters = 'baby_status=' + status
// set up XHR object
var xhr = new XMLHttpRequest()
xhr.open('POST', 'update_baby.php', true)
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded')
// handle response
xhr.onload = function () {
console.log(this.responseText)
alert(this.responseText)
}
xhr.send(parameters)
}
// hook up baby status function to form submit
document.getElementById('baby_form').addEventListener('submit', function(evt){
evt.preventDefault()
update_baby()
})
</script>
page2.php
<?php
// execute baby update code and get current status
include 'update_baby.php';
echo '<center> <h1>Baby Jacob Dailey Is Currently ' . ucwords($status) . '</h1>';
if ($status == "playing") {
echo '<img src="http://cdn.sheknows.com/articles/2013/02/baby-playing-with-blocks.jpg"/>';
}
elseif ($status == "awake") {
echo '<img src="http://www.westheimphoto.com/lightbox/gallery/TaiwanStockPhotos/TWNhw1221.jpg"/>';
}
elseif ($status == "sleeping") {
echo '<img src="http://www.babycare.onlymyhealth.com/imported/images/neonatal/2012/July/19_Jul_2012/6-Months-Old-ssl.jpg"/>';
}
?>
<br>
Update Baby Status
</center>
update_baby.php
<?php
if (isset($_POST['baby_status'])) {
$status = $_POST['baby_status'];
// prepare php script text for baby status file
$status_write = <<<EOT
<?php
\$status = '$status';
?>
EOT;
// write status to baby_status.php
if ( $baby_status_file = fopen('baby_status.php', 'w') ) {
fwrite($baby_status_file, $status_write);
fclose($baby_status_file);
}
echo 'Baby status updated.';
}
else {
if ( is_readable('baby_status.php') ) {
include 'baby_status.php';
}
$status = ( $status )? $status: '??';
}
?>
Note: To use this option the directory these files are in must be writeable by the web server.