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Hiding embed links?

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bbalegere

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I have added many videos to my site using youtube and other video sharing sites using the code they have provided.Anybody who views my site can view that code by looking at the source.I want to hide that code.I do not want my viewers to know from which video sharing sites i have embedded videos.
How do i do this?

Please help.
 
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The views expressed on this page by users and staff are their own, not those of NamePros.
GoDaddyGoDaddy
youtube video logo always shows
you might want to get hosted at magnify.net and create your own logo and embed/place it to replace youtube logo.

or you need this script
http://www.clip-share.com/
 
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Actually i am not bothered about the you tube logo.
I just do not want them find the embed link.
For example i have embedded a video from stage6.com
When a user sees the source code he gets the id of the video and embeds it in his site saying he uploaded it.I want to prevent that.
 
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any free alternative?
 
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Thanks.
Please reply if anyone knows more about such techniques.
 
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This script will parse whatever is in the file FILE_NAME.php invisibly onto the page using javascript.


<script language="javascript" type="text/javascript">
window.onload = initJavaScript;

function initJavaScript() {
getPage();
}

function getPage(){
var xmlhttp1=false; //Clear our fetching variable
try {
xmlhttp1 = new ActiveXObject('Msxml2.XMLHTTP'); //Try the first kind of active x objectโ€ฆ
} catch (e) {
try {
xmlhttp1 = new
ActiveXObject('Microsoft.XMLHTTP'); //Try the second kind of active x object
} catch (E) {
xmlhttp1 = false;
}
}
if (!xmlhttp1 && typeof XMLHttpRequest!='undefined') {
xmlhttp1 = new XMLHttpRequest(); //If we were able to get a working active x object, start an XMLHttpRequest
}
var file1 = 'FILE_NAME.php'; //This is the path to the file we just finished making *

xmlhttp1.open('GET', file1, true); //Open the file through GET, and add the page we want to retrieve as a GET variable **
xmlhttp1.onreadystatechange=function() {
if (xmlhttp1.readyState==4) { //Check if it is ready to recieve data
var content1 = xmlhttp1.responseText; //The content data which has been retrieved ***
if( content1 ){ //Make sure there is something in the content variable
document.getElementById('video').innerHTML = content1; //Change the inner content of your div to the newly retrieved content ****
}
}
}
xmlhttp1.send(null) //Nullify the XMLHttpRequest
return;
}
</script>

<div id="video"></div>
 
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Thanks a lot for the code.It is working.
I just want the code to put in FILE_NAME.php so that it displays content only when it parsed by the above code.
If a user directly tries to access FILE_NAME.php then he should not be able to view the content.
 
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smagik, please remember to use [html ][ /html] and [code ][ /code] and [ php][ /php]
Thanks.

http://www.namepros.com/programming-announcements.html
SecondVersion said:
Use the appropriate code tag (e.g. [ php][/php ] for PHP code, [ html][/html] for html code, and [ code][/code ] for any other code - without spaces of course) to make your code easy to read.
 
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