I'm writing a game in html5 and using wordpress for the rest of the site. I want users to be able to save their progress, but I'm not clear on how the architecture looks.
I would just do an ajax call to a php page that access the database, but I feel like using the existing wordpress database and API is the correct thing to do. Not only that, but using the wordpress api will give me access to things like nonce.
That said, do I write a wordpress plugin to do this? How do I properly execute the save request from the javascript that runs my game?
In your js file, do your ajax request like this:
jQuery.post('path/to/wp-admin/admin-ajax.php', {action:'save_progress',other_parameter:'other_value'}, function(response) {
// let your user know if his progress is saved or not
});
add this to your theme's functions.php:
function save_progress() {
global $wpdb;
// do the saving job with received parameters here
die(); // this is necessary
}
add_action('wp_ajax_save_progress', 'save_progress');
If you're interested in a more generalizable solution than what the wordpress API allows (for more overall understanding), below is a simple, but complete demo:
Here's the HTML:
<input id="posted_data" ></input>
<button id="saveBtn">Save</button>
Here's the JS:
$(document.body).on('click', '#saveBtn', function(){
var posted_data=$('#posted_data').val();
var url="/submit_page.php";
$.ajax({url: url, data: {posted_data:posted_data }, type: "POST",
success: function(response){
//do something with the `response` data
} //success
}); //ajax
}); //.saveEditBtn
Here's the submit_page.php:
$posted_data=$_POST['posted_data'];
$posted_data; //do something with this variable, like insert it into a database
echo "Success"; // this will be the `response` variable in the JS above.
Related
I've recently changed my searching page to a searchable datatable page due to my employer's request for easier data management. The problem is that it is taking too long to load.
I'm wondering it there is a way to only load like a portion of the table and finish loading the page first. Then finish off loading the rest of the table after that, e.g. while the user actually attempt to search for the data.
This was requested because the user might want to navigate to other parts of the page instead of using the datatable.
Extra info : The page is in .php and the data is loaded using php do-while loop. Maybe we can do a workaround using php functions?
Using the AJAX method recommended in the comments, the following is similar to how you could handle the page-load. You would need the jQuery library for the below.
Initial page
<script type="text/javascript">
// when the page is done loading,
// let's send a call to load more data
$(document).ready(function(){
myFunction();
});
// function to handle AJAX request to gather data
function myFunction(){
$.ajax({
type: "POST",
url: "./linkToMyPHP.php?loadData=1",
success: function(data){
// handle the data using the "data" variable
}
});
}
</script>
AJAX Page
<?php
if(isset($_GET["loadData"])){
// call query here and echo information
}
It may be recommended, to actually use a PHP function called json_encode() to echo back the information from your AJAX page in JSON form. This would allow you to transmit an array of information, instead of raw data. You would then need to update your AJAX request function similar to below.
$.ajax({
type: "POST",
url: "./linkToMyPHP.php?loadData=1",
dataType: "json",
success: function(data){
$("#myDivToChange").html(data);
}
});
You can read up on JSON at this highly rated question.
i am developing a site with html and php ...
what i have done so far for like button on my page is as follow(concept is same only name is different on facebook it is like and on my site it is Points Up )
<form method="post">
<input type="hidden" value="<?php echo $posts[postid]; ?>" name="postid">
<input type="submit" name="pointsup" value="Points Up" />
</form>
the above will create the button with name Points Up.
if(isset($_POST['pointsup']))
{ in this if block i have written all queries to update database and user interface and all
}
what i want is instead of that button there should be some link witch will run my sql code.
i also tried JavaScript but it doesn't help anything
Thanks in advance for your answers!
When you are working with a single anchor tag (hyperlink), you should send your parameters by GET method. in case, /your/address/rate.php?id= and check $_GET['id'] in PHP (validate/sanitize/...)
also, you can send your requests as POST by using AJAX.
this is a sample for jQuery:
$.ajax({
type: "POST",
url: "some.php",
data: { name: "John", location: "Boston" }
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
take a look here
but if you want to use pure JavaScript, you should work with XMLHttpRequest object. take a look here.
You should do something like:
Find where you should point your link to and what parameters are needed, i will call it $url. It can involve such things like creating additional files, refactoring your code, etc...
Write js method which will send request to $url. I will call this method sendPointsUp;
Create link which will react to onClick event and call sendPointsUp with desired parameters from there.
What you can do is just add a link with an id so you can reach it easily with jquery.
Create a php-page where you do the query and other stuff (security check, cookie, whatever you want to do), and finally attach an ajax event to the link to call your php-page.
Print the link in your page:
Vote up
Create php
<?php
// do the query and other stuff, return the result in json format
// if you want to do something with the result (for example display the votes)
Javascript
$('#voteup').click(function(e){
$.ajax({
url: "your/url/to/phpfile",
}).done(function(data) {
// rewrite link to undo vote or whatever you want
// do something with the returned data
});
});
I have a page with multiple forms that do the same thing, acting as a like button for each post in the page, and right next to it the number of likes inside a div named "likes".$id, so I can identify where to write the likes count after the ajax call. I was trying to use jQuery ajax function, but I couldn't set what div to write the results of the function.
$.ajax({
type:'POST',
url: 'likepost.php',
data:$('#like').serialize(),
success: function(response) {
$('#like').find('#likediv').html(response);
}
});
And how would I access the data on likepost.php? I am terrible with javascript, so I hope someone could help me and explain how the jQuery function really works, because I've been copying and pasting it without really knowing what I was doing.
Would this work?
$(function () {
$("#likebutton").click(function () {
var id = $('input[name=id]'); // this is me trying to get a form value
$.ajax({
type: "POST",
url: "likepost.php",
data: $("#like"+id).serialize(), // the form is called like+id e.g. like12
success: function(data){
$("#likes"+id).html(data); // write results to e.g. <div id='likes12'>
}
});
});
});
I put this in the code but when the button is clicked, the usual post refreshing page is done. Why is that?
Making a mini-form, serializing it, and POSTing it seems like a lot of heavy lifting when all you really want to do is send the ID to the likepost.php script.
Why not just retrieve the ID and post it to the script?
First let's break down your function:Type is the type of the request we're making, you specified POST here. This means in your PHP file you'll access the data we're sending using $_POST. Next up is URL which is just the url of where you're sending the data, your php file in this case.
After that is data, that is the data we're sending to the url (likepost.php). You're serializing whatever has a ID of "like" and sending it to the php file. Finally success is a function to run once the request is successful, we get a response back from the PHP and use it in the function to output the response.
As for the multiple forms I'd recommend doing something like:
http://www.kavoir.com/2009/01/php-checkbox-array-in-form-handling-multiple-checkbox-values-in-an-array.html
Here's documentation on the stuff we talked about, if you're every confused about jquery just break it down and search each part.
http://api.jquery.com/serialize/
http://api.jquery.com/jQuery.ajax/
you can try :
function submitform(id) {
var jqxhr = $.post('./likepost.php',$("#"+id).serialize(), function(data) {
$("#"+id).find('#likediv').html(data);
}, "json")
return false;
}
in form:
<form method="post" id="likeForm" onsubmit="return submitform(this.id);">
<input..... />
<input type="submit" value="Submit" />
</form>
in likepost.php add first line:
if ($_SERVER['HTTP_X_REQUESTED_WITH'] != "XMLHttpRequest") {
header("location: " . $_SERVER['HTTP_REFERER']);
exit();
}
you can see more : http://api.jquery.com/serialize/
working for me.
I'm building a website which has a page that users can add content to, and they can rearrange the divs to whichever position and size they want. I'd like to have a save button which saves the current position of each div; however, I don't want the page to refresh each time (I'm also going to have an auto-save, which will have to save the information in the background).
I can't figure out how to post the data to the server though, without causing the page to reload. I figure I need some kind of AJAX request, but can't find anything that tells me how to do that (all the AJAX examples I can find seem to be about reading data from the server). I think I'm just starting to go round in circles now, but I can't get my head around this at all - I know it's probably not a hard thing to do, but I keep getting confused by the different examples.
So, first of all, is this the best way to do it? And, if so, can someone point me to a straightforward example of posting data via AJAX? I'm already using jQuery, so can use that for the Ajax as well.
Thanks.
Super simple AJAX with jQuery:
$.ajax({
url: '/save-the-stuff-url',
type: 'POST',
data: {
// information about your divs, etc.
'foo' : 'bar'
},
success: function(response) {
// if the AJAX call completes successfully, this function will get called.
alert('POST successful!');
}
});
Give it a shot!
Here, try this for AJAX:
$.post("example.php", {
from : "ajax", // put some info in these - they are the params
time : "2pm",
data : "save"
},
function(data) { // callback function - data always passed to it
$("#success").html(data); // do something with that data
}
);
And put this somewhere:
<span id='success'></span>
And then, try this example for example.php:
<?php
if(isset($_POST['from']) and $_POST['from'] == 'ajax'){
echo "<span style='color: green;'>Saved!</span>";
}
else {
echo "<span style='color: red;'>Failure!</span>";
}
?>
And then just modify these to fit your needs, probably changing the file of target. Whatever the script outputs is what is given to the ajax request. This means that if this was my PHP script:
<?php echo "Aloha!"; ?>
And this was my javascript:
$("#output").load("myScript.php");
Then #output would have "Aloha!" in it.
Hope this helps!
Please go through the Jquery site for various examples of post.
HTH
and the jQuery docs pages are a great way to learn jQuery.. the page for post is http://docs.jquery.com/Post
you may also want to look at jQuery draggables if you're not using that yet..
http://docs.jquery.com/UI/API/1.8/Draggable
you can fire a save tied to your draggable object being let go rather easily with
$( ".selector" ).draggable({
stop: function(event, ui) { ... }
});
I'd like to add a simple functionality to my pages, where a user will see a "follow" button and by clicking it a db record will be created (userID and pageID). I'll handle query on the backend, I suppose. I think I need to do it in AJAX, but I havebn't done much with AJAX. I was also thinking that updating the button status from FOLLOW to FOLLOWING (or something similar) I could do with jQuery, with some sort of toggle, while the request is being processed on the background.
Am I on the right track with this?
You're on the right track.
I've created an example which uses a button like <input type="image" class="follow">. When I user clicks on it it sends a request to the server (url). On success it updates the button image.
$('input[type=image].follow').click(function() {
var button = $(this);
var current_img = $(button).attr('src');
var current_alt = $(button).attr('alt');
$(button).attr('src', '/style/icons/ajax-loader.gif');
$(button).attr('alt', 'Requesting data from the server...');
$.ajax({
url: url of script the processes stuff (like db update),
type: 'POST',
data: {},
dataType: "json",
error: function(req, resulttype, exc)
{
$(button).attr('src', '/style/error.png');
$(button).attr('alt', 'Error while updating!');
window.setTimeout(function() {
$(button).attr('src', current_img);
$(button).attr('alt', current_alt);
}, 3000);
},
success: function(data)
{
$(button).attr('src', '/style/followed.png');
$(button).attr('alt', 'Followed');
}
});
return false;
});
Above is just some example code. Change it at your will. Have fun with it.
AJAX is right, jQuery makes ajax easy.
//Post with jQuery (call test.php):
$.post('test.php', function(data) {
//Do something with result data
});
It sounds like you are on the right track here. If you're working with a smaller application then using an AJAX request and creating your record would be easiest using a Java servlet and putting for example some JDBC code in your doGet or doPost method to perform the database operations.
At the same time your onSuccess method for your AJAX request can call the jQuery code necessary to update your button. Good Luck!