Easiest Way To Make A Form Submit Without Refresh - php

I have been trying to create a simple calculator. Using PHP I managed to get the values from input fields and jump menus from the POST, but of course the form refreshes upon submit.
Using Javascript i tried using
function changeText(){
document.getElementById('result').innerHTML = '<?php echo "$result";?>'
but this would keep giving an answer of "0" after clicking the button because it could not get values from POST as the form had not been submitted.
So I am trying to work out either the Easiest Way to do it via ajax or something similar
or to get the selected values on the jump menu's with JavaScript.
I have read some of the ajax examples online but they are quite confusing (not familiar with the language)

Use jQuery + JSON combination to submit a form something like this:
test.php:
<script type="text/javascript" src="jquery-1.4.2.js"></script>
<script type="text/javascript" src="jsFile.js"></script>
<form action='_test.php' method='post' class='ajaxform'>
<input type='text' name='txt' value='Test Text'>
<input type='submit' value='submit'>
</form>
<div id='testDiv'>Result comes here..</div>
_test.php:
<?php
$arr = array( 'testDiv' => $_POST['txt'] );
echo json_encode( $arr );
?>
jsFile.js
jQuery(document).ready(function(){
jQuery('.ajaxform').submit( function() {
$.ajax({
url : $(this).attr('action'),
type : $(this).attr('method'),
dataType: 'json',
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
return false;
});
});

The best way to do this is with Ajax and jQuery
after you have include your jQuery library in your head, use something like the following
$('#someForm').submit(function(){
var form = $(this);
var serialized = form.serialize();
$.post('ajax/register.php',{payload:serialized},function(response){
//response is the result from the server.
if(response)
{
//Place the response after the form and remove the form.
form.after(response).remove();
}
});
//Return false to prevent the page from changing.
return false;
});
Your php would be like so.
<?php
if($_POST)
{
/*
Process data...
*/
if($registration_ok)
{
echo '<div class="success">Thankyou</a>';
die();
}
}
?>

I use a new window. On saving I open a new window which handles the saving and closes onload.
window.open('save.php?value=' + document.editor.edit1.value, 'Saving...','status,width=200,height=200');
The php file would contain a bodytag with onload="window.close();" and before that, the PHP script to save the contents of my editor.
Its probably not very secure, but its simple as you requested. The editor gets to keep its undo-information etc.

Related

Resetting a PHP $_SESSION array with jquery function

I am trying to reset a session array in php with a function in jquery using a button. I would use a submit but I don't want the page to refresh. I tried to send a $.post request leaving the variables and return blank, and then sending a variable so I could use $_session[''] = array() but none of it worked. I have searched and can't find much about it just a lot on sending strings.
OK this is very simple to stop the page from refreshing you need to tell js to disable the default event i use jquery for this here is my code
Html & js
<html>
<head>
<title>Reseting a PHP $_SESSIO array with jquery function</title>
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
<script>
function sessRest(){
$.post("rest.php", {x: "9845621"}).done(function(data){
alert("States: " + data);
});
}
$(document).ready(function(){
$("#target").click(function(event){
event.preventDefault();
sessRest();
});
});
</script>
</head>
<body>
<div id="main">
Click to rest me
</div>
</body>
</html>
php code rest.php
<?php
session_start();
(string)$data = $_POST['x'];
if($data == "9845621"){
$_SESSION['gx'] = array();
return $_SESSION['gx']; //return the empty array to js
}else(
return "error";
)
?>
I hope this helps .
User below jquery to submit to php code
var requestData = { param: "value"};
$.ajax({
url: your_url/session_change.php,
type: "post",
dataType: "json" or what ever,
data: your_data,
success: function (data) {
}
});
You can end the session successfully on server side with an ajax call, but apart from reloading the page, you're not going to clear what information was loaded already on client side. The session information wont be there once you do reload, but there is no way around that.
You can, however, emulate what you want to do with javascript.
When you load your session information, echo it to the page as javascript variables, then you have full control on client side. Just beware of echoing sensitive information like passwords, obviously.
try this:
your html file should contain this jQuery file:
$('#button').click(function(e){
e.preventDefault();
jQuery.ajax({
url: 'http://yourwebsite.com/session.php'
}).done(function(data){
if(data=='reseted'){
//do anything...
}
else {
//do anything...
}
})
});
and in your session.php file:
<?php
session_start();
session_unset();
if($_SESSION == FALSE){
echo 'reseted';
}
else echo 'no';
?>
the answer was
jquery $.post('reset.php');
in reset.php
$_SESSION['products'] = array();
?>
this reset my session array when the reset button was clicked with no page refresh...
I had done this originally and forgot to include my core.php in the reset.php which contained my start session()..
Thank you all for the help though.... great suggestions

Codeigniter AJAX Example

I've been looking for a week now for a decent full working example of how to use AJAX with Codeigniter (I'm an AJAX novice). The posts / tuts I've seen are old - all the programming languages have moved on.
I want to have an input form on a page which returns something to the page (e.g. a variable, database query result or html formatted string) without needing to refresh the page. In this simple example is a page with an input field, which inserts the user input into a database. I'd like to load a different view once the input is submitted. If I could understand how to do this I'd be able to adapt it to do whatever I needed (and hopefully it would help others too!)
I have this in my 'test' controller:
function add(){
$name = $this->input->post('name');
if( $name ) {
$this->test_model->put( $name );
}
}
function ajax() {
$this->view_data["page_title"] = "Ajax Test";
$this->view_data["page_heading"] = "Ajax Test";
$data['names'] = $this->test_model->get(); //gets a list of names
if ( $this->input->is_ajax_request() ) {
$this->load->view('test/names_list', $data);
} else {
$this->load->view('test/default', $data);
}
}
Here is my view, named 'ajax' (so I access this through the URL www.mysite.com/test/ajax)
<script type="text/javascript">
jQuery( document ).ready( function() {
jQuery('#submit').click( function( e ) {
e.preventDefault();
var msg = jQuery('#name').val();
jQuery.post("
<?php echo base_url(); ?>
test/add", {name: msg}, function( r ) {
console.log(r);
});
});
});
</script>
<?php echo form_open("test/add"); ?>
<input type="text" name="name" id="name">
<input type="submit" value="submit" name="submit" id="submit">
<?php echo form_close(); ?>
All that happens currently is that I type in an input, updates the database and displays the view "test/default" (it doesn't refresh the page, but doesn't display "test/names_list" as desired. Many thanks in advance for any help, and for putting me out of my misery!
Set unique id to the form:
echo form_open('test/add', array('id'=>'testajax'));
I assume that you want replace a form with a view:
jQuery(document).ready(function(){
var $=jQuery;
$('#testajax').submit(function(e){
var $this=$(this);
var msg = $this.find('#name').val();
$.post($this.attr('action'), {name: msg}, function(data) {
$this.replace($(data));
});
return false;
});
better way if you return url of view in json response:
$.post("<?php echo base_url(); ?>test/add", {name: msg}, function(data) {
$this.load(data.url);
},"json");
from your last comment - I strongly not suggest to replace body, it will be very hard to support such code.
but here is anser:
$.post("<?php echo base_url(); ?>test/add", {name: msg}, function(data) {
$('body').replace(data);
});

jQuery/JavaScript ajax call to pass variables onClick of div

I am trying to pass two variables (below) to a php/MySQL "update $table SET...." without refreshing the page.
I want the div on click to pass the following variables
$read=0;
$user=$userNumber;
the div Basically shows a message has been read so should then change color.
What is the best way to do this please?
here's some code to post to a page using jquery and handle the json response. You'll have to create a PHP page that will receive the post request and return whatever you want it to do.
$(document).ready(function () {
$.post("/yourpath/page.php", { read: "value1", user: $userNumber}, function (data) {
if (data.success) {
//do something with the returned json
} else {
//do something if return is not successful
} //if
}, "json"); //post
});
create a php/jsp/.net page that takes two arguments
mywebsite.com/ajax.php?user=XXX&secondParam=ZZZZ
attache onClick event to DIV
$.get("ajax.php?user=XXX&secondParam=ZZZZ". function(data){
// here you can process your response and change DIV color if the request succeed
});
I'm not sure I understand.
See $.load();
Make a new php file with the update code, then just return a json saying if it worked or not. You can make it with the $.getJSON jQuery function.
To select an element from the DOM based on it's ID in jQuery, just do this:
$("#TheIdOfYourElement")
or in your case
$("#messageMenuUnread")
now, to listen for when it's been clicked,
$("#messageMenuUnread").click(function(){
//DO SOMETHING
}
Now, for the AJAX fun. You can read the documentation at http://api.jquery.com/category/ajax/ for more technical details, but this is what it boils down to
$("#TheIdOfYourImage").click(function(){
$.ajax({
type: "POST", // If you want to send information to the PHP file your calling, do you want it to be POST or GET. Just get rid of this if your not sending data to the file
url: "some.php", // The location of the PHP file your calling
data: "name=John&location=Boston", // The information your passing in the variable1=value1&variable2=value2 pattern
success: function(result){ alert(result) } // When you get the information, what to do with it. In this case, an alert
});
}
As for the color changing, you can change the CSS using the.css() method
$("#TheIdOfAnotherElement").css("background-color","red")
use jQuery.ajax()
your code would look like
<!DOCTYPE html>
<head>
</head>
<body>
<!-- your button -->
<div id="messageMenuUnread"></div>
<!-- place to display result -->
<div id="frame1" style="display:block;"></div>
<!-- load jquery -->
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
//attach a function to messageMenuUnread div
$('#messageMenuUnread').click (messageMenuUnread);
//the messageMenuUnread function
function messageMenuUnread() {
$.ajax({
type: "POST",
//change the URL to what you need
url: "some.php",
data: { read: "0", user: "$userNumber" }
}).done(function( msg ) {
//output the response to frame1
$("#frame1").html("Done!<br/>" + msg);
});
}
}
</script>
</body>

AJAX Post to self in PHP

I feel like this is something that I should have learned by now, and I'm sure it's something small I'm missing, but I could use clarification to make sure my approach is correct.
I'm using AJAX to post data to self which is a file that contains php and html. I can write the php fine, but after a successful ajax post, how do I only return the data that is processed via php and not the remaining html? Is it better to just post to a separate script?
If you have the PHP handling the POST request in the beginning of the file, you can just do something like this:
<?php
if (isset($_POST['somevar'])) {
/* do something */
exit(0);
}
?>
exit() will stop the loading of the page at that line.
I, for one, think it's better to be utilizing a separate script to deal with dynamic AJAX requests.
You can scrape changed parts of the resulting document and insert them into the original page. This way you can also make your page work for a user with JavaScript disabled not doing anything specially.
Example:
<html><title>Unobtrusive AJAX Example</title>
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script><script type="text/javascript">
$("form.ajax[id]").live('submit', function() {
$(this).find("input[type='submit']").attr("disabled", true);
$.ajax({
type: $(this).attr('method') || 'POST',
url: $(this).attr('action') || window.location.pathname,
data: $(this).serialize(),
context: $(this),
success: function(data) {
$(this).html(
$(data).find("#" + $(this).attr("id")).html()
);
}
});
return false;
});
</script>
</head><body>
<div><form method="post" class="ajax" id="main">
<p><?php echo date('H:i:s'); ?></p>
<p><input type="submit"></p>
</form></div>
<!-- keep the div: you got to have at least one div to make it work -->
</body></html>
It always depends on what are your needs, but if using the same script is enough for you then do it.
If you want the script not to send anything more than your answer to an XML HTTP Request, after sending the data, use an exit(); in PHP, which will make the script finish at that point.
Put to the of the script:
if($_POST['id']) {
$data = array('return'=>'returnValue');
$data = json_encode($data);
exit($data); }
Javascript:
$.ajax({
url: 'frmSelf.php',
data: $("#frmSelf").serialize(),
dataType: 'json',
type : 'post',
success : function(returnData) {
console.log(returnData);
}
});

jquery ajax request firebug error

I am using php/ajax to submit a form without page refresh. Here are my files-
coupon.js
jQuery(document).ready(function(){
jQuery(".appnitro").submit( function(e) {
$.ajax({
url : "http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php",
type : "post",
dataType: "json",
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
//return false or
e.preventDefault();
});
});
sms.php
<?php
//process form
$res = "Message successfully delivered";
$arr = array( 'mess' => $res );
echo json_encode( $arr );//end sms processing
unset ($_POST);
?>
and here is code for my html page -
<form id="smsform" class="appnitro" action="http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php" method="post">
...
</form>
<div id="mess" style="background:green;"></div>
Now when i click on submit button nothing happens and firebug shows following under console panel -
POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
404 Not Found 1.29s `jquery.min.js (line 130)`
Response
Firebug needs to POST to the server to get this information for url:
http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
This second POST can interfere with some sites. If you want to send the POST again, open a new tab in Firefox, use URL 'about:config', set boolean value 'extensions.firebug.allowDoublePost' to true
This value is reset every time you restart Firefox This problem will disappear when https://bugzilla.mozilla.org/show_bug.cgi?id=430155 is shipped
When i set 'extensions.firebug.allowDoublePost' to true then following results show up -
POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
404 Not Found 1.29s `jquery.min.js (line 130)`
Response -
{"mess":"Message successfully delivered"}
CaN anyone help me in fixing this firebug error of 404 not found. And why is it showing jquery.min.js (line 130) along side?
P.S -do not worry about http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street this is my base url
You may want to try putting the e.preventDefault() statement before the $.ajax call.
EDIT:
My x.html, corresponds to your HTML page
<!DOCTYPE html>
<html>
<head>
<title>x</title>
<script
type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
</script>
<script type="text/javascript" src="/so/x.js"></script>
</head>
<body>
<form id="smsform" class="appnitro" action="/so/x.php">
<input type="text" name="zz">
<input type="submit">
</form>
<div id="mess" style="background:green;"></div>
</body>
</html>
My x.js, corresponds to your coupon.js
jQuery(document).ready(function(){
jQuery(".appnitro").submit( function(e) {
e.preventDefault();
$.ajax({
url : "/so/x.php",
type : "post",
dataType: "json",
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
//return false or
//e.preventDefault();
});
});
My x.php, corresponds to your sms.php
<?php
$res = "Message successfully delivered.";
$arr = array('mess'=>$res);
echo json_encode($arr);
unset($_POST);
?>
This actually works in my environment, although I do not have the rest of the HTML markup or the additional PHP form processing code. The "Message successfully delivered." shows up in green directly below the input field.
When inside the Ajax call this refers to the Ajax object you need to do this
var __this = this;
Before going into the Ajax call, then it would be
data : __this.serialize()
Or look up the use of context within an Ajax call in Google. Or serialise your data into a variable before going into the Ajax call.

Categories