Php - Ajax like / dislike button

Question

I found this script from internet search by using like / dislike button for my script

http://wcetdesigns.com/view.php?dtype=tutorials&category=php&id=22

Everythings work good when using for single post but,

I want to use this rating script for all documents come from while cycle

For example, I have news script and all news are listed, and near to title you see like/dislike buttons. (like stackoverflow web page) and I want to rate all this content separately. In this script, I couldn't separate article id's in same page

Can you please help me about that?

sure. What do you have, what have you tried, and what doesn't work for you? — Rene Pot, Sep 28 '12 at 13:17
I did successfully rate single post, I didn't rate posts in listed one page separately. script just rate first post. I want to do same think like stackoverflow. posts and answers are listed in page and everybody can rate every post by click like or dislike button — Seyhan, Sep 28 '12 at 13:24

score 8 · Answer 1 · answered Sep 28 '12 at 14:04

First, you need to specify for each button which article is it for, you can easily do it with data- attribute

<button class="votebutton" data-article="1" data-vote="1">Up vote</button>
<button class="votebutton" data-article="1" data-vote="-1">Down vote</button>
<span class="votes" data-article="1"><?=//print the current number of votes of this article//?>

Now you need to do something when the button is clicked, so your javascript would look something like that:

$('.votebutton').on("click", function(){
  vote = $(this).attr("data-vote");
  article = $(this).attr("data-article");
  rateArticle(vote, article);
})

now you need the rateArticle() function that would look something like this:

function rateArticle(vote, article){ 
  var data = 'vote='+rating+'&articleID='+article;
  $.ajax({
     type: 'POST',
     url: 'rateArticle.php', 
     data: data,
     success: function(votes){
         $('.vote[data-article='+article+']').html(votes); 
     }
   });
}

on the server side, the rateArticle.php file that you'll post to would look something like this:

$vote = $_POST['vote'];
$articleID = $_POST['articleID'];
$currentVotes = ; //fill it with the votes that the article currently have

if($vote != "-1" || vote != "1"){
  //goddamn script kids messing with us again, let's put them back where they belong
  header("Location: http://www.lemonparty.com"); //probably not this, but you can think of something
}

if(userHasAlreadyVoted()){ //if user has already voted we just return the current number of votes
  echo $currentVotes;
}else{
  $newVotes = $currentVotes + $vote;
  //save newVotes to db...
  echo $newVotes;
}

And that's about it...

score 0 · Answer 2 · answered Apr 03 '13 at 04:43

i have write this code and it works well for like-dislike using ajax in p.h.p-code :)

i have divided it into three pages, named test1.php , test1.js, test1php.php

test1.php

<?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
?>
<html>
    <head>
        <script type="text/javascript" src="test1.js"></script>
    </head>
    <body>
        <form action="" method="post" enctype="">
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>                    
        </form>
    </body>
</html>

test1.js

function change_status(country)
{
  // alert("hellow");
if (country.length==0)
  {
   //alert("hellow"); 
  document.getElementById("status_id").innerHTML="";
  return;
  }
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("status_id").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","test1php.php?status_id="+country,true);
xmlhttp.send();
}

test1php.php

    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
    $id = $_GET['status_id'];
    $q = mysql_query("SELECT * from s_user WHERE u_id = $id");
    $r = mysql_fetch_array($q);
    $fetch_status = $r['u_status'];
    if($fetch_status == 1){
        mysql_query("UPDATE s_user SET u_status = 0 WHERE u_id = '$id'");
    }else{
        mysql_query("UPDATE s_user SET u_status = 1 WHERE u_id = '$id'");
    }
?>
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>

Php - Ajax like / dislike button

2 Answers2