Искам да използвам променливата $thumb и в друг file, но не знам от къде идва. Не мога да разбера по кода... Ако може помогнете.
Код:
<script language='javascript'>
function showthumb(divid,thumb) {
document.getElementById(divid).style.display = 'block';
document.getElementById(divid).innerHTML='<img src="'+thumb+'" border=0 width=130 height=97>';
}
function hidethumb(divid) {
document.getElementById(divid).style.display = 'none';
}
</script>
<style>
.thumbs {
border: 1px solid #CCCCCC;
background: #EAEAEA;
width: 130px;
height: 97px;
position: absolute;
display: none;
}
</style>
<div align='center'>
<table width='600' border=0>
<tr>
<td>
<div align='left'>
<table width='100%' border=1 bordercolor='#FFFFFF'>
<?
$i=0;
foreach($videos as $index=>$video) {
$type = $video['type'];
$title = $video['title'];
$link = $video['link'];
if(trim($title)&&trim($link)) {
if(($i%2)==0)
echo "<tr style='background: #EAEAEA' height='25'>";
else
echo "<tr height='25'>";
$i++;
if($type=='dm')
$type = "<img src='dm.png' border=0 title='daily motion video' alt='daily motion video'>";
if($type=='yt')
$type = "<img src='yt.png' border=0 title='youtube video' alt='youtube video'>";
if($type=='mc')
$type = "<img src='mc.png' border=0 title='metacafe video' alt='metacafe video'>";
if($type=='gv')
$type = "<img src='gv.jpg' border=0 title='google video' alt='google video'>";
$url = urlencode($link);
$title2 = urlencode($title);
$thumb = $video['thumb'];
$link = "<a href=\"#\" onclick=\"watchvideo('$url','$title2');\" ><img src='button2.png' border='0'></a>";
echo "<td width='400'><div style='cursor: pointer' onmouseout=\"hidethumb('div$i');\" onmouseover=\"showthumb('div$i','$thumb');\"><img src='tick.gif'> $title</div><div class='thumbs' id='div$i'></div></td><td align='center' width='100'>$type</td><td align='center' width='100'>$link</td>";
echo "</tr>";
}
}
?>
</table>
</div>
</td>
</table>
</div>