获取不带某些 $_GET 变量的当前 URL/URI

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 
);

Answer 1

Yii 1

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

For Yii2:

Yii::$app->request->url

Answer 2

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

这将输出以下格式的内容：

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

Answer 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

Answer 4

要获得绝对当前请求URL（完全如地址栏中所示，使用GET参数和http：//），我发现以下情况很有效：

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

Answer 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

Answer 6

So, you may use

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

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

Answer 7

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.

Answer 8

你肯定在寻找这个

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

Answer 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.

Answer 10

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

Answer 11

Yii2

Url::current([], true);

or

Url::current();

Answer 12

对于Yii2：这应该是更安全的Yii::$app->request->absoluteUrl而不是Yii::$app->request->url

Answer 13

For Yii1

I find it a clean way to first get the current route from the CUrlManager, and then use that route again to build the new url. This way you don't 'see' the baseUrl of the app, see the examples below.

Example with a controller/action:

GET /app/customer/index/?random=param
$route = Yii::app()->urlManager->parseUrl(Yii::app()->request);
var_dump($route); // string 'customer/index' (length=14)
$new = Yii::app()->urlManager->createUrl($route, array('new' => 'param'));
var_dump($new); // string '/app/customer/index?new=param' (length=29)

Example with a module/controller/action:

GET /app/order/product/admin/?random=param
$route = Yii::app()->urlManager->parseUrl(Yii::app()->request);
var_dump($route); // string 'order/product/admin' (length=19)
$new = Yii::app()->urlManager->createUrl($route, array('new' => 'param'));
var_dump($new); string '/app/order/product/admin?new=param' (length=34)

This works only if your urls are covered perfectly by the rules of CUrlManager:)

Answer 14

For Yii2

I was rather surprised that the correct answer for Yii2 was in none of the responses.

The answer I use is:

Url::to(['']),

You can also use:

Url::to()

...but I think the first version makes it more obvious that your intention is to generate a url with the curent request route. (ie "/index.php?admin%2Findex" if you happened to be running from the actionIndex() in the AdminController.

You can get an absolute route with the schema and domain by passing true as a second parameter, so:

Url::to([''], true)

...would return something like "https://your.site.domain/index.php?admin%2Findex" instead.

Answer 15

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.

Answer 16

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

Answer 17

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