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zivenday / learning

知识点脑图，js实现，算法实现。
MIT License
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写一个获取当前url查询字符串中的参数的方法 #40

Open zivenday opened 5 years ago

zivenday commented 5 years ago 
url.toString().split('?')[1].split('&')
zivenday commented 5 years ago 
function urlParam(){
    const param = {};
    location.search.replace(/([^&=?]+)=([^&]+)/g,(m,$1,$2)=> param[$1] = $2);
    return param;
}