I have two versions of my project. For one i use a different CSS and index page and for another i use different. Rest of the things that is controller, models and components are same. The only difference is in view(one or two files) and CSS.
Is there any way to manage this? Like when the URL is URL1 then use CSS1/View1 folder and when url is URL# use CSS2/view2 folder. I have gone through the modules section of Yii but i don't think they are what i need here.
So now I started to use themes. My folder structure is like:
WebRoot
- assests
- css
- images
- protected
- themes
- theme1
-views
-site
-layout
-template
- theme1
-theme2
-views
-site
-layout
-template
In my controller I have done this:
public function init() {
if (SITE_TITLE == 'xxxxx')
Yii::app()->theme = 'theme1';
else
Yii::app()->theme = 'theme2';
parent::init();
}
Which sets theme correctly. but i keep getting file not found as renderer is looking in protected.
I think, you need use themes. Here is documentation: http://www.yiiframework.com/doc/guide/1.1/en/topics.theming
UPDATED after discussion
Trouble in ETwigViewRenderer and it working with themes
If you want to change entire layout, perhaps this is a good way to do:
Setting Layout in Yii
In case you want only to change css they why don't you rely on request uri or domain name?
Yii::app()->getBaseUrl(true)
Related
Is it possible to display a view of a component without iframe and plugin?
(That is to say, if possible with a few lines of PHP and maybe SQL queries?)
EDIT:
To be more clear: I'd like to do it directly in the PHP-Template!
(Would be fine to do it in an article as well, as I have written a
PHP-function showArticle(mixed $ident))
(I'm using Joomla 3.5)
I'd like to do something like
<jdoc:include type="component" view="example" name="position-x" />
or
<?php
show_component('component-name', 'view-name');
?>
you can use this component http://extensions.joomla.org/extension/components-anywhere
Install the plugin and enable it.
Then you can call the component this way {component url/of/the/component}
{component index.php?component=com_example&form=1}
Try to use non-sef urls in the url but sef url will still work.
There is another way to achieve this by calling the model into your controller file this way
JModelLegacy::addIncludePath(JPATH_SITE . '/components/com_example/models', 'ExampleModel');
What this does is it searches the model class starting with ExampleModel in the folder specified. here you can eneter just a path string or array of the directories as the first parameter. Next you have to call the method inside the views file this way
$exmodel = JModelLegacy::getInstance('Something', 'ExampleModel', array('ignore_request' => true));
So here you create an instance of the class object which can be used to get the items from the model this way
$items = $exmodel->getitem();
$this->assignRef('items', $items);
next you can copy the default.php file in the tmpl folder of that component and place it anywhere you like inside your layout file. Basically instead of copying the entire component you are calling the model and getting the data which you can use in your layouts.
In Yii application I used own theme(demirTheme) have modules(retail, corporate), and layouts(main, layout_retail, layout_corporate).
When rendering one page(ex index) I want my app to render page, related layout (retail vs corporate) and then main accordingly. How can I achieve this?
I read layout tutorials and questions no help. They only mention about changing layout path, setting default layout for module, and so on.
I tried
to put all layouts in themes layouts folder,
to put main layout in themes layouts folder and other two in modules layouts folder respectively by their module,
to put all layouts in protected/view/layouts folder,
to put all layouts in modules and one copy of main layout for each module
But can't get it work. I tried changing layout path and other settings.
Is there any way? Have you done layout rendering like this? Any suggestions welcome.
add public $layout='//layouts/layout_retail'; in the controller you want to apply the layout to (so in every controller in the retail module for example). The views should be in "protected/view/layouts"
http://www.yiiframework.com/doc/api/1.1/CController#layout-detail
I am also sure you can put the layouts in the modules map. I however use the above methode for our admin module.
http://www.yiiframework.com/doc/api/1.1/CWebModule#layout-detail
EDIT:
You retail or corporate layout should be then like the code underneath. So the retail or corporate layout would be inside the main layout. Clearly it should contain more then just this, but the content of you layouts should be within $this->beginContent('//layouts/main'); and $this->endContent();
<?php $this->beginContent('//layouts/main'); ?> //main
<div id="content">
<?php echo $content; ?> //viewOfRetail
</div>
<?php $this->endContent(); ?>
For the newcomers: in many help forums of the Internet, when someone asks about theming a module, everyone suggests a path alias to the themes folder, and I disagree with those other forums. I think this is wrong, because it implies modules to be splitted, and modules are supposed to be a black-box that can be used across projects. The advice given in such forums would only be valid if a theme is shared among several modules. If someone wants to "package" a theme inside a module, she can:
-add an init function to the controller of the module
-inside that init, use the class attribute layout and a path alias, like this, supose a module whose id is "Sample":
then you add, to SampleController.php:
public function init() {
//BELOW: it will use the layouts/main.php inside the module.
$this->layouts = "sample.views.layouts.main";
}
Yo can check about path alias here:
http://www.yiiframework.com/doc/guide/1.1/en/basics.namespace
In my Laravel 4 application's root directory, I have a folder themes. Inside the themes folder, I have default and azure.
How can I access view from this themes/default folder in a specific route.
Route::get('{slug}', function($slug) {
// make view from themes/default here
});
My directory structure:
-app
--themes
---default
---azure
I need to load views from localhost/laravel/app/themes/default folder. Please explain this.
This is entirely possible with Laravel 4. What you're after is actually the view environment.
You can register namespace hints or just extra locations that the finder will cascade too. Take a look here
You'd add a location like so:
View::addLocation('/path/to/your/views');
It might be easier if you namespace them though, just in case you have conflicting file names as your path is appended to the array so it will only cascade so far until it finds an appropriate match. Namespaced views are loaded with the double colon syntax.
View::addNamespace('theme', '/path/to/themes/views');
return View::make('theme::view.name');
You can also give addNamespace an array of view paths instead of a single path.
Here I am not accessing my project from public folder. Instead of this I am accessing from project root itself.
I have seen a forum discussion about Using alternative path for views here. But I am little confused about this.The discussed solution was,
You'd add a location like,
View::addLocation('/path/to/your/views');
Then add namespace for theme,
View::addNamespace('theme', '/path/to/themes/views');
Then render it,
return View::make('theme::view.name');
What will be the value for /path/to/ ?
Can I use the same project in different operating system without changing the path?
Yes, we can do this using the following,
Put the following in app/start/global.php
View::addLocation(app('path').'/themes/default');
View::addNamespace('theme', app('path').'/themes/default');
Then call view like the default way,
return View::make('page');
This will render page.php or page.blade.php file from project_directory/app/themes/defualt folder.
I've developed a theme package for laravel 5 with features like:
Views & Asset seperation in theme folders
Theme inheritence: Extend any theme and create Theme hierarcies
Try it here: igaster/laravel-theme
\View::addLocation($directory); works fine but the new right way to do it is using loadViewsFrom($path, $namespace) (available on any service provider).
I have a website with many scripts written in "pure" PHP, i.e. no specific framework has been used to write the files. Furthermore, all the URLs are custom using .htaccess and specific PHP scripts.
For a smooth transition, I would like to start using CodeIgniter for new pages without disrupting access to the old pages, but all the documentation I've seen on CodeIgniter gives the impression that the whole website (perhaps with a few exceptions) needs to be based on the framework.
Would it be possible to use the framework for single pages here and there while leaving old URLs and code intact?
Short answer, yes.
You could access the CI framework from a subfolder, for instance, leaving the existing site untouched.
i.e
www.site.com/my_new_app/controller/method/
where my_new_app is the renamed application folder.
I'm going to go on the assumption that you already have a basic template system in place, and are able to render full pages with your existing site. Since Codeigniter is really just a framework, there's nothing to stop you from using vanilla php, like include, or additional libraries and classes. So, one thing you can do is dump your site into a sub directory in your views folder, then create a "master" controller which does nothing but load full html pages.
class Master extends CI_Controller {
function __construct()
{
parent::__construct();
}
function index()
{
// We're expecting something like "registration/how-to-apply" here
// Whatever your URL is. The .php extension is optional
$args = func_get_args();
$path = 'path_to_my_old_site/'.explode('/', $args);
$this->load->view($path);
}
}
// Then use this in config/routes.php
$route['(:any)'] = 'master/index/$1';
This will route all pages through the master controller. So, yoursite.com/pages/faq will load the file application/views/old_site/pages/faq.php. You can apply different routes as you see fit.
This way, you can take your time migrating to use Codeigniter conventions, one page at a time.
I have two modules, default and mojo.
After the initial bootstraping code which is the same for both of the modules, I want, for example, to use different layouts for each module (Or use different credentials check etc).
Where do I put this: IF(module=='mojo') do this ELSE do that
If you are using Zend_Application (in ZF1.8) then you should be able to use the module specific configuration options to provide this functionality with a as explained in the relevant section in the documentation.
This would require you to set the layout in the config so it looked something like
mojo.resources.layout.layout = "mojo"
anothermodule.resources.layout.layout = "anotherlayout"
The layout would then be set automatically by the bootstrap.
The other alternative is to use a front controller plug-in that implements the preDispatch() method to set the layout based on the module name.
hmm i havent tried this
http://www.nabble.com/Quick-Guide-How-to-use-different-Layouts-for-each-module-to23443422.html#a24002073
the way i did that now was thru a front controller plugin
something like
switch ($request->getModuleName()) {
case "":
// set layout ...
}
I've looked into the subject a couple of days ago, trying to get it to work on bootstrap config alone. The big problem is that all the bootstrap files are loaded, so it gives some weird results in which layout is used.
My conclusion was that you can have the config in place, but you need to work with FrontController plugins or ActionController helpers. If you want to use config set in the application.ini and you want to load the config trough the bootstrap, helpers is the only way to go. From the helper, you can then load the ActionController and on that execute the getInvokeArgs to load the bootstrap. A lot of hastle... :)
Anyway, I've done a small implementation as an example in a blog post: http://blog.keppens.biz/2009/06/create-modular-application-with-zend.html
Goodluck,
Jeroen