2017-03-14 16:10
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I need to do an url address that looks like website/user/username where username is coming from database. I've tried to reach it by placing a $user variable in my action function as a parameter and result looks like this website/action?user=username. But it looks kind of hinty and ugly. How can I get a desirable result?

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我需要做一个看起来像 website / user / username 的网址< em> username 来自数据库。 我试图通过在我的动作函数中放置一个$ user变量作为参数来达到它,结果看起来像 website / action?user = username 。 但它看起来有点简洁和丑陋。 我怎样才能得到理想的结果?

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  • dongre6227 2017-03-14 16:45

    First you need to configure url rules in config/web.php.

    'components' => [
        'urlManager' => [
            'enablePrettyUrl' => true,
            'showScriptName' => false,
            'enableStrictParsing' => false,
            'rules' => [
                "user/<username:\w+>"=> "controller/action"


    And add url match condition in rules array like "user/<username:\w+>"=> "controller/action"

    Create. htaccess file with below url condition in web folder

    RewriteEngine on
    # If a directory or a file exists, use it directly
    RewriteCond %{REQUEST_FILENAME} !-f
    RewriteCond %{REQUEST_FILENAME} !-d
    # Otherwise forward it to index.php
    RewriteRule . index.php

    And generate urls with Url helper class or Html class like below

    echo Url::to(['controller/action', 'username' => 'jack']);
    echo Html::a('Profile', ['controller/action', 'username' => 'jack'], ['class' => 'profile-link']) 

    Note:- controller name, action name and username should match with Url condition. Which we define in rules array.

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