I created a code for like on my website, but when ever anybody likes a comment , the page is totally refresh and it takes the users to the top of the site, but does anyone have idea how nairaland did their likes e.g? www.nairaland.com/do_likepost?session=BBD453600902D4CBAE23FBAA58F25CEB193CFE59E48B17F480E3876BF71443BE&redirect=%2F2957092%2Foliseh-resigns-super-eagles-coach%2343259579&post=43259579
it doesn't reload a page, it act like a button, .. actually am writing with php language,

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Liljeez:
Send an e-mail to seun...

thanks, but I don't think he can help because nairaland is written in python

Lexxyfund:
thanks, but I don't think he can help because nairaland is written in python

yeah but he culd hav an idea

I think u can use increment-storage style.... such dat for every click on d "likes" a code fires increasing the existing value and saving it in a database....

Eg ur like will have 0 has default but will not show... so when a user clicks on it, the number value will increase to 1 and 1 will b saved in a database and show .. Also if anoda person click it... it will increase to 2 which will change d saved number in d database to d current. ...

As regards d issue of having d change instantly in d view without reloading d page... U will av to use these MVC frameworks like Angular, Backbone, React etc...

Then whatever html code u av used 4 ur likes as regards gui ensure it does not av a default way of behaviour and if it does then prevent d default behaviour - u can use client side js 4 dat...


.... Just my idea anyway....

Lexxyfund:
I created a code for like on my website, but when ever anybody likes a comment , the page is totally refresh and it takes the users to the top of the site, but does anyone have idea how nairaland did their likes e.g? www.nairaland.com/do_likepost?session=BBD453600902D4CBAE23FBAA58F25CEB193CFE59E48B17F480E3876BF71443BE&redirect=%2F2957092%2Foliseh-resigns-super-eagles-coach%2343259579&post=43259579
it doesn't reload a page, it act like a button, .. actually am writing with php language,

a. Use ajax. You send the event in the background and update the ui with response from the server.
b. use an iframe. You load the like button in an iframe and when the user clicks it just update with the required image etc.

you will have to use ajax to implement the feature

in your html:

<button onclick='increment_like(1, 3)'> Like </button> <!-- '1' being the id of the post and '3' being its current number of likes -->


Now in your javascript:

<script type='text/javascript'>

//

function increment_like(postId, currentLikes){

var variables = {id: postId, current_value: currentLikes};

$.post("http://ursite.com/increment_likes.php", variables , function(data, status){

alert('updated'); //this happens when the value has been changed

});

}
</script>

IN your increment_likes.php


<?php

$id = $_POST['id'];
$currentValue = intval($_POST['current_value']);

$currentValue++; // add 1 to the no of likes

//connect to database here (you know what i mean)

mysqli_query($conn, "UPDATE likes_table SET no_of_likes = $currentValue WHERE id = $id" or die("something went wrong"; // update the entry by changing the value of likes
mysqli_close($conn) //close the connection
?>

larisoft:

you will have to use ajax to implement the feature

in your html:

<button onclick='increment_like(1, 3)'> Like </button> <!-- '1' being the id of the post and '3' being its current number of likes -->


Now in your javascript:

<script type='text/javascript'>

//

function increment_like(postId, currentLikes){

var variables = {id: postId, current_value: currentLikes};

$.post("http://ursite.com/increment_likes.php", variables , function(data, status){

alert('updated'); //this happens when the value has been changed

});

}
</script>

IN your increment_likes.php


<?php

$id = $_POST['id'];
$currentValue = intval($_POST['current_value']);

$currentValue++; // add 1 to the no of likes

//connect to database here (you know what i mean)

mysqli_query($conn, "UPDATE likes_table SET no_of_likes = $currentValue WHERE id = $id" or die("something went wrong"; // update the entry by changing the value of likes
mysqli_close($conn) //close the connection
?>

This will require jquery of course.

If this is too complicated, then you can use html to go to the post of the page where the like button is located after a reload.

http://mypage.php#postSection1.



You will have to assign a different post section id to each post like this


<section id='postSection1'>
<!-- post here -- >

<! -- post button here -- >

</section



Phew... I really hope any of this makes sense

thanks to everybody that contributed... Liljeez Olumyco seunthomas larisoft

Lexxyfund:
thanks to everybody that contributed... Liljeez



Olumyco



seunthomas



larisoft

Though i didn't do anytin compared to those oda guys oh