How do you echo out current URL in Cake\'s view?
for CakePHP 3.x You can use UrlHelper
:
$this->Url->build(null, true) // output http://somedomain.com/app-name/controller/action/params
$this->Url->build() // output /controller/action/params
Or you can use PaginatorHelper
(in case you want to use it in javascript or ...):
$this->Paginator->generateUrl() // returns a full pagination URL without hostname
$this->Paginator->generateUrl([],null,true) // returns a full pagination URL with hostname
I use $this->here
for the path, to get the whole URL you'll have to do as Juhana said and use the $_SERVER
variables. There's no need to use a Cake function for this.
for CakePHP 3:
$this->Url->build(null, true) // full URL with hostname
$this->Url->build(null) // /controller/action/params
for cakephp3+:
$url = $this->request->scheme().'://'.$this->request->domain().$this->request->here(false);
will get eg: http://bgq.dev/home/index?t44=333
I prefer this, because if I don't mention "request" word, my IDE gives warning.
<?php echo $this->request->here; ?>
API Document: class-CakeRequest
Edit: To clarify all options
Current URL: http://example.com/en/controller/action/?query=12
// Router::url(null, true)
http://example.com/en/controller/action/
// Router::url(null, false)
/en/controller/action/
// $this->request->here
/en/controller/action/
// $this->request->here()
/en/controller/action/?query=12
// $this->request->here(false)
/en/controller/action/?query=12
// $this->request->url
en/controller/action
// $_SERVER["REQUEST_URI"]
/en/controller/action/?query=12
// strtok($_SERVER["REQUEST_URI"],'?');
/en/controller/action/
Getting current URL for CakePHP 3.x ?
In your layout :
<?php
$here = $this->request->here();
$canonical = $this->Url->build($here, true);
?>
You will get the full URL of the current page including query string parameters.
e.g. http://website.example/controller/action?param=value
You can use it in a meta tag canonical if you need to do some SEO.
<link rel="canonical" href="<?= $canonical; ?>">