js中的get请求中，将json格式的对象拼接成复杂的url参数

consturl="/mock/products"constquery={pageIndex:1,pageSize:5}方法一constserialize=function(obj){变量=[];for(varpinobj)if(obj.hasOwnProperty(p)&&obj[p]){ary.推(encodeURIComponent(p)+'='+encodeURIComponent(obj[p]));}返回ary.join('&');};方法二使用antd，当form有undefined时，map方法也会将undefined拼接。需要处理查询。constqueryStr=Object.keys(query).map(key=>query[key]&&`${encodeURIComponent(key)}=${encodeURIComponent(query[key])}`).join('&');方法三constqueryStr=Object.keys(query).reduce((ary,key)=>{if(query[key]){ary.push(encodeURIComponent(key)+'='+encodeURIComponent(query[key]));}returnary;},[]).join('&');url+=`?${queryStr}`;