PHP: how to determine if the browser supports javascript in PHP?

Question

I am building an AJAX deep-linked site.

I want PHP to load all the HTML code of the page if the user is trying to access the site with a Javascript non-supported browser or if it is a search crawler. Basically PHP will return the whole page.

On the contrary, when the user is trying to access the site with Javascript supported browser, I want PHP to return only the template code, and let Javascript (AJAX) take care of the rest. Basically PHP will only load design elements and let Javascript populate them with content.

I looked into PHP's get_browser() function, however it seems it is not such a reliable tool. What is the industry's practice see if the browser supports Javascript or it is a search crawler using PHP?

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Background:

Why I want the site to have this behavior.

Since I want the home page to load just by loading the address: example.com, which does not send any query to PHP, PHP returns the HTML code of the home page. This however causes issues when the user tries to load the following page: example.com#foo. So, for this example, PHP will return the home page and once the home page is loaded, Javascript (AJAX) will change the content around so that it shows proper content for #foo. This will make the user to see the home page, therefore load time will be slower and user-experience will not be so nice. However if my PHP script can figure out that if the use with Javascript supported browser is trying to load the page, it will only return the template of the web site, which has no content) and the javascript will populate that template with content whatever is supposed to be displayed for #foo. On the other hand, if the Javascript non-separated browser or a crawler will try to access the page example.com#foo, home page will be returned.

I am using SWFaddress (http://www.asual.com/swfaddress/) library for the deep-linking.

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Edit

Thank you guys. I did not think of using <noscript></noscript> before.

Here is what I decided to do. PHP by default will load pages such as example.com or example.com#foo (which is essentially the same as example.com from PHP's point of view since fragments by definition are not sent to the server) blank (just visual template) with <noscript> tag inside for the content of the home page. This way users with javascript will not see the home page and AJAX will populate the content of the page according to the #foo fragment. On the other hand, search crawlers and users without javascript will see a home page.

Thank you again. I think this is pretty simple and elegant solution. If you have any further suggestions, please post a comment or another answer.

Answer 1

You can't do this using PHP. What you can do though is use a noscript tag to redirect to another php page if they don't have javascript:

<noscript>
<meta http-equiv="refresh" content="0; URL=nojavascript.php">
</noscript>

Answer 2

It's not possible to accomplish this in the way you're trying to do it.

It's rare that someone has JS turned off and doesn't know it.

PHP doesn't get passed anything after #, only javascript can do anything with that. So even if PHP could determine if the browser has javascript turned on then it still couldn't read # anyways.

Answer 3

You could include a link inside some <NOSCRIPT> tags that point the user to something like example.com#foo?javascript=disabled.

Unfortunately, browsers do not report whether JS is enabled or not, so there's no way to know from a simple HTTP GET whether or not you should send JS reliant pages.