手写实现getUrlParams方法

1.只传递一个参数url

function getUrlParams(url) {
    const params = {};
    const queryString = url.split('?')[1] || '';
    
    if (!queryString) {
        return params;
    }

    const pairs = queryString.split('&');
    return pairs.reduce((acc, pair) => {
        const [key, value] = pair.split('=');
        acc[decodeURIComponent(key)] = decodeURIComponent(value || '');
        return acc;
    }, params);
}

// 示例用法
const url = 'https://example.com/?name=JohnDoe&age=25&city=NewYork';
const params = getUrlParams(url);
console.log(params);
// 输出: { name: 'JohnDoe', age: '25', city: 'NewYork' }

2. 传递两个参数，（url,id）

function getUrlParam(url, id) {
    const queryString = url.split('?')[1] || '';
    if (!queryString) {
        return null;
    }

    const pairs = queryString.split('&');
    const params = pairs.reduce((acc, pair) => {
        const [key, value] = pair.split('=');
        acc[decodeURIComponent(key)] = decodeURIComponent(value || '');
        return acc;
    }, {});

    return params[id] || null;
}

// 示例用法
const url = 'https://example.com/?name=JohnDoe&age=25&city=NewYork';
const id = 'city';
const paramValue = getUrlParam(url, id);
console.log(paramValue); // 输出: 'NewYork'