Like system using AJAX, jQuery and PHP - javascript

I am trying to make a like/dislike system on the posts of my social networking website with each post having an pid(auto-increment). I run a while loop that fetches the post from database and show it and also give an option to like/dislike below each post. I am not using any input radio or checkbox button, instead I have used an icon for which the colour should change when clicked and also an AJAX function is called using onclick event on the same icon.
The AJAX function should go to a file likes.php where the a search is done for the row having that post id and the likes column of that row is updated or incremented by 1 and then the result is returned by echo. All this should work but not working.
The AJAX function takes post id (pid) as a parameter and pass it to likes.php where likes field corresponding to that post id is incremented. The like icon for each post has been assigned an id = post id so as to select that like button which is clicked and for which the post id is passed in AJAX function. The result is returned to a span element having an id = "like"+ post pid.
My code
index.php - to show posts and like button
$q = $conn->prepare("SELECT * FROM posts ORDER BY pid DESC");
$q->execute();
while($row = $q->fetch(PDO::FETCH_ASSOC)) {
#my post representation using div and all...
#like/dislike button
<span class=\"right floated\" id=\"like".$row['pid']."\" >
<i id=\"".$row['pid']."\" class=\"heart icon\" onclick=\"like(".$row['pid'].")\"></i>
</span>
}
ajax-function
'a' is a parameter-local variable
<script>
$(document).ready(function(){
function like(a) {
$("#"+a).css('color','red');
$.ajax({
url: "likes.php",
data: ({pid: a}),
type: "POST",
success:function(data){
$("#like"+a).html(data);
},
error:function (){}
});
}
});
}
</script>
likes.php file
<?php
session_start();
include 'db.php';
$j =$_POST['pid'];
$sql = "UPDATE posts SET likes = likes +1 WHERE pid ='" . $j . "'";
$r = $conn->prepare($sql);
$r->execute();
$sql = "SELECT * FROM posts WHERE pid ='" . $j . "'";
$r = $conn->prepare($sql);
$r->execute();
$ry=$r->fetch();
if($r)
{
echo $ry['likes'];
}
?>
Please tell me why doesn't it work; at least it should detect the icon click using onclick but I think even that's not working. The AJAX function is not getting executed as nothing happens on clicking.

The problem here, that you declare the like function inside the document ready so outside of that scope your function is not defined.
Posible solutions:
Declare your function before the document ready scope
Create a var outside of document ready
Use jQuery .click (or .on('click',) instead of the onClick html property to call your function
Declare your function before the document ready scope
script:
function like(a) {
$("#" + a).css('color','red');
$.ajax({
url: "likes.php",
data: {pid: a},
type: "POST",
success:function(data){
$("#like" + a).html(data);
},
error:function (){}
});
}
Create a var outside of document ready
var like;
$(document).ready(function(){
like = function(a) {
$("#" + a).css('color','red');
$.ajax({
url: "likes.php",
data: {pid: a},
type: "POST",
success:function(data){
$("#like"+a).html(data);
},
error:function (){}
});
}
});
(This can cause some problem, because until the document is not ready this function not gonna work)
Use jQuery .click (or .on('click',) instead of the onClick html property to call your function
index.php
<span class=\"right floated\" id=\"like".$row['pid']."\" >
<i id=\"".$row['pid']."\" class=\"heart icon action-like\"></i>
</span>
script
$('.action-like').on('click', function(){
var pid = $(this).attr('id');
$("#" + pid).css('color','red');
$.ajax({
url: "likes.php",
data: {pid: pid},
type: "POST",
success:function(data){
$("#like" + a).html(data);
},
error:function (){}
});
});
This should work. But use mysqli instead mysql and do not pass parameters to a query like you do (read about sql injection). And do not echo long static strings instead close php tag and open when needed to echo out a dynamic data.

Related

Trying to post a javascript variable to a php file using the ajax method : POST but getting an undefined index in $POST array within php file

I have a php file with the directory "JqueryPHP/HighestBid.php". All I want to do is be able to post javascript variables from one file "views/AuctionPage.php" to another file "JqueryPHP/HighestBid.php".
I then want to echo a value from "JqueryPHP/HighestBid.php" into the span tags with the id "price" into "views/AuctionPage.php".
The problem is that when I load the page "views/AuctionPage.php" it shows me the alert with the returned value "hi" but where the text is supposed to be outputted between the span tags, it is telling me that the index within my $POST array is undefined.
//JS views/AuctionPage.php
<script>
$(document).ready(function() {
var auc = "hi";
$.ajax({
url: "JqueryPHP/HighestBid.php",
method: "POST",
data: {'auctionid': auc },
success: function (result) {
alert("result: " + result);
}
});
});
</script>
<script type="text/javascript">
$(document).ready(function() {
setInterval(function () {
$('#price').load('JqueryPHP/HighestBid.php')
}, 333);
});
</script>
//HTML views/AuctionPage.php
<h4 class="price">Highest bid : <span id="price"></span></h4>
//PHP FILE "JqueryPHP/HighestBid.php"
<?php
$auctionid = $_POST['auctionid'];
echo $auctionid;
?>
When I get rid of the $POST array in "JqueryPHP/HighestBid.php" and just assign auctionid with a normal string.
<?php
$auctionid = "hi";
echo $auctionid;
?>
the text gets outputted between the span tags like it's supposed too so I am having a problem posting the variables to another page and I have no idea why. I have tried many ways to get this to work following examples on stack overflow but to no success.
This line:
$('#price').load('JqueryPHP/HighestBid.php')
loads the JqueryPHP/HighestBid.php script using GET, and that's another, completely independent request from your AJAX one, that's why the $_POST superglobal is empty. What you need to do is to change the span inside the success function of your AJAX call:
$.ajax({
url: "JqueryPHP/HighestBid.php",
method: "POST",
data: {'auctionid': auc },
success: function (result) {
// alert("result: " + result);
$('#price').html(result);
}
});
That gives the result you need. Delete the whole setInterval code, you don't need it at all.

how to fetch data from sql server database in php without refreshing the page

I am trying to get some data from the database. I create a function that is located in functions.php file that return a value. On another page, I create a variable and just get that value. I was trying to use the onkey to check the database but then I realize that i need to know the amount of tickets even if they don't type anything.
Here is the function:
function.php
function is_ticket_able($conn){
$query = "select number_of_tickets from [dbo].[TICKETS] " ;
$stmt = sqlsrv_query($conn, $query);
while ($row = sqlsrv_fetch_array($stmt)) {
$amount_of_tickets = $row['number_of_tickets'];
}
return $amount_of_tickets;
}
And, I am trying to check the database (without refreshing the page) and get the value on this page:
application.php
$amount_of_tickets = is_ticket_able($conn);
Then, I just check that $amount_of_tickets is not 0 or 1. Because if is one then some stuff have to change.
I am doing this (inside application.php):
if($amount_of_tickets !=0){
//show the form and let them apply for tickets.
//also
if($amount_of_tickets == 1){
//just let them apply for one ticket.
}
}
EDIT: I saw that AJAX would be the right one to use, but I am so confuse using it.
UPDATE:
function.php
function is_ticket_able($conn){
$query = "select number_of_tickets from [dbo].[TICKETS_LKUP] " ;
$stmt = sqlsrv_query($conn, $query);
while ($row = sqlsrv_fetch_array($stmt)) {
$ticket = $row['number_of_tickets'];
}
return $ticket;
}
application.php
$amount_of_tickets = is_ticket_able($conn);
<script type="text/javascript">
var global_isTicketAble = 0;
checkTicket();
function checkTicket()
{
$.ajax(
{
url: "application.php",
method: 'GET',
dataType: 'text',
async: true,
success: function( text )
{
global_isTicketAble = text;
alert(global_isTicketAble);
if( global_isTicketAble == 0 ){
window.location.replace("http://www.google.com");
}
setTimeout( checkTicket, 5000 ); // check every 5 sec
}
});
}
</script>
So, now the problem is that when I alert(global_isTicketAble); it doesn't alert the value from the database but it does alert everything that is inside application.php...Help plzzz
Server side
Assuming you need to check $amount_of_tickets periodically and this can be computed into application.php, inside that file you'll have
<?php
// $conn is defined and set somewhere
$amount_of_tickets = is_ticket_able($conn);
echo $amount_of_tickets;
exit(0);
?>
This way when the script is invoked with a simple GET request the value is returned in the response as simple text.
Client Side
ajax is the way to go if you want to update information on page without reloading it.
Below is just a simple example (using jQuery) that may be extended to fit your needs.
The code below is a JavaScript snippet. A global is used to store the value (globals should be avoided but it's just for the purpose of the example)
Then a function is invoked and the updated value is fetched from function.php script.
The function -prior termination- schedules itself (with setTimeout) to be re-invoked after a given amount of milliseconds (to repeat the fetch value process).
var global_isTicketAble = 0;
checkTicket();
function checkTicket()
{
$.ajax(
{
url: "application.php",
method: 'GET',
dataType: 'text',
async: true,
success: function( text )
{
global_isTicketAble = text;
// eventually do something here
// with the value just fetched
// (ex. update the data displayed)
setTimeout( checkTicket, 5000 ); // check every 5 sec
}
}
}
Note that $.ajax() sends the request but does not wait for the response (as async is set to true). When the request is received the function specified as success is executed.
Complete jQuery ajax function documentation can be found here
http://api.jquery.com/jquery.ajax/
I assume that you have a page (application.php) that displays a table somewhere.
And that you wish to fill that table with the data found in you database.
I'm not sure about WHEN you want these data to be refreshed.
On button click or periodically (like ervery 5 seconds)... But it doesn't matter for what I explain below.
In application.php:
Assemble all your page as you already know how.
But inside it, somewere, just insert an empty div where your table should show:
<div id="dynamicContent"></div>
Also add this script at the bottom of the page:
<script>
function getData(){
PostData="";
$.ajax({
type: "POST",
url: "function.php",
data: PostData,
cache: true,
success: function(html){
$(Destination).html(html);
}
});
}
getData(); // Trigger it on first page load !
</script>
There is 2 variables here... I named it "PostData" and "Destination".
About PostData:
You can pass data collected on the client side to your PHP function if needed.
Suppose you'd need to pass your user's first and last name, You'd define PostData like this:
Fname=$("#Fname").val(); // user inputs
Lname=$("#Lname").val();
PostData="Fname="+Fname+"&Lname="+Lname;
In your function.php, you will retreive it like this (like any normal POST data):
$Fname=$_POST['Fname'];
$Lname=$_POST['Lname'];
If you do not need to pass data from your client side script to you server side PHP... Just define it empty.
PostData="";
Then, about Destination:
This is the place for the empty "dynamic div" id ( I named it "dynamicContent" above).
Don't forget about the hashtag (#) for an id or the dot for a class.
This is a jQuery selector.
So here, PostData would be defined like this:
Destination="#dynamicContent";
The result of the ajax request will land into that "dynamic div".
This WILL be the result of what's defined in function.php..
So, if you follow me, you have to build your table in function.php...
I mean the part where you do your database query and your while fetch.
echo "<table>";
echo "<tr><th>column title 1</th><th>column title 2</th></tr>"
while ($row = sqlsrv_fetch_array($stmt)){
echo "<tr><td>" . $row['data1'] . "</td><td>" . $row['data2'] . "</td></tr>";
}
echo "</table>";
So if you have no data, the table will be empty.
You'll only get the table and table headers... But no row.
There is then no need for a function that checks if there is data or not.
Finally... About the trigger to refresh:
In application.php, you may place a button that fires getData()... Or you may define a setInterval.
It's up to you.
This is how I use ajax to refresh part of a page without reloading it completly.
Since ajax is new to you, I hope this answer will help.
;)
------------------------
EDIT based on Ariel's comment (2016-05-01)
Okay, I understand! Try this:
In application.php:
<div id="dynamicDiv"></div>
<script type="text/javascript">
// timer to trigger the function every seconds
var checkInterval = setInterval(function(){
checkTicket();
},1000);
function checkTicket(){
$.ajax({
type: "POST",
url: "function.php",
data: "",
cache: true,
success: function(html){
$("#dynamicDiv").html(html);
}
});
}
function noMoreTikets(){
clearInterval(checkInterval);
window.location.replace("http://www.google.com");
}
</script>
In function.php:
// Remove the "function is_ticket_able($conn){" function wrapper.
// Define $conn... Or include the file where it is defined.
// I assume that your query lookup works.
$query = "select number_of_tickets from [dbo].[TICKETS_LKUP] " ;
$stmt = sqlsrv_query($conn, $query);
while ($row = sqlsrv_fetch_array($stmt)) {
$ticket = $row['number_of_tickets'];
}
// Add this instead of a return.
if($ticket>0){
echo "There is still some tickets!"; // Text that will show in "dynamicDiv"
}else{
?>
<script>
$(document).ready(function(){
noMoreTikets();
});
</script>
<?php
}
Remember that your PHP scripts are executed server-side.
That is why your "return $ticket;" wasn't doing anything.
In this ajax way to call function.php, its script is executed alone, like a single page, without any relation with application.php, which was executed long ago.
It produces text (or javascript) to be served to the client.
If you want to pass a PHP variable to the client-side javascript, you have to echo it as javascript.
So here, if the PHP variable $ticket is more than zero, some text saying that there is still tickets available will show in "dynamicDiv" and the application page will not be refreshed. I suppose it shows a button or something that allows students to get a ticket.
Else, it will be the javascript trigger to "noMoreTikets()" that will land in the "dynamicDiv".

Wordpress get current page name or id within ajax request callback

I need to get current page id or name from ajax request callback. Initially at loading a page i made an ajax request. In its callback method i need to get the current page id or name. I used following code for ajax request.
$.ajax({
type: "POST",
url: my_site.home_url + '/wp-admin/admin-ajax.php',
data: {
action: "notes_select_page"
},
dataType: "html",
success: function (Response) {
if (Response == "OK") {
Notes.renderBoardList();
} else {
}
},
async: true
});
I took the request from action hook.
add_action('wp_ajax_nopriv_notes_select_page', 'Notes::select_page');add_action('wp_ajax_optimal_notes_select_page', 'Notes::select_page');
And the callback i used several code but doesn't work. Try 1.
public static function select_page(){
global $pagename;
die($pagename);
}
Try 2
public static function select_page(){
global $wp_query;
$pagename = get_query_var( 'pagename' );
if ( !$pagename) {
$post = $wp_query->get_queried_object();
$pagename = $post->post_name;
}
die($pagename);
}
Try 3
public static function select_page(){
global $post;
die($post->ID);
}
But unfortunately any of them doesn't work to get current page ID or name. Callback is working fine with other values.
Thanks in advance.
function get_current_page_id() {
var page_body = $('body.page');
var id = 0;
if(page_body) {
var classList = page_body.attr('class').split(/\s+/);
$.each(classList, function(index, item) {
if (item.indexOf('page-id') >= 0) {
var item_arr = item.split('-');
id = item_arr[item_arr.length -1];
return false;
}
});
}
return id;
}
You don't need ajax for this.
Add this function to your code.
You can now get the page id by using:
var id = get_current_page_id();
To retrieve the post details you have to send the data yourself
data:{
action: "notes_select_page",
post_id: current_post_id, //current_post_id should either parsed from DOM or you can write your ajax in PHP file
}
You can either use a hidden box for current post id and get in the Js file using class or id or write the ajax in you php file itself.
Then you can retrieve via POST
public static function select_page(){
$post_id = $_POST['post_id'];
}
I'm getting post ID from the default WordPress post editing form, like so :
var post_ID = jQuery('[name="post_ID"]').val()*1;
Tje *1 converts the ID into an integer, otherwise it's interpreted as a string.
First take page id by this function
either
<div id="current_page_id"> <?php get_the_ID(); ?> </div>
or
<body page-id="<?php get_the_ID(); ?>">
Now In jquery ajax take following
var page_id = $('current_page_id').html();
OR
var page_id = $('body').attr("page-id");
$.ajax({
type: "POST",
url: my_site.home_url + '/wp-admin/admin-ajax.php',
data: {
action: "pageid="+page_id,
},
dataType: "html",
success: function (Response) {
if (Response == "OK") {
Notes.renderBoardList();
} else {
}
},
async: true
});
There is a solution to solve the issue in Wordpress. Adding ajax code in wp_footer hook, where using php code current page id can be retrieved and pass as ajax value.
You can obtain alternatively by the hidden field the post/page id in the following manner. This code is inserted in the template file (and then the value will be send to your ajax action hook as indicated above):
<?php
echo '<input type="hidden" name="activepost" id="activepost"
value="'.get_the_ID().'" />'
;?>
Check out this for reference: https://developer.wordpress.org/reference/functions/get_the_id/

how can I make one button correspond to different clicked div html jquery php?

You can see my code below. I face a challenge that I don't know how to use one button to correspond different click. On the php, if I put the button inside the foreach loop, it will create a lot of button, that's not what I want. In the js, if I put the on.click button inside the foreach elements loop, it will also create a lot of on.click button, so I click one button, it will run many times depends on the number of label_name. I think about addClass, if I clicked the heart div, I use js to add a class, and then get the attr('id') inside button.on.(click), so I can differentiate them in my server php and mysql can request the correspond data. But the problem is that if a user click every div, then every div add classes, then problem again.
var current_page = 1;
var elements_body = {
"heart": "1",
"eye": "2",
"ear_nose_throat": "3",
"hand_foot_mouth": "4"
};
jQuery.each(elements_body, function (label_name, label_num) {
var disease_label = $('#' + label_name + '_d');
disease_label.on('click', function () {
var data = {
action: 'body_part_keyword', //wordpress loading url
postSearchNonce: MyAjaxSearch.postSearchNonce,
current_page: current_page,
label_name: label_name //this label_name will differentiate data that need to request from mysql in the action.php
};
$.ajax({
url: MyAjaxSearch.ajaxurl,
type: 'POST',
cache: false,
data: data,
success: function (data) {
disease_menu_result.append(data);
current_page++
}
}); //ajax
});
}); //jQuery.each
$('#loadmorebutton_body').on('click', function () {
//I dont know how can I make this button to correspond above code
});
<div id="disease_menu">
<?php
$arr = Array(
'heart'=>'heart',
'eye'=>'eye',
'ear_nose_throat'=>'ear nose throat',
'hand_foot_mouth'=>'hand foot mouth'
);
foreach ($arr as $key=>$value) {
?>
<div class="disease_li" id="disease_li_<?php echo $key;?>">
<span class="disease_span" id="<?php echo $key;?>_d"><label>(<?php echo $value;?>)</label>diseases</span>
</div>
<!--disease_li-->
<?php }?>
</div>
<!--disease_menu-->
<button id="loadmorebutton_body">Load More</button>
Use javascript functions :
function MyFunction() {
jQuery.each( elements_body, function( label_name, label_num) {
var disease_label= $('#'+ label_name + '_d');
disease_label.on('click',function(){
var data={
action: 'body_part_keyword',//wordpress loading url
postSearchNonce : MyAjaxSearch.postSearchNonce,
current_page:current_page,
label_name:label_name//this label_name will differentiate data that need to request from mysql in the action.php
};
$.ajax({
url: MyAjaxSearch.ajaxurl,
type:'POST',
cache: false,
data: data,
success: function(data){
disease_menu_result.append(data);
current_page++
}
});//ajax
});
});
}
$('#loadmorebutton_body').on('click',function(){
MyFunction();
}

Ajax POST is not posting onclick to current page

Alright so this has been bugging me for a long time now... I have tried everything but I cant get it to work!
So what I want to have is a link that acts as a button, and once you click it, it POSTs an ID number of the button in the form "{ 'id' : id }"
edit-homepage.php:
<script>
$(function() { // document ready
$('a.inactive').on('click', function(event) {
event.preventDefault(); // instad of return false
var id = $(this).data('id');
// use $.post shorthand instead of $.ajax
$.post('edit-homepage.php', {id: id}, function(response) {
// after you get response from server
editSlide(id);
});
});
});
</script>
The a href button is created using PHP and I want it to call the ajax function postID( id ) which will post the id so that later I can populate a form via PHP using the posted id.
edit-homepage.php:
echo '<li><a class="inactive" id="slide-'.$info["id"].
'" onClick="postID('.$info["id"].'); editSlide('.$info["id"].'); return false;">'
.'<img src="../images/'.$info["img"].'" width="175"/><p>Edit Slide '
. $info["id"] .'</p></a></li>';
Currently, when I click the link, it opens the alert but it is EMPTY or Undefined. It is supposed to display "ID: 1" for example if the link clicked has a ID of 1.
edit-homepage.php:
<script>
function editSlide($id) {
<?PHP
if (isset ($_POST['id'])) {
echo "alert('success!2');";
}$id = !empty($_POST['id']) ? $_POST['id'] : '';
$data = mysql_query("SELECT * FROM slider WHERE id='$id'") or die(mysql_error());
$info = mysql_fetch_array( $data );?>
document.getElementById("edit-slide-id").innerHTML="Edit Slide #"+$id;
document.getElementById("edit-form").style.display = "block";
document.getElementById("short-title").value="<?PHP echo $info['s_title']; ?>";
}
</script>
Thanks!
With jquery, you don't need to use attributes to attach events, like that:
$(function() { // document ready
$('a.inactive').on('click', function(event) {
event.preventDefault(); // instad of return false
var id = $(this).data('id');
// use $.post shorthand instead of $.ajax
$.post('edit-homepage.php', {id: id}, function(response) {
alert('ID:' + response);
// after you get response from server
editSlide(id);
});
});
});
As of server side, try replacing raw
<?PHP echo $_POST['id']; ?>
With
<?php echo !empty($_POST['id']) ? $_POST['id'] : '' ?>
You likely get notice about Undefined index id, which breaks javascript if there is no post data.
UPDATE
edit-homepage.php shold be separated something like that:
if(!empty($_POST)) {
// here you process your post data and return
// only wenever you want to pass to script
// not all the html
} else {
// here you output html and scripts, but don't do request processing
}
You should always remember, that your HTML rendering must always be separated from your logic. It is better to put views in separate files from logic, though it is not required, it is much easier to debug and maintain.
You can not include PHP code that is supposedly to run after the ajax call. The PHP code will be run only to generate the page. Anything you want to include in alert should be provided in the ajax response, in your case the data variable.
You need to use alert('ID: ' + id).
The $_POST['id'] part of the script does not react to the AJAX request. It is whatever the $_POST['id'] value is when the script is output to the browser (i.e. when the page is first loaded).
You will see this if you view the source.
alert ("ID:"+data);
then only you will get response
or
alert("ID"+id);
this will alert the id passes to function
http://jsfiddle.net/U54ME/
$(".checkthisclass").click(function() {
$.ajax({
type: "POST",
url: "edit-homepage.php",
data: { 'id' : $(this).attr("slideid"); },
success: function(data) {
alert(data);
}
});
}
});
--
<ul>
<li><a class="inactive checkthisclass" id="slide-5" slideid = "5" ><img src="http://blog.entelo.com/wp-content/uploads/2013/04/stackoverflow-logo.png" width="175"/><p>Edit Slide 5</p></a></li>
</ul>

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