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javascript - Checking if a website doesn't permit iframe embed

I am writing a simple lightbox-like plugin for my app, and I need to embed an iframe that is linked to an arbitrary page. The problem is, many web sites (for example, facebook, nytimes, and even stackoverflow) will check to see if is being embedded within a frame and if so, will refresh the page with itself as the parent page. This is a known issue, and I don't think there's anything that can be done about this. However, I would like the ability to know before hand if a site supports embed or not. If it doesn't, I'd like to open the page in a new tab/window instead of using an iframe.

Is there a trick that allows me to check this in javascript?

Maybe there is a server-side script that can check links to see if they permit an iframe embed?

I am developing a browser extension, so there is an opportunity to do something very creative. My extension is loaded on every page, so I'm thinking there's a way to pass a parameter in the iframe url that can be picked up by the extension if it destroys the iframe. Then I can add the domain to a list of sites that don't support iframe embed. This may work since extensions aren't loaded within iframes. I will work on this, but in the meantime....

Clarification:

I am willing to accept that there's no way to "bust" the "frame buster," i.e. I know that I can't display a page in an iframe that doesn't want to be in one. But I'd like for my app to fail gracefully, which means opening the link in a new window if iframe embed is not supported. Ideally, I'd like to check iframe embed support at runtime (javascript), but I can see a potential server-side solution using a proxy like suggested in the comments above. Hopefully, I can build a database of sites that don't allow iframe embed.

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Check x-frame-options header by using following code

$url = "http://stackoverflow.com";
$header = get_headers($url, 1);
echo $header["X-Frame-Options"];

If return value DENY, SAMEORIGIN or ALLOW-FROM then you can't use iframe with that url.


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