I had one page in Views in CakePHP, it have normal javascript block,
Just inserted:
<script language="JavaScript" type="text/javascript">
---code---
</script>
Inside page, and it was all working okay...
But now... It doesn't show...
How can I change configuration or something to enable showing javascript blocks without CakePHP commands.
Javascript needs data from that page so I can't use outer file,
and it's too long to use $javascript->codeBlock
Is there any way to reconfigure stupid CakePHP to start showing those blocks?
Some files are showing javascript, and it's working all okay, but some of them won't show...
Please help...
If you mean that you want to view the code when the page is displayed, try surrounding it with <pre>...</pre>
If you mean you want the browser to process the code, then provided you are
actually going to that view file and
the code isn't commented out (<!-- ... --> or <?php /* ?> ... <?php */ ?> etc.) and
the code isn't being obliviated by a php conditional (if ... then ... else... endif)
then it will be there. Try Firefox ctrl-u to view the source.
Also try posting the view code here so that we can give you some sort of informed solution.
Related
I'm trying to insert some HTML and JS into an existing dynamic PHP page.
I felt that a switch() statement would be my best alternative.
Allow me to explain;
I have a switch statement embedded within a PHP page and when a certain condition is met I'm including an include().
Here's what my code looks like. (abbreviated)
Within my PHP file that displays the page;
switch($color)
{
case "Red":
include ('red_widgets.php');
break;
}
Then red_widgets.php looks like;
?>
<div class="Shadow">
<script language="JavaScript" src="AScriptThatDisplaysWidgets"></script></div>
<?php
But when I run the page, I'm getting "?>" outputted and not the JS.
This may be obvious to some of you but if you could share with me what I'm doing wrong I'd be most appreciative and I thank you all in advance.
When you include, you're in HTML-mode. Just omit the leading ?> and the trailing <?php and that will work fine.
You do not need to "stop PHP". Just update the appropriate PHP/JS/HTML file and the next time you request it from the server it will deliver the updated file/script output to your browser.
As to not receiving the correct output you have an error in your code.
Is the include file surrounded by <?php and ?> ?
How would I put this in my cakephp default.ctp file?
Im pretty novice at cakephp as ive just started using it.
<script>
$(function()
{
$('#slider-id').codaSlider();
});
</script>
Thanks, in advance.
Although your question is too vague to answer in its current state, you should have a look at the JsHelper,especially Js->buffer(). This allows you to append script in your views and output them all at once in your layout.
http://book.cakephp.org/2.0/en/core-libraries/helpers/js.html#working-with-buffered-scripts
open the app/View/Layout/default.ctp file in your favorite code editor
add the following between your <head> and </head>:
<script>
$(function()
{
$('#slider-id').codaSlider();
});
</script></li>
In CakePHP 2.x there are now things called blocks, and blocks are just chunks of output stored in memory until they are fetched for output.
There use to be a $scripts_for_layout variable that you could put your script into, but now we use the $this->fetch('scripts'); to get any JavaScript needed for the Html.
There are a few ways to inject JavaScript into the scripts block using the HtmlHelper.
To add it to the block, and this can be done in Views or Layouts. Just run this code before you fetch the block.
$this->Html->scriptBlock("$('#slider-id').codaSlider();",array('inline'=>false));
To output the scripts in your layout is easy.
$this->fetch('scripts');
The advantage of this approach is you can add JavaScript from multiple places in CakePHP, but they will be outputted in the layout at the location you desire.
I'm working with my JS files, what i have now is a unique php file with JS header, if a variable is set it includes the real js file, which is fine.
The "home" page has the script tag for the php-js file:
<head>
<script type="text/javascript" language="javascript" src="bootstrap.php"></script>
</head>
the bottstrap.php file has something like:
if(isset($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!!!
I'm learning all this web programming stuff after years writing .EXE Windows programs so bear with me.
I developed a basic .php and mysql website that works fine.
But I went to add javascript code to my index.php and I don't think the javascript code is executing.
My index.php has the form:
<?php
require_once blah blah
call_my_php_functionBlah();
?>
Then I added this code inside the php blocks of the '<\?\php' and "\?>" as follows:
<script type="text/javascript">
// some known-good javascript code that displays an image
</script>
Nothing showed up.
So I thought "ah-HAH, I blew it, all I need to do is -- move the javascript code outside
of the php block, at the bottom of index.php, and surely I'm good to go."
And still, Nothing showed up.
I checked the source of my 'known-good' javascript code and it said 'embed this javascript code
in your HTML file' so I thought "wow, I guess I need an index.html or something here."
So my questions:
1) should my index.php be able to run the javascript block of code?
I'm guessing 'No because index.php executes on the server and javascript runs on the client machine.'
2) How should I architect this if I want to keep my index.php, whose code works fine and I don't want to mess with it?
I'm thinking that this is an extremely basic client/server, php and javascript script organization issue that every web programmer knows how to handle, but like I said, I'm new to all this. I read in the archives about .htaccess etc. etc. but I
bet there's an easier way, and I'm not sure if the stuff I read applies.
the file name extension is completely irrelevant
PHP executes on the server and doesn't care at all about any Javascript
code inside <?php ?> tags must of course be valid PHP code to be executed by PHP
your browser receives whatever the result of your PHP execution is
you can use PHP code to output Javascript or simply have Javascript on the same page outside of <?php ?> tags
only whatever the browser receives matters, so use View Source
look at the browser's Javascript Console to debug client-side Javascript problems
Then I added this code inside the php blocks of the '" as follows:
Dont add your script inside the php block bring it outside php block.
After you are done with script you can reopen php block and write php again
index.php can run javascript, just that You need to echo the javascript code to put it in the page.
Anything that appears inside your php open/close tags has to be echoed or printed to be rendered to the html page. Anything outside your php open/close tags should appear in your html page but whether it works correctly or not is another matter not necessarily related to your php. The php interpreter doesn't run your javascript code, however, so it can't just sit inside your php tags.
Javascript will run inside .php file.
But you have to write outside the tags.
Eg:
index.php
<?php
echo "Helloooooo";
?>
<script>
function TestingMyFirstScript()
{
alert(1)
}
</script>
Javascript will execute in a PHP file but not inside of a PHP block. It executes in the server, yes and anything coming from PHP should be printed out to see. You should have the JS code outside of the PHP block and it can be anywhere in the page e.g.
It depends how to mix/match the code but of course keep it clean and easy to read (and debug).
<?php
// code here
?>
<script type="text/javascript">
// JS here
</script>
<?php
// some more code here
?>
Answer to both of your question is that you dont have to create a separate html file to execute your JS code. You can have HTML, JS, and PHP code in the same file. PHP code inside the PHP tags will be processed on the server and replace with HTML. The server generated HTML will be combined with other HTML present on the .php file and sent to the browser as one HTML.
There must be some error in the JS code which is causing the script to fail.
<?php
require_once blah blah
call_my_php_functionBlah();
?>
<script type="text/javascript">
// some known-good javascript code that displays an image
</script>
<?php
// other php code
?>
Most of the above comments should help you with your PHP + JS problem. However, if you are still getting errors with your output, try using:
alert("breakpoint 1");
//some code
alert("breakpoint 2");
throughout your Javascript function (it will show you where the code is failing). Good for beginners debugging. Also check out http://www.jslint.com/
I have a wordpress theme that I like to duplicate. To make things easier on myself, I'm trying to use the bloginfo function to call my jquery file..like this:
<script type="text/javascript" src="<?php echo bloginfo("template_url"); ?>/js/jquery-1.4.2.min.js"></script>
I just can't get it to work. When I check my source file it looks exactly like the code above.
Can this even be done, or should I just forget it? Thanks guys!
Are you sure the above code is actually in a PHP-file and gets parsed by the server? I can't think of a different reason why PHP-code should just be printed and not executed.
See the Referencing Files From a Template chapter in http://codex.wordpress.org/Theme_Development
Whats the name of this file? It should end in .php.
What does the source look like when you view it from the browser?
Are there other places in the file where you use and is that data correct?
Make sure there are no blank lines at the end of the source file. PHP doesn't like blanks lines at the end of the code.
Two problems: No need for the echo in the WP template tag; should be
<?php bloginfo("template_url"); ?>
Template Tags/info « WordPress Codex
And, why are you trying to include jQuery from the template directory, when it resides in wp-includes/js? You should be using Function Reference/wp enqueue script, if your theme doesn't already include jQuery.
You should try to edit your theme's header.php file directly because custom fields of your theme may not be executed by PHP interpreter.
As I've never used PHP, this is only a guess...
PHP is a server side language, and you're tyring to use it on the client side?