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Editorial Team
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Asked: 2026-06-07T05:57:09+00:00

I’m working with my JS files, what i have now is a unique php

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I’m working with my JS files, what i have now is a unique php file with JS header, if a variable is set it includes the real js file, which is fine.

The “home” page has the script tag for the php-js file:

<head>
    <script type="text/javascript" language="javascript" src="bootstrap.php"></script>
</head>

the bottstrap.php file has something like:

if(isset($hostData) && !empty($hostData)) {
    include('bootstrap.js');
}else {
    echo "document.write('<center><bold>PLEASE DO SOMETHING...!</bold></center>');";
}

all that seems to be fine, however when viewing the source code (CTRL+U) the browser shows the “bootstrap.php” part as a link, if clicked it obviously redirects to http://mydomain/bootstrap.php and the js code can be easily seen, which is exactly what i don’t want…

So my question is, is there any php-way to know if the file is being loaded from browser’s “rendering view” or being loaded from browser’s “source code view” ???

Any help is truly appreciated =)

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1 Answer

  1. Editorial Team

    I’m really sorry for disappearing from here…
    The best solution I decided to implement is quite simple: don’t show ANY URL or PHP files within JS code; so during last months I’ve used a unique PHP file to do all necessary database queries, a stored procedure generates dynamically all the URL’s needed from JS.
    In that way URL’s vary every time and what I’ve named “poor logic” goes free for users to view/copy I don’t mind that while server data is secure.

    THANKS ALL FOR YOUR VALUABLE ANSWERS!!!

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