Get current URL/URI without some of $_GET variables

Question

How, in Yii, to get the current page's URL. For example:

http://www.yoursite.com/your_yii_application/?lg=pl&id=15

but excluding the $GET_['lg'] (without parsing the string manually)?

I mean, I'm looking for something similar to the Yii::app()->requestUrl / Chtml::link() methods, for returning URLs minus some of the $_GET variables.

Edit: Current solution:

unset $_GET['lg'];

echo Yii::app()->createUrl(
  Yii::app()->controller->getId().'/'.Yii::app()->controller->getAction()->getId() , 
  $_GET 
);

Solution 1

Yii 1

Yii::app()->request->url

For Yii2:

Yii::$app->request->url

Solution 2

Yii::app()->createAbsoluteUrl(Yii::app()->request->url)

This will output something in the following format:

http://www.yoursite.com/your_yii_application/

Solution 3

Yii 1

Most of the other answers are wrong. The poster is asking for the url WITHOUT (some) $_GET-parameters.

Here is a complete breakdown (creating url for the currently active controller, modules or not):

// without $_GET-parameters
Yii::app()->controller->createUrl(Yii::app()->controller->action->id);

// with $_GET-parameters, HAVING ONLY supplied keys
Yii::app()->controller->createUrl(Yii::app()->controller->action->id,
    array_intersect_key($_GET, array_flip(['id']))); // include 'id'

// with all $_GET-parameters, EXCEPT supplied keys
Yii::app()->controller->createUrl(Yii::app()->controller->action->id,
    array_diff_key($_GET, array_flip(['lg']))); // exclude 'lg'

// with ALL $_GET-parameters (as mensioned in other answers)
Yii::app()->controller->createUrl(Yii::app()->controller->action->id, $_GET);
Yii::app()->request->url;

When you don't have the same active controller, you have to specify the full path like this:

Yii::app()->createUrl('/controller/action');
Yii::app()->createUrl('/module/controller/action');

Check out the Yii guide for building url's in general: http://www.yiiframework.com/doc/guide/1.1/en/topics.url#creating-urls

Solution 4

To get the absolute current request url (exactly as seen in the address bar, with GET params and http://) I found that the following works well:

Yii::app()->request->hostInfo . Yii::app()->request->url

Solution 5

In Yii2 you can do:

use yii\helpers\Url;
$withoutLg = Url::current(['lg'=>null], true);

More info: https://www.yiiframework.com/doc/api/2.0/yii-helpers-baseurl#current%28%29-detail

View more solutions

Comments

[T.Todua]

possible duplicate of How to remove the querystring and get only the url?

[Sarkhan]

stackoverflow.com/questions/8413062/…

[Sebastian]

This is really helpful. My only consideration is that Yii is using either the 'normal' (?var=value) format, or PATH (/var/value), they are toggled in a config file. That's why links in Yii are constructed using Chtml::link() with an array of $_GET variables.

[Sebastian]

Oh ! So I can... give the whole $_GET array as an input parameter, after unsetting one value :). [leaves for a minute to try it].

[DaveRandom]

@Sebastian This could be easily re-worked to use it in /var/value format - the exact details of how it would be done depend on the situation, but in effect you are just reformatting strings from the input data, so it's not that difficult. You can just do something like foreach ($get as $key => val) $myUrl .= "/$key/$val"; (may need slight alteration depending on exactly how your URLs are formatted).

[Sebastian]

elegant trick with the "/$key/$val"! Another solution: Yii::app()->createUrl(Yii::app()->controller->getId().'/'.Yi‌​i::app()->controller‌​->getAction()->getId‌​(), $_GET ); seems to work (after unsetting $_GET['lg'] :)!

[GusDeCooL]

This should be the correct answer, since the asker want to be done with Yii

[marcovtwout]

This is not what the poster asked for. See the answer below: stackoverflow.com/questions/8413062/…

[marcovtwout]

No need to write code that Yii already provides. Read this: yiiframework.com/doc/api/1.1/CHttpRequest

[marcovtwout]

Hardcoding the url is a bad practise. See this topic about creating url's in general: yiiframework.com/doc/guide/1.1/en/topics.url#creating-urls

[marcovtwout]

Relying on one particular url style is bad and unnessecary. See this topic about creating url's in general: yiiframework.com/doc/guide/1.1/en/topics.url#creating-urls

[Michael Butler]

This is the wrong answer. User needs to EXCLUDE some GET parameters.

[Mahomedalid]

This should be the correct answer, sometimes the url includes not just controller and action, but view, and depends on the route methods.

[Flion]

Yes, great answer! (and indeed the only correct one to the question) Could use an update it for Yii2 too though..

[Flion]

found it: Yii2: Yii::$app->urlManager->createUrl(array_merge([Yii::$app->req‌​uestedRoute], $getParams));

[Vladimir]

This anwer has a lot of up votes but it is misleading, @MichaelButler answer is very good but you can see shorter yii2 specific answer stackoverflow.com/questions/8413062/…

[RN Kushwaha]

This one is the correct answer. But for yii2 it should be Yii::$app->request->pathInfo

[Sukma Saputra]

Yii2: Yii::$app->request->url

[Alexandru Trandafir Catalin]

I think this is wrong. Because createAbsoluteUrl expects a route not a URL. The author's original solution is quite right but a more correct one would be: $this->createUrl($this->getRoute(), $_GET) and before calling it, unset from $_GET the params that you don't wish.

[rob006]

This is terrible answer, don't use it! It will give you a bunch of bugs when your URLs does not match routes. For example if /news/index should display route news/view for News model with slug index. With this answer you will get incorrect URL - it will redirect you to URL for /news/index route instead /news/view.

[Nico Haase]

How does this remove "some" of the variables from the URL?