I've created my own templating/viewing engine to use with Codeigniter. In it I'm able to specify certain css/js files to use with a specific view. I assign the file names in an array, which will then get looped through while echoing the necessary <link href="X"..., <script type="X"..., etc for the respective file type in the header file of the template.
The problem is that I can't seem to use the resources I'm trying to include. The CSS/JS files aren't working even though they're being included and embedded and everything looks right in terms of the syntax in the HTML source code.
My theory is that because I'm using echo to actually print the link/script object into the HTML, that it's actually not really an object that HTML can recognize? Kind of like trying to echo an object in PHP - it doesnt work.
Any advise?
It does not matter if you use plain HTML of php generated code. For the browser it is all the same.
You need to check your source code, and check if the scripts you include are accessible. So copy/paste the src="blabla" from your source code from your browser, and paste it in the address-bar and see what happens.
It is definitely not PHP's fault.
Related
I downloaded a Template + CSS File for a Website that I'm Building, the template worked well until I tried to break it down and put every code in its own file (for easy modification and editing in the future).
So, when I cut the head part which included (Title + Meta Data .. etc ), and put it in its own file, and replaced it (for sure) with an include() function, I lost the CSS styles and returned to the basic & standard style (Black & white with no extra format .. etc)
Where did I Go wrong? Knowing that here is the include function that I've used:
<?php
include 'files/head.php';
?>
With an URL like file:///C:/xampp/htdocs/test6/index.php PHP is NOT executed. You must run it with apache being involved. Currently you are opening your PHP script as a regular txt or html file - it is just passed to browser without processing.
In order to make include function work you must run it with apache. As you are using xamp, I think you should simply open it with URL like http://localhost/test6/index.php In this case, apache will get that request and pass it to PHP. PHP engine will interpret your PHP script and "replace" include files/head.php with a content of head.php.
If everything is Ok, after pressing Ctrl+U (or looking at HTML with Developer Tools or Firebug) you should see a content of head.php instead of <?php include ....
Please note that css files should be linked with relative URL like css/screen.css. Or absolute URL like http://localhost/test6/css/screen.css. like Search for relative and absolute URLs in google for more info.
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 =)
In short, no. You can't hide your script source from your users. The best you can do is obfuscate it using tools like YUICompressor.
There's no way you can hide the javascript code. It needs to be executed by the client, and even if you try to hide it by formatting your code badly, tools like firebug can easily introspect the code and pull out the code.
To be honest I don't think you can actually hide it like that. I'm assuming the best thing you've got to go on is the useragent string but I'm assuming if you "view source" in a browser it would still send the regular headers.
The only way I can think of adding the JS include without it appearing when in view source mode is to actually load the external file via javascript (you could even break the path of the js file into variables so it isn't really human readable) which I would not advise.
If someone wants to get at your javascript they will there no is way of avoiding it.
and the js code can be easily seen, which is exactly what i don't want...
You don't want the JS to be seen, but you do want to use it???
There IS something wrong with your code though if you want the js file to be used in your page.
You need to include / require the file:
<script type="text/javascript" language="javascript" src="<?php include bootstrap.php ?>"></script>
Otherwise the browser will load the contents of the bootstrap file, but you want to run the code inside it (which can only be done at the server).
Also:
change:
include('bootstrap.js');
to
echo bootstrap.js;
EDIT
by re-reading your question (and other answers) that's exactly what you want: make your JS code invisible (correct me if wrong).
The answer to that is: No cannot be done.
You can try to obfuscate the code but it will take someone who wants to see it seconds to 'decode'.
Try using the $_SERVER["HTTP_referer"], which have the url that called this file.
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!!!
Before I ask this question I must point out that I have tried to search for EVERYTHING!
My question is how can I run javascript from an external file instead of inside my php / html. What I'm trying to do is.
function ClearForm() {
document.form.message.value="";
}
function comeBack(){
if (document.form.message.value == "") {
document.form.message.value="What's on your mind?";
}
}
I have included<script type="text/javascript" src="javascript.js"></script> in the <head> and I have a file in the root called javascript.js and my php file is in the root too so that shouldn't be the problem! But how do I run that pieces of code you see above in the javascript.js file instead of in my php file. It work's fine if I have it in the php file but I want to separate things!
I have also tried to give the form / input field an id and then use getElementById in the external JavaScript file.
But as you can see and hear I'm kinda new to JavaScript so I'm apparently doing something wrong here.
If the above code is the only thing in your Javascript.js file, then you need to call the functions to run the code.
You've included the external Javascript file correctly - however, because all of your JS is included within functions, these function/s must be called before the code will run.
A call to 'ClearForm()' or 'comeBack()' from within your PHP file should run the code.
That JS file will have to be in the same folder as your PHP page.
Test whether the file is found or not by adding this line at the top of the js file
alert('js file found OK!');
document.form is an array, so if you only have one form use:
document.forms[0]
Also depending on which browser you use, find and install some Developer Tools to help you identify these errors.
You have declared those functions in the <head>. All fine.
The question is when do you want to call/run those functions?
If you simply want to run them at the end of the page, then you can add another external javascript file and include it using <script src="my_external_file.js"> right before the </body> tag.
Otherwise, you have to declare onXXX handlers, like onLoad() for the document, onClick() for certain elements, onSubmit() for forms, etc. These, too, can be declared in an external file, but specified after the relevant elements are loaded.
I am using a WP template that allows me to incorporate arbitrary HTML. Unfortunately, I have to use this particular widget and can't use other WP widgets.
I have on my webserver /some/path/serve_image.php that spits out a random HREF'd IMG SRC with a caption and some other info from a MySQL query.
Now...how can I say "take that output and treat it as HTML"? If I just put "/some/path/serve_image.php" I get that literal string.
I tried:
<script type="javascript" src="/some/path/serve_image.php"></script>
but that didn't work. I tried changing everything in serve_image.php to be document.write() calls and that didn't seem to work either. I'm not the world's greatest JS guy...
So if I have a URL on the net that spits out some HTML and I want to include that HTML in my web page, what's the best way to do that? Sort of like what Google does with Adsense - you source their show_ads.js.
Why no? Add
header('Content-Type: application/javascript');
And output JavaScript Like:
echo("var image = \"".$images[array_rand($images)]."\";");
echo("$('img.randim').attr('src', image);
No. JavaScript and PHP are two completely separate languages. In fact, if it was JavaScript, you aren't even loading it the right way.
<script type="text/javascript"></script>
The way you're trying to do it would throw a parse error, because it would try to use the PHP as JavaScript. Some browsers would even reject it, because PHP files have a text/html MIME type, while JavaScript should be application/javascript.
PHP has to be done server side, so loading it in the client just doesn't work.
What I think you're after is this:
<?php
require('/some/path/serve_image.php');
?>
Just place that wherever you want the image to be.
PHP files can be used as external javascript files. Basically make a php file output valid javascript and use that php file as your javascript file: http://www.javascriptkit.com/javatutors/externalphp.shtml . Can this be done with cakephp since we don't specify php files in the browser but rather a directory based on controllers and their actions?
Late answer, but anyway, this is how I did it.
When linking to external javascript file, don't forget to set inline to false like the one shown below:
$this->Html->script('scriptname', array('inline' => false));
Sure, as long as you output valid JS, id does not matter what the URL looks like and what is behind that URL.
When you link a javascript file with
$this->Html->script('scriptname');
all that happens is that a tag is created in the HTML
<script type="text/javascript" src="path/to/webroot/js/scriptname.js"></script>
So, you can link whatever you'd like.