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I've created a like button for my users posts. I'm trying to print out the amount of likes. At the moment when I click the LIKE button nothing shows beside the like button unless I refresh the page then it shows the amount of like clicks the post has (4).

The problem I face is I have a LIKE_DO.php page that collects the data then sends it to a function which is in another page called function_user.php. Is there anyway I can use Ajax to grab the printed value of likes and how would I go about it? I would really like to just build ontop of the ajax I already have, to insert the data and make a call for the printed likes to save confusion and having to re-write all my code again.

Like Button 

echo "<div class='stream_option'><a id='likecontext_".$streamitem_data['streamitem_id']."' style='cursor:pointer;' onClick=\"likestatus(".$streamitem_data['streamitem_id'].",this.id);\">";

CURRENT AJAX

function likestatus(postid,contextid){
var obj = document.getElementById(contextid);
if(obj.innerHTML=="Like"){
obj.innerHTML="Unlike";
}else{
obj.innerHTML="Like";
}
$.post("../include/like_do.php", { streamitem_id: postid} );
}

LIKE_DO.PHP

<?php
session_start();
require"rawfeeds_load.php";
if(isset($_POST['streamitem_id'])){

$check = user_core::check_liked($_SESSION['id'],$_POST['streamitem_id'],1);

user_core::do_like($_SESSION['id'],$_POST['streamitem_id'],1);
if($check==0){?>
<?php }else{ ?>
<?php }
}else{
echo "<script>alert('Error liking post');</script>";
}
?>

USER_CORE - (With my printed likes that I need to get with ajax)

    function print_like_count($streamid){
        $check      =   "SELECT feedback_id FROM streamdata_feedback WHERE feedback_streamid='$streamid' AND feedback_rating=1";
        $check1     =    mysql_query($check);
        $check2     =    mysql_num_rows($check1);
        if($check2>0){
        return "(".$check2.")";
        }
}
dave
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2 Answers2

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Assuming user_core::do_like() adds like data to the database and user_core::check_liked() returns number of likes made for that posts.

(Correct me if I'm wrong)

if(user_core::do_like($_SESSION['id'],$_POST['streamitem_id'],1)){
echo $check; // as your user_core::check_liked() function returns number of likes.
}else{
echo "<script>alert('Error liking post');</script>";
}

ALSO : Since you are calling say X.php and from X.php you call a function in Y.php. Function in Y.php actually know the value you are looking for.

Since you are calling X.php values that are echoed in Y.php will not be visible to your script.

Solution - If you are calling X.php then make Y.php return the values to X.php and echo your values in X.php only.

+++++++++++++++++++++

UPDATE As per current latest question:

Change your $.post handler to

$.post("../include/like_do.php", { streamitem_id: postid},function(data){
//see the parameter data in this function. Data contains whatever that you echo in your php file.

$("#yourLikeDisplayDivId").html(data+" Likes");

} );
AdityaParab
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  • That sounds pretty on spot. How would I update my Ajax to show the value of the echo $check to show it in a div on my home.php page? – dave Jul 28 '12 at 10:23
  • updated! PS : Please read documentation about the functions you use in your code. 1) That will help you save lot of time. 2) That will actually add up in your useful knowledge. 3) That will help sites like this to have genuine problems to be discussed :) – AdityaParab Jul 28 '12 at 10:30
  • I'd recommend using $.ajax API instead of using $.post and similar versions of it. – AdityaParab Jul 28 '12 at 10:33
  • Thank you Aditya, the code and help you provided is more than useful and I have now got it working. Thank you for your help, time and advise. – dave Jul 28 '12 at 10:36
  • I have a slight issue. When I refresh the page, the like dissapears. And also it only works on the top post..So other posts it doesn't seem to print at all. – dave Jul 28 '12 at 11:01
  • Have you loaded the like data in your $(document).ready() ? Because if you haven't loaded it in the beginning it will not be visible. So retrieve the data while document is being loaded. Also, have you made the ajax call for all the posts? If not I'd recommend retrieving like data of all the posts in single JSON. I am free today, You can send me the code and I can help you in better way. Its really tough to tell you what exactly needs to be done without seeing your code. – AdityaParab Jul 28 '12 at 11:09
  • let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/14568/discussion-between-dave-and-aditya-parab) – dave Jul 28 '12 at 14:12
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If you need just to know how many people liked the page: Use XML or JSON to retrieve the data from scripts. Doing so, you can retrieve many pieces of data from scripts. Example:

<?php 
$xml = '<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<likes>
  <like>
    <number>'.getNumOfLikes().'</number>
  </like>
</likes>' ;
echo $xml ; 
?>

Create a simple ajax request:

var req = new XMLHttpRequest() ;
var url = "../include/like_do.php" ;

function execute(data){
  req.open("POST", url, true) ;
  req.setRequestHeader("Content-type", "application/x-www-form-urlencoded") ;
  req.onreadystatechange = function(){
    if (req.readyState == 4 and req.status == 200){
      var xml = req.responseXML ; //Get the xml response
      document.getElementById("numOfLikes").innerHTML = xml.getElementsByTagName()("number")[0].firstChild.nodeValue ; //get the value of the first "number" tag
      req.abort ;
    }
  req.send(data)
  }
}

execute("likeId=1234&anything=123") ;

Other things you do in your PHP scripts (what to display, how much stuff). I hope this helps as you can see how to use an AJAX request.

sybear
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