Yii2 create user friendly URL with parameters

Question

Hi I've to create user friendly URL but when using with parameters it's not working.

Url:

Url::to(['site/index', 'id' => 1]);

url looks like :

localhost/testApplication/frontend/web/index.php/site/index?id=1

/forntend/config/main.php

'urlManager' => [
        'class' => 'yii\web\UrlManager',
        'enablePrettyUrl' => true,
        //'showScriptName' => false,
        'rules' => [

        ],
    ],

I want a output like

localhost/testApplication/frontend/web/index.php/site/index/id/1

and after that how to access id value in controller.

Do you want this for site/index only?

arogachev
– arogachev

2015-03-17 10:56:43 +00:00
Commented Mar 17, 2015 at 10:56 — arogachev
– arogachev, Commented Mar 17, 2015 at 10:56
yes it's only for site/index

Santosh Gaikwad
– Santosh Gaikwad

2015-03-17 11:00:19 +00:00
Commented Mar 17, 2015 at 11:00 — Santosh Gaikwad
– Santosh Gaikwad, Commented Mar 17, 2015 at 11:00
possible duplicate of url management in Yii 2

Alex
– Alex

2015-03-18 09:38:08 +00:00
Commented Mar 18, 2015 at 9:38 — Alex
– Alex, Commented Mar 18, 2015 at 9:38

Timothée Planchais · Accepted Answer · 2015-03-17 10:57:52Z

1

'rules' => [
  'site/index/id/<id:\d+>' => 'site/index'
  //'site/index/<id:\d+>' => 'site/index' Better solution
  //'<controller>/<action>/<id:\d+>' => '<controller>/<action>' Will be applied for all controllers and actions
],

Routing Doc.

And after in your action :

public function actionIndex($id)
{
  ...
}

answered Mar 17, 2015 at 10:57

Timothée Planchais

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3 Comments

Santosh Gaikwad Over a year ago

how can write a rule for more than one parameters?

Timothée Planchais Over a year ago

Url::to(['site/index', 'id' => 1, 'param2' => 'test']); // 'site/index/<id:\d+>/<param2>' => 'site/index' // public function actionIndex($id, $param2)

Santosh Gaikwad Over a year ago

is any rule like site/index/*=>site/index

Luiz Rossi · Accepted Answer · 2021-04-15 18:04:27Z

0

Is really strange, for me using the parameter 'id' always show errors, I needed to change the parameter to 'user_id', but in other parts of the code I could use, really don't know why, but try to rename the parameter.

answered Apr 15, 2021 at 18:04

Luiz Rossi

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