javascript - 如何像框架一样检查url参数？

Question

如何制作checkUrl类似于 vue、angular、symfony、laravel 的函数？我需要检查域名主机和路径

有图书馆吗？

if (checkUrl('myShop.com/product/:id',url)) {
    console.log('matches myShop.com/product/pepsi')
}

if (checkUrl('myShop.com/product/:id/**',url)) {
    console.log('matches myShop.com/product/pepsi and myShop.com/product/pepsi/reviews/1/comments')    
}

if (checkUrl('ozon.ru/product/:id/**',url)) {
    console.log('matches ozon.ru/product/pepsi and ozon.ru/product/pepsi/reviews/1/comments')    
}

标签： javascript

Answer 1

找到库https://github.com/visionmedia/page.js 示例用法

function checkUrl(pattern, url) {
    return new Route(pattern).match(url)
}

if (checkUrl('*ozon.ru/product/:id/**',location.href)) {
    console.log('matches ozon.ru/product/pepsi and ozon.ru/product/pepsi/reviews/1/comments')    
}