I want to enhance my tool's page where as soon use click a button. Request goes to server and depending upon return type (fail/pass) i change color of button. No Refresh/page reload
Page has multiple buttons : some what like below.
Name 9-11 - 11-2 2-5
Resource1 - Button - Button - Button
Resource2 - Button - Button - Button
Resource1 - Button - Button - Button
I am a c++ programmer so you might feel i asked a simple question
Here's a sample of jQuery Ajax posting a Form. Personally, I'm unfamiliar with PHP but Ajax is the same no matter what. You just need to post to something that can return Success = true or false. This POST happens asynchronously so you don't get a page refresh unless you do something specific in the success: section.
$("document").ready(function () {
$('form').submit(function () {
if ($(this).valid()) {
$.ajax({
url: yourUrlHere,
dataType: "json",
cache: false,
type: 'POST',
data: $(this).serialize(),
success: function (result) {
if(result.Success) {
// do nothing
}
}
});
}
return false;
});
});
Of course you don't have to be doing a POST either, it could be a GET
type: 'GET',
And if you don't need to pass any data just leave data: section out. But if you want to specify the data you can with data: { paramName: yourValue },
The cache: false, line can be left out if you want to cache the page. Seeing as how you aren't going to show any changes you can remove that line. jQuery appends a unique value to the Url so as to keep it from caching. Specifying type: "json", or whatever your specific type is, is always a good idea but not necessary.
Try using the $.post or $.get functions in jquery
$.post("url",$("#myform").serialize());
Adding a callback function as FabrÃcio Matté suggested
$.post("url",$("#myform").serialize(),function(data){alert(data);$("#myform").hide()//?Do something with the returned data here});
Here you go. You will find an example of a form, a button a the necessary ajax processing php page. Try it out and let us know how it goes:
<form action="" method="post" name="my_form" id="my_form">
<input type="submit" name="my_button" id="my_button" value="Submit">
</form>
<script type="text/javascript">
$("document").ready(function () {
$('#my_form').submit(function () {
$.ajax({
url: "ajaxpage.php",
dataType: "json",
type: "POST",
data: $(this).serialize(),
success: function (result)
{
//THere was an error
if(result.error)
{
//So apply 'red' color to button
$("#my_button").addClass('red');
}
else
{
//there was no error. So apply 'green' color
$("#my_button").addClass('green');
}
}
});
return false;
});
});
</script>
<?php
//ajaxpage.php
//Do your processing here
if ( $processed )
{
$error = false;
}
else
{
$error = true;
}
print json_encode(array('error' => $error));
die();
?>
Related
I am using this jQuery plugin for making toggle, but I have an issue that when I make multiple toggles that have same ids and class so in that case I am not able to identify particular toggle for applying auto load ajax on changing value.
I would to ask that how I make same toggle with this same plugin but different ids or class or name so I make ajax function like when I click toggle it will update in PHP without submitting submit button.
The plugin I am using is this one
The code I am using is this:
HTML
<p>Default: <span class="easyswitch"></span></p>
<p>Checked: <span class="easyswitch" data-default="1"></span></p>
SCRIPT
<script>
$('.easyswitch').easyswitch();
</script>
AJAX
$('MY_CLASS_NAME').change(function(){
var mode= $(this).prop('checked');
$.ajax({
type:'POST',
dataType:'JSON',
url:'test.php',
data:'mode='+mode,
success:function(data)
{
$("body").html('Operation Saved');
}
});
You can not handle easyswitch's change event. you need to create click event of it, and from it you can get the status of current toggle.
$('.easyswitch').easyswitch();
$('.easyswitch').click(function () {
var mode = $(this).hasClass('on');
toogleStatus(mode);
});
// for all controlls.
$(".easyswitch").each(function() {
var mode = $(this).hasClass('on');
toogleStatus(mode);
});
function toogleStatus(mode)
{
if (!mode) {
alert('checked')
}
else {
alert('unchecked')
}
}
Try using callback option
$('.easyswitch').easyswitch({
callback: function(val, ele) {
$.ajax({
type: 'POST',
dataType: 'JSON',
url: 'test.php',
data: { mode: val },
success: function(data) {
$("body").html('Operation Saved');
}
});
}
});
I'm learning AJAX by reading some online tutorials, so please understand I am very new to AJAX and programming in general. I have managed to do the following with 3 selectboxes:
populates selectbox 2 based on selection from selectbox 1
populates selectbox 3 based on selection from selectbox 2
Everything is working perfectly
Here is my code:
$(document).ready(function()
{
$(".sport").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_sport.php",
dataType : 'html',
data: dataString,
cache: false,
success: function(html)
{
$(".tournament").html(html);
}
});
});
$(".tournament").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_round.php",
data: dataString,
cache: false,
success: function(html)
{
$(".round").html(html);
}
});
});
});
</script>
Here is an Example
What I want to do
I would like to send the value of the 3 selectboxes to 3 php variables without the form reloading.
My Problem
When the user clicks submit:
The form reloads (which I dont want)
The selectbox values does not get send to my php variables
my code to get the values after submit is clicked is as follows:
if(isset($_POST['submit'])){
$a = $_POST['sport'];
$b = $_POST['tournament'];
:
}
However my code is flawed as I mentioned above.
If any one can help me to explain how to send my form data to the 3 php variables without the form reloading it will be greatly appreciated
If you don't want to submit your form when you click the button, you need to set that input as button and not submit. You can, also, attach the submit event handler to the form and prevent it to submit:
$("form").on("submit", function(e){
e.preventDefault(); //This is one option
return false; //This is another option (and return true if you want to submit it).
});
So, being said this, you could probably do something like:
$("form").on("submit", function(e) {
var formData = $(this).serialize();
e.preventDefault();
$.ajax({
url: 'yoururl',
data: formData,
type: 'post', //Based on what you have in your backend side
success: function(data) {
//Whatever you want to do once the server returns a success response
}
});
});
In your backend:
if (isset($_POST["sport"])) {
//Do something with sport
}
if (isset($_POST["tournament"])) {
//Do something with torunament
}
echo "Successfull response!"; //You have to "write" something in your response and that is what the frontend is going to receive.
Hope this helps!
Try using the javascript function preventDefault().
See this SO question.
Use a <button>Submit</button> element instead of <input type="submit"/> since the submit automatically submits the form.
Edit: And you would have to use on.('click') instead of looking for submit event in your jQuery.
edit - the info appears to be posting, but on form_data.php it doesn't seem to be retrieving the posted values
Here's the AJAX
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$("#submit_boxes").submit(function() { return false; });
$('input[type=submit]').click(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: $(this).serialize(),
success: function(data) {
$('#view_inputs').html(data); //view_inputs contains a PHP generated table with data that is processed from the post. Is this doable or does it have to be javascript?
});
return false;
});
};
</script>
</head>
Here is the form I'm trying to submit
<form action="#" id = "submit_boxes">
<input type= "submit" name="submit_value"/>
<input type="textbox" name="new_input">
</form>
Here is the form_data page that gets the info posted to
<?php
if($_POST['new_input']){
echo "submitted";
$value = $_POST['new_input'];
$add_to_box = new dynamic_box();
array_push($add_to_box->box_values,$value);
print_r($add_to_box->box_values);
}
?>
Your form is submitting because you have errors which prevents the code that stops the form from submiting from running. Specifically dataType: dataType and this.html(data) . Firstly dataType is undefined, if you don't know what to set the data type to then leave it out. Secondly this refers to the form element which has no html method, you probably meant $(this).html(data) although this is unlikely what you wanted, most likely its $(this).serialize() you want. So your code should look like
$('form#submit_boxes').submit(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: $(this).serialize(),
success: success
})
return false;
});
Additionally if you have to debug ajax in a form submit handler the first thing you do is prevent the form from submitting(returning false can only be done at the end) so you can see what errors occurred.
$('form#submit_boxes').submit(function(event) {
event.preventDefault();
...
});
You can use jQuery's .serialize() method to send form data
Some nice links below for you to understand that
jquery form.serialize and other parameters
http://www.tutorialspoint.com/jquery/ajax-serialize.htm
http://api.jquery.com/serialize/
One way to handle it...
Cancel the usual form submit:
$("#submit_boxes").submit(function() { return false; });
Then assign a click handler to your button:
$('input[type=submit]').click(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: this.html(data),
success: success,
dataType: dataType
})
return false;
});
I have a PHP populated table from Mysql and I am using JQuery to listen if a button is clicked and if clicked it will grab notes on the associated name that they clicked. It all works wonderful, there is just one problem. Sometimes when you click it and the dialog(JQuery UI) window opens, there in the text area there is nothing. If you are to click it again it will pop back up. So it seems sometimes, maybe the value is getting thrown out? I am not to sure and could use a hand.
Code:
$(document).ready(function () {
$(".NotesAccessor").click(function () {
notes_name = $(this).parent().parent().find(".user_table");
run();
});
});
function run(){
var url = '/pcg/popups/grabnotes.php';
showUrlInDialog(url);
sendUserfNotes();
}
function showUrlInDialog(url)
{
var tag = $("#dialog-container");
$.ajax({
url: url,
success: function(data) {
tag.html(data).dialog
({
width: '100%',
modal: true
}).dialog('open');
}
});
}
function sendUserfNotes()
{
$.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data:
{
'nameNotes': notes_name.text()
},
success: function(response) {
$('#notes_msg').text(response.the_notes)
}
});
}
function getNewnotes(){
new_notes = $('#notes_msg').val();
update(new_notes);
}
// if user updates notes
function update(new_notes)
{
$.ajax({
type: "POST",
//dataType: "json",
url: '/pcg/popups/updateNotes.php',
data:
{
'nameNotes': notes_name.text(),
'newNotes': new_notes
},
success: function(response) {
alert("Notes Updated.");
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
});
}
/******is user closes notes ******/
function closeNotes()
{
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
Let me know if you need anything else!
UPDATE:
The basic layout is
<div>
<div>
other stuff...
the table
</div>
</div>
Assuming that #notes_msg is located in #dialog-container, you would have to make sure that the actions happen in the correct order.
The best way to do that, is to wait for both ajax calls to finish and continue then. You can do that using the promises / jqXHR objects that the ajax calls return, see this section of the manual.
You code would look something like (you'd have to test it...):
function run(){
var url = '/pcg/popups/grabnotes.php';
var tag = $("#dialog-container");
var promise1 = showUrlInDialog(url);
var promise2 = sendUserfNotes();
$.when(promise1, promise2).done(function(data1, data2) {
// do something with the data returned from both functions:
// check to see what data1 and data2 contain, possibly the content is found
// in data1[2].responseText and data2[2].responseText
// stuff from first ajax call
tag.html(data1).dialog({
width: '100%',
modal: true
}).dialog('open');
// stuff from second ajax call, will not fail because we just added the correct html
$('#notes_msg').text(data2.the_notes)
});
}
The functions you are calling, should just return the result of the ajax call and do not do anything else:
function showUrlInDialog(url)
{
return $.ajax({
url: url
});
}
function sendUserfNotes()
{
return $.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data: {
'nameNotes': notes_name.text()
}
});
}
It's hard to tell from this, especially without the mark up, but both showUrlInDialog and sendUserfNotes are asynchronous actions. If showUrlInDialog finished after sendUserfNotes, then showUrlInDialog overwrites the contents of the dialog container with the data returned. This may or may not overwrite what sendUserfNotes put inside #notes_msg - depending on how the markup is laid out. If that is the case, then it would explains why the notes sometimes do not appear, seemingly randomly. It's a race condition.
There are several ways you can chain your ajax calls to keep sendUserOfNotes() from completing before ShowUrlInDialog(). Try using .ajaxComplete()
jQuery.ajaxComplete
Another ajax chaining technique you can use is to put the next call in the return of the first. The following snippet should get you on track:
function ShowUrlInDialog(url){
$.get(url,function(data){
tag.html(data).dialog({width: '100%',modal: true}).dialog('open');
sendUserOfNotes();
});
}
function sendUserOfNotes(){
$.post('/pcg/popups/getNotes.php',{'nameNotes': notes_name.text()},function(response){
$('#notes_msg').text(response.the_notes)
},"json");
}
James has it right. ShowUrlInDialog() sets the dialog's html and sendUserOfNotes() changes an element's content within the dialog. Everytime sendUserOfNotes() comes back first ShowUrlInDialog() wipes out the notes. The promise example by jeroen should work too.
I have a form: <form id="form" action="updatescore.php" method="post"> and a php file:updatescore.php which contains the code to update the database using the input values from the form. This all works when using a submit button.
Now I want to remove the submit button and submit the form if a javascript statement is true.
The js code part is:
if (document.getElementById('uhs').innerHTML > 0) { //this is true because the div gone is hidden
$('#gone').hide();
$.ajax({
type: "POST",
data: $("#form").serialize(),
cache: false,
url: "updatescore.php",
success: function () { //if submit to db is done
getUsers(1); //a function to reload a page overview
}
});
}
But nothing happens if the statement is true and the database is not updated. Any ideas on this one?
Kind regards,
are you sure the condition is TRUE?
if (document.getElementById('uhs').innerHTML > 0) { //this is true because the div gone is hidden
alert('its true');
also that is not right:
if (document.getElementById('uhs').innerHTML > 0)
perhaps:
if (document.getElementById('uhs').innerHTML.length > 0)
or:
if ($('#gone')[0].style.display=="none") {
The submit button is what triggers the event that runs the code:
$('#gone').hide();
$.ajax({
type: "POST",
data: $("#form").serialize(),
cache: false,
url: "updatescore.php",
success: function () { //if submit to db is done
getUsers(1); //a function to reload a page overview
}
});
The document.getElementById('uhs').innerHTML > 0 is not a event so there is not way to run your code .
Try this.....
document.addEventListener('keyup', function (e) {
//
if (document.getElementById('uhs').innerHTML > 0) {
$('#gone').hide();
$.ajax({
type: "POST",
data: $("#form").serialize(),
cache: false,
url: "updatescore.php",
success: function () { //if submit to db is done
getUsers(1); //a function to reload a page overview
}
});
}
});
Thanks for the replies. The code worked, but it seemed that a bug in another part of the code was messing with the AJAX call. So in the end there was nothing wrong with the code at all.
Again, thanks for the replies and suggestions!