Get current URL/URI without some of $_GET variables

问题

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 
);

回答1:

Yii 1

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

For Yii2:

Yii::$app->request->url

回答2:

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

This will output something in the following format:

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

回答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

回答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

回答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

回答6:

I don't know about doing it in Yii, but you could just do this, and it should work anywhere (largely lifted from my answer here):

// Get HTTP/HTTPS (the possible values for this vary from server to server)
$myUrl = (isset($_SERVER['HTTPS']) && $_SERVER['HTTPS'] && !in_array(strtolower($_SERVER['HTTPS']),array('off','no'))) ? 'https' : 'http';
// Get domain portion
$myUrl .= '://'.$_SERVER['HTTP_HOST'];
// Get path to script
$myUrl .= $_SERVER['REQUEST_URI'];
// Add path info, if any
if (!empty($_SERVER['PATH_INFO'])) $myUrl .= $_SERVER['PATH_INFO'];

$get = $_GET; // Create a copy of $_GET
unset($get['lg']); // Unset whatever you don't want
if (count($get)) { // Only add a query string if there's anything left
  $myUrl .= '?'.http_build_query($get);
}

echo $myUrl;

Alternatively, you could pass the result of one of the Yii methods into parse_url(), and manipulate the result to re-build what you want.

回答7:

So, you may use

Yii::app()->getBaseUrl(true)

to get an Absolute webroot url, and strip the http[s]://

回答8:

You are definitely searching for this

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

回答9:

Something like this should work, if run in the controller:

$controller = $this;
$path = '/path/to/app/' 
  . $controller->module->getId() // only necessary if you're using modules
  . '/' . $controller->getId() 
  . '/' . $controller->getAction()->getId()
. '/';

This assumes that you are using 'friendly' URLs in your app config.

回答10:

Yii2

Url::current([], true);

Url::current();

回答11:

$validar= Yii::app()->request->getParam('id');

回答12:

For Yii2: This should be safer Yii::$app->request->absoluteUrl rather than Yii::$app->request->url

回答13:

Try to use this variant:

<?php echo Yii::app()->createAbsoluteUrl('your_yii_application/?lg=pl', array('id'=>$model->id));?>

It is the easiest way, I guess.

回答14:

Most of the answers are wrong.

The Question is to get url without some query param .

Here is the function that works. It does more things actually. You can remove the param that you don't want and you can add or modify an existing one.

/**
 * Function merges the query string values with the given array and returns the new URL
 * @param string $route
 * @param array $mergeQueryVars
 * @param array $removeQueryVars
 * @return string
 */
public static function getUpdatedUrl($route = '', $mergeQueryVars = [], $removeQueryVars = [])
{
    $currentParams = $request = Yii::$app->request->getQueryParams();

    foreach($mergeQueryVars as $key=> $value)
    {
        $currentParams[$key] = $value;
    }

    foreach($removeQueryVars as $queryVar)
    {
        unset($currentParams[$queryVar]);
    }

    $currentParams[0] = $route == '' ? Yii::$app->controller->getRoute() : $route;

    return Yii::$app->urlManager->createUrl($currentParams);

}

usage:

ClassName:: getUpdatedUrl('',[],['remove_this1','remove_this2'])

This will remove query params 'remove_this1' and 'remove_this2' from URL and return you the new URL

回答15:

echo Yii::$app->request->url;

来源：https://stackoverflow.com/questions/8413062/get-current-url-uri-without-some-of-get-variables

标签

php

frameworks

uri

Yii

current-page