How to minify view HTML in Codeigniter 4 (CI 4) - codeigniter

I know we can minify HTML in CI3 through the hook but in CI 4
I have done it by adding a minify function before return on every method.
minifyHTML(view('admin/template/template',$this->data));
Any other way I can do it without using the minify function everywhere?
I also figure out another solution which is to add a template function in BaseConroller which renders all the views. but I have already used view() in many places in the project and its not feasible for me but can be work for others.

In CodeIgniter4 You can use Events to Minify the output (Instead of using Hooks like what we did in CodeIgniter3) , by adding the following code inti app/Config/Events.php file
//minify html output on codeigniter 4 in production environment
Events::on('post_controller_constructor', function () {
if (ENVIRONMENT !== 'testing') {
while (ob_get_level() > 0)
{
ob_end_flush();
}
ob_start(function ($buffer) {
$search = array(
'/\n/', // replace end of line by a <del>space</del> nothing , if you want space make it down ' ' instead of ''
'/\>[^\S ]+/s', // strip whitespaces after tags, except space
'/[^\S ]+\</s', // strip whitespaces before tags, except space
'/(\s)+/s', // shorten multiple whitespace sequences
'/<!--(.|\s)*?-->/' //remove HTML comments
);
$replace = array(
'',
'>',
'<',
'\\1',
''
);
$buffer = preg_replace($search, $replace, $buffer);
return $buffer;
});
}
});
see https://gitlab.irbidnet.com/-/snippets/3

You need the extend your core system classes to be able to do that in a system wide scope.
Every time CodeIgniter runs there are several base classes that are initialized automatically as part of the core framework. It is possible, however, to swap any of the core system classes with your own version or even just extend the core versions.
Two of the classes that you can extend are these two:
CodeIgniter\View\View
CodeIgniter\View\Escaper
For example, if you have a new App\Libraries\View class that you would like to use in place of the core system class, you would create your class like this:
The class declaration must extend the parent class.
<?php namespace App\Libraries;
use CodeIgniter\View\View as View;
class View implements View
{
public function __construct()
{
parent::__construct();
}
}
Any functions in your class that are named identically to the methods in the parent class will be used instead of the native ones (this is known as “method overriding”). This allows you to substantially alter the CodeIgniter core.
So in this case you can look at your system view class and just change it to return the output already compressed.
In your case You might even add an extra param so that the view function can return the output either compressed or not.
For more information about extending core classes in CodeIgniter 4 read this:
https://codeigniter.com/user_guide/extending/core_classes.html#extending-core-classes

In CodeIgniter 4, you can minify the HTML output of your views by using the built-in output compression library.
First, you need to enable output compression in your application's configuration file, which is located at app/Config/App.php.
Then, you need to set the compression level. You can set the compression level by changing the value of compress_output option. The possible values are:
0 : Off (Do not compress the output)
1 : On (Compress output, but do
not remove whitespace)
2 : On (Compress output, and remove
whitespace)
Finally, you need to load the Output Library in your controller, before the output is sent to the browser. You can load it using the following line of code:
$this->load->library('output');

Related

Laravel 8 multilanguage routes and how get the translated link of the same page

I'm trying to make my test app in multilanguage way.
This question has two correlated questions:
First question:
I followed the second answer in How to create multilingual translated routes in Laravel and this help me having a multilanguage site and the route cached, but I've a question and some misunderstanding.
It's a good practice overwrite an app config as they do int the AppServiceProver.php, making:
Config::set('app.locale_prefix', Request::segment(1));
Isn't better to work with the Session::locale in any case?
Second question:
In my case I've two languages, and in the navbar I want to print just ENG when locale is original language, and ITA when session locale is English.
If I'm in the Italian page, the ENG link in the navbar should point to the same English translated page.
Working with the method used in the other question, I hade many problems caused by the:
Config::set('app.locale_prefix', Request::segment(1));
We overwrite the variable in the config file local_prefix, and every time I switch to English language the locale_prefix will change to 'eng' and this sounds me strange, another thing I did is this:
if ( $lang && in_array($lang, config('app.alt_langs')) ){
return app('url')->route($lang . '_' . $name, $parameters, $absolute);
}
We use the alt_langs where are defined only the alternative languages, and this is a problem cause if I pass the local lang, in my case 'it', like lang parameter, this will not be found cause, from the description, the alt_lang should not contain the locale language and you will be able to get only the translated string.
If I change the:
if ( $lang && in_array($lang, config('app.alt_langs')) ){
return app('url')->route($lang . '_' . $name, $parameters, $absolute);
}
in:
if ( $lang && in_array($lang, config('app.all_langs')) ){
return app('url')->route($lang . '_' . $name, $parameters, $absolute);
}
Now using app.all_langs I'm able to choose which URL you want and in which language I want.
How do I get the translated URL?
In the blade file I need to get the translated URL of the page, and if read the other question, we used the $prefix for caching the routes and giving to the route a new name ->name($prefix.'_home'); in this way I can cache all the route and I can call the routes using blade without prefix {{ route('name') }} but, needing the translated url of the actual page a made this on the top of the view:
#php
$ThisRoute = Route::currentRouteName();
$result = substr($ThisRoute, 0, 2);
if ($result =='it' ){
$routeName = str_replace('it_', '', $ThisRoute);
$url = route($routeName,[],true,'en');
} else {
$routeName = str_replace('en_', '', $ThisRoute);
$url = route($routeName,[],true,'it');
}
#endphp
Doing this I get the actual route name that should be it_home I check if start with it_ or en_, I remove the it_ or en_ prefix and I get the translated URL, now you can use the $url as <a href="{{ $url" }}>text</a> cause if I call the {{ route('page') }} I get the link, with the locale language.
This code is not very good, I know, but I written in 5 minutes, need more implementation, and check, but for the moment is just to play with Laravel.
It's a good way?? How can I do it better (except the blade link retrieving)?? Many solution I found used middleware, but I would like to avoid a link in the navbar like mysite.com/changelang?lang=en
Is a good approach overriding the app.locale_prefix?
First
according to your question, it's a bad practice to save the preferences into .env or session because as soon as the session is finished the saved language will be removed also it's common when you need to store any preferences related to your website such as (Color, Font, Language, ...etc) you must store any of them into the cache.
Second
honestly, your code is a very strange and NOT common way and there are two ways to handle what do you need
First
There is a very helpful and awesome package called mcamara it'll help you too much (I recommend this solution).
Second
you can do it from scratch using the lang folder located in the resource folder and you must create files with the same count of the needed languages then use the keys that you'll define into these files into views and you can prefix your routes with the selected language you can use group method like so
Route::group(['prefix' => 'selected_lang'], function() {
Route::get('first_route', [Controller::class, 'your_method']);
});
or you can add the selected language as a query string like so localhost:8000/your_route?lang=en you can follow this tutorial for more info.

Get segment from url CodeIgniter but not first

I know i can get all segments from url like this
Lets say i have this example link
www.example.com/de/products.html
Using url_helper like this:
$data['url'] = $this->uri->uri_string();
I will get value like this
de/products
But i dont need first segment de, only products, the problem is that
i dont know how many segments it will be, i only need to remove the first
Is there possible to forget first segment with url helper in CI?
Try like this...
Use the php's explode() function to make the url string as array.Then apply array's array_shift() function which always removes the first element from array.
Code is looks like as below
$data= $this->uri->uri_string();
$arr=explode('/', $data);
array_shift($arr);
//print_r($arr);
Then use the php's implode() method to get the URI without first segment.Hope it will works...
$uri=implode('/',$arr);
echo $uri;
There is no URL helper in the CI to forget the first segment. However you can easily make a custom one and put #Hikmat's answer below it in the application/helpers/MY_url_helper.php in the Core folder.
e.g.
function my_forget_first_segment() {
$data= $this->uri->uri_string();
$arr=explode('/', $data);
array_shift($arr);
$uri=implode('/',$arr);
return $uri;
}
Before Edit answer.
You need to try this
$second_segment = $this->uri->segment(2);
From Codeigniter documentation -
$this->uri->segment(n);
Permits you to retrieve a specific segment. Where n is the segment number you wish to retrieve. Segments are numbered from left to right. For example, if your full URL is this:
http://example.com/index.php/news/local/metro/crime_is_up
The segment numbers would be this:
1. news
2. local
3. metro
4. crime_is_up
The optional second parameter defaults to NULL and allows you to set the return value of this method when the requested URI segment is missing. For example, this would tell the method to return the number zero in the event of failure:
$product_id = $this->uri->segment(3, 0);
example:
<?php
$data=$this->uri->segment(2);
$val=explode('.', $data);
echo $val[0];
?>

WordPress Pagination not working with AJAX

I'm loading some posts though AJAX and my WordPress pagination is using the following function to calculate paging:
get_pagenum_link($paged - 1)
The issue is that the pagination is getting created through AJAX so it's making this link look like: http://localhost:1234/vendor_new/wp-admin/admin-ajax.php
However the actual URL that I'm trying to achieve is for this:
http://localhost:1234/vendor_new/display-vendor-results
Is there a way to use this function with AJAX and still get the correct URL for paging?
I can think of three options for you:
To write your own version of get_pagenum_link() that would allow you to specify the base URL
To overwrite the $_SERVER['REQUEST_URI'] variable while you call get_pagenum_link()
To call the paginate_links() function, return the whole pagination's HTML and then process that with JS to only take the prev/next links.
#1 Custom version of get_pagenum_link()
Pros: you would have to change a small amount of your current code - basically just change the name of the function you're calling and pass an extra argument.
Cons: if the function changes in the future(unlikely, but possible), you'd have to adjust your function as well.
I will only post the relevant code of the custom function - you can assume everything else can be left the way it's in the core version.
function my_get_pagenum_link( $pagenum = 1, $escape = true, $base = null ) {
global $wp_rewrite;
$pagenum = (int) $pagenum;
$request = $base ? remove_query_arg( 'paged', $base ) : remove_query_arg( 'paged' );
So in this case, we have one more argument that allows us to specify a base URL - it would be up to you to either hard-code the URL(not a good idea), or dynamically generate it. Here's how your code that handles the AJAX request would change:
my_get_pagenum_link( $paged - 1, true, 'http://localhost:1234/vendor_new/display-vendor-results' );
And that's about it for this solution.
#2 overwrite the $_SERVER['REQUEST_URI'] variable
Pros: Rather easy to implement, should be future-proof.
Cons: Might have side effects(in theory it shouldn't, but you never know); you might have to edit your JS code.
You can overwrite it with a value that you get on the back-end, or with a value that you pass with your AJAX request(so in your AJAX request, you can have a parameter for instance base that would be something like window.location.pathname + window.location.search). Difference is that in the second case, your JS would work from any page(if in the future you end-up having multiple locations use the same AJAX handler).
I will post the code that overwrites the variable and then restores it.
// Static base - making it dynamic is highly recommended
$base = '/vendor_new/display-vendor-results';
$orig_req_uri = $_SERVER['REQUEST_URI'];
// Overwrite the REQUEST_URI variable
$_SERVER['REQUEST_URI'] = $base;
// Get the pagination link
get_pagenum_link( $paged - 1 );
// Restore the original REQUEST_URI - in case anything else would resort on it
$_SERVER['REQUEST_URI'] = $orig_req_uri;
What happens here is that we simply override the REQUEST_URI variable with our own - this way we fool the add_query_arg function into thinking, that we're on the /vendor_new/display-vendor-results page and not on /wp-admin/admin-ajax.php
#3 Use paginate_links() and manipulate the HTML with JS
Pros: Can't really think of any at the moment.
Cons: You would have to adjust both your PHP and your JavaScript code.
Here is the idea: you use paginate_links() with it's arguments to create all of the pagination links(well - at least four of them - prev/next and first/last). Then you pass all of that HTML as an argument in your response(if you're using JSON - or as part of the response if you're just returning the HTML).
PHP code:
global $wp_rewrite, $wp_query;
// Again - hard coded, you should make it dynamic though
$base = trailingslashit( 'http://localhost:1234/vendor_new/display-vendor-results' ) . "{$wp_rewrite->pagination_base}/%#%/";
$html = '<div class="mypagination">' . paginate_links( array(
'base' => $base,
'format' => '?paged=%#%',
'current' => max( 1, $paged ),
'total' => $wp_query->max_num_pages,
'mid_size' => 0,
'end_size' => 1,
) ) . '</div>';
JS code(it's supposed to be inside of your AJAX success callback):
// the html variable is supposed to hold the AJAX response
// either just the pagination or the whole response
jQuery( html ).find('.mypagination > *:not(.page-numbers.next,.page-numbers.prev)').remove();
What happens here is that we find all elements that are inside the <div class="mypagination">, except the prev/next links and we remove them.
To wrap it up:
The easiest solution is probably #2, but if someone for some reason needs to know that the current page is admin-ajax.php while you are generating the links, then you might have an issue. The chances are that no one would even notice, since it would be your code that is running and any functions that could be attached to filters should also think that they are on the page you need(otherwise they might mess something up).
PS: If it was up to me, I was going to always use the paginate_links() function and display the page numbers on the front-end. I would then use the same function to generate the updated HTML in the AJAX handler.
This is actually hard to answer without specific details of what and how is being called. I bet you want to implement that in some kind of endless-sroll website, right?
Your best bet is to get via AJAX the paginated page itself, and grab the related markup.
Assume you have a post http://www.yourdomain.com/post-1/
I guess you want to grab the pagination of the next page, therefore you need something like this:
$( "#pagination" ).load( "http://www.yourdomain.com/post-1/page/2 #pagination" );
This can easily work with get_next_posts_link() instead of get_pagenum_link().
Now, in order for your AJAX call to be dynamic, you could something like:
$( "#pagination" ).load( $("#pagination a").attr('href') + " #pagination" );
This will grab the next page's link from your current page, and load its pagination markup in place of the old.
It's also doable with get_pagenum_link() however you'd need to change the $("#pagination a").attr('href') selector appropriately, in order to get the next page (since you'd have more than one a elements inside #pagination

Output of getPaymentHtml() in Magento

I haven't been able to find where does the output of getPaymentHtml() comes from.
Its defined as:
public function getPaymentHtml() {
return $this->getChildHtml('payment_info');
}
I couldn't find out the template for payment_info block.
Basically I want to be able to retrieve credit card type and credit card number in the progress block of checkout.
How do I find out the method names? Something like $this->getCreditCardType()
Edit: OK! I understand that Magento figures out the payment method first which has their corresponding templates which are used to render output. But in progress.phtml of checkout, var_dump( $this instanceof Mage_Payment_Block_Info_Cc ); returns false, so how do I access that in current context?
The Progress block doen't have it's own template for Payment info. Mage_Checkout_Block_Onepage_Payment_Info block uses the selected Payment Method block to output html. Look at the Mage_Checkout_Block_Onepage_Payment_Info::_toHtml() method:
protected function _toHtml()
{
$html = '';
if ($block = $this->getChild($this->_getInfoBlockName())) {
$html = $block->toHtml();
}
return $html;
}
To find the actual template and block for the specific Payment method you use, you need to perform next steps:
First - get model alias for current payment method Mage::getStoreConfig('payment/'.$yourMethod.'/model') and instantiate it using Mage::getModel(alias)
then get block type using $model->getInfoBlockType() - so you'll be able to find the actual Block by it's type
For example for ccSave payment method the info block is Mage_Payment_Block_Info_Ccsave, and template for it is app\design\frontend\base\default\template\payment\info\default.phtml. You'll be able to find all data inside those.
Good luck ;)
For the sake of completeness, exact functions to fetch CC type and last 4 digits of CC number are:
echo Mage::getSingleton('checkout/session')->getQuote()->getPayment()->getCcType();
echo Mage::getSingleton('checkout/session')->getQuote()->getPayment()->getCcLast4();
The block class is declared in layout update XML; see the onepage checkout and multishipping directives from checkout.xml. The actual child block which is used depends on the payment model which is being used, but there is a common template that will be used unless overridden.
Example:
See the generic CC method model Mage_Payment_ModelMethod_Cc
From that see its info block Mage_Payent_Block_Info_Cc...
...which will lead you to the "base" payment info block Mage_Payment_Block_Info which sets a default template.

Does anyone know how to stop T4 (tt) templates from regerating every single file? Is there a way to flag certain files?

So the tt templates will regenerate every file whenever you save. Now, great, it generates files. However, I am making partial classes to extend other classes, but I only need the files that dont already exist for me generated. The ones that exist, I'd like to preserve. So far, I am finding not one solid solution googling the globe...
In my code below, the exception for finding existing files doesnt matter, because the template will start by deleting all files first. Then it regenerates.
It there a method like "onsave" that I can override?
// BEGIN CODE TO GENERATE EXTENSIONS
<#
foreach (EntityType entity in ItemCollection.GetItems<EntityType>().OrderBy(e => e.Name))
{
string fileName = entity.Name + ".Extension.cs";
string filePath = this.Host.TemplateFile.Substring(0,this.Host.TemplateFile.LastIndexOf(#"\"));
filePath = filePath + #"\Extensions\" + fileName;
if((File.Exists(filePath) && PreserveExistingExtensions == false) || !File.Exists(filePath))
{
fileManager.StartNewFile(fileName);
BeginNamespace(namespaceName, code);
bool entityHasNullableFKs = entity.NavigationProperties.Any(np => np.GetDependentProperties().Any(p=>ef.IsNullable(p)));
#>
<#=Accessibility.ForType(entity)#>
<#=code.SpaceAfter(code.AbstractOption(entity))#>partial class
<#=code.Escape(entity)#><#=code.StringBefore(" : ", code.Escape(entity.BaseType))#>
{
}
<#
EndNamespace(namespaceName);
}
}
fileManager.Process();
#>
I do something similar (partial classes) where I have one that is always generated, but the custom one will only be generated if it doesn't exist. This second one is created as starting class for customizations. I'll output two files like so:
MyClass.generated.cs
MyClass.cs
MyClass.cs will never be recreated, unless it doesn't exist. MyClass.generated.cs will always be recreated.
I use the T4toolbox to do this, Oleg Sych has actually made this quite easy.
You can check out some sample T4 Templates I built here. Specifically have a look at this one, it's a good example for generated partial classes where one needs to be created every time, and one is only created if it doesn't exist.
The main thing to look at is this line in the code:
var requestBaseMessageCustom = new MessageTemplate(rootNamespace, serviceName + "Request");
requestBaseMessageCustom.Output.File = "Messages/" + serviceName + "Request.cs";
requestBaseMessageCustom.Output.PreserveExistingFile = true;
requestBaseMessageCustom.Render();
Notice the property called PreserveExistingFile, that's the key.

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