Onmouseover Current URL Snapshot On This PHP Script

Question Description

I don’t know how to handle this Programming question and need guidance.

Hi folks, ok I have this script below which allows users of channels to show up as offline or online.

It works flawlessly, on the webpage is here, its in a module on he right hand side called channels.

Ok so I am wanting there to be a popout image of the twitch channel relating to the users link, so people can hover over the links to the different channels etc and they can see the current image, the online users would really be the most important ones.

Ok, this site has an example of this: (top right hand corner)

Ok guys, can this be done, is there something simple I can adde for the current code to pickup the image, you will see a line where the users are added to the code below, as in their username on twitch:





font-family:Verdana, Arial, Helvetica;

font-size: 70%;

color: #825a15;






<style type="text/css">

a:link {



a:visited {



a:hover {




<FONT color=#825a15 face="tahoma" size="2pt">



* @version   0.1

* @copyright (C) 2012/2013 Cees Middel (Avon)

* @Inspired by Syi

* @license   GNU/GPL v3



* array with all stream names i.e."iksf".

* To add more users just add their channel names with comma seperation

* This is all there is, no futher editing required


$members = array("aoe2hdozclan","manbeast_aoe","brannin","pain1800","usk_apocalipce","resonance22","williamskovhansen","ninzie90","snezzey","tyrant_theviper");

// This variable becomes one long url with the channel names stringed up behind it

// This url then fetches a json file from twitch with all the selected channels information

$userGrab = "";

//I use this array to compare with the members array. All users in this arrat are substracted from the members array and hence are //"offline"

$checkedOnline = array (); 

foreach($members as $i =>$value){

$userGrab .= ",";

$userGrab .= $value;



//grabs the channel data from streams

$json_file = file_get_contents($userGrab, 0, null, null);

$json_array = json_decode($json_file, true);

//get's member names from stream url's and checks for online members

foreach($members as $i =>$value){

$title = $json_array[$i]['channel']['channel_url'];

$array = explode('/', $title);

$member = end($array);

$viewer = $json_array[$i] ['stream_count'];

onlinecheck($member, $viewer);

$checkedOnline[] = signin($member);




//checks if player streams are online

function onlinecheck($online, $viewers)


//If the variable online is not equal to null, there is a good change this person is currently streaming

if ($online != null)


                    echo '<a href="'.$online.'" target="_blank" title="'.$value.' IS STREAMING LIVE & HAS '.$viewers.' WATCHING RIGHT NOW">> <strong>'.$online.'</strong></a>';

echo '&nbsp <img src="/images/online.png"><strong></strong> Online</br>'; 

                    echo '<img src="/images/viewers.png"><strong> Viewers:</strong> '.$viewers.'</br>';



//This funcion add's online channel names to the checked online array

function signin($person){

if($person != null){

return $person;



return null;






<FONT color=#825a15 face="tahoma" size="1.8pt">


//This part list all the people currently offline. Here the array with online users is compared with the total users.

//online users are then removed from the total users array.

foreach ($members as $i => $value1) {

foreach($checkedOnline as $ii => $value2){

if($value1 == $value2){





//print a nice list with people that can't currently be bothered with streaming their games

foreach ($members as $i => $value) {

echo '<a href="'.$value.'" target="_blank" title="'.$value.' IS OFFLINE RIGHT NOW"> <strong>'.$value.'</strong></a>';

echo '&nbsp<img src="/images/offline.png"> <strong></strong> Offline</br>';






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Final Answer

OK. so you want to get a picture from user's current link? Well. I would suggest the following way(Maybe not the best):
1. Create a database with the following columns:

2. Every time the user go to a spefic action, say they open the css.php page. You want to show a css picture in this case.
Get the image(from any html tag if you want or just hard code one to get) and save it to the database.

3. Ajax call/JSON from database and get the picture to show in the pop-over box

henry (1070)
University of Virginia

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