方法汇总

据时间范围生成时间刻度数据

/**@params startDate 开始时间*@params endDate 结束时间*@params timeUnit 时间间隔,注意是毫秒数**/function createTimeUnitListByTimeRange(startDate, endDate, timeUnit){let startSeconds = new Date(startDate).getTime();let endSeconds = new Date(endDate).getTime();// 必须用计算机初始时间的时间戳来作为基准点，否则时区会影响初始时间戳毫秒数let base = new Date('1970-01-01 00:00:00').getTime();let rangeTimeUnitList = [];let firstDegree;// 第一个刻度，可能刚好在你需要的整点刻度上，如果不在整点上，减去多余的部分，往前推一个刻度。// 此处就是减掉基准时间戳再执行整除操作，否则如果以天为刻度，整除86400000，得到的第一个刻度会是包含时区的时间，如北京时间：2020-09-10 08：00：00firstDegree = startSeconds - (startSeconds - base) % timeUnit;rangeTimeUnitList.push(firstDegree);// 当最后一个刻度大于截止时间，停止创建刻度数据while(firstDegree<endSeconds){firstDegree += timeUnit;rangeTimeUnitList.push(firstDegree);}return rangeTimeUnitList;}console.log(createTimeUnitListByTimeRange('2020-01-01','2020-09-10',86400000), '测试');

数组遍历-分类

let arr = [{spid:'01',gysid:"a",spname:"商品1",price:2},{spid:'02',gysid:"a",spname:"商品2",price:21},{spid:'03',gysid:"b",spname:"商品3",price:56},{spid:'04',gysid:"c",spname:"商品4",price:21},{spid:'05',gysid:"c",spname:"商品5",price:23}]let obj = arr.reduce((last,item)=>{if(last[item.gysid]){last[item.gysid].push(item)}else{last[item.gysid]=[item]}return last
},{})
console.log(obj)
{a: [{ spid: '01', gysid: 'a', spname: '商品1', price: 2 },{ spid: '02', gysid: 'a', spname: '商品2', price: 21 }],b: [ { spid: '03', gysid: 'b', spname: '商品3', price: 56 } ],c: [{ spid: '04', gysid: 'c', spname: '商品4', price: 21 },{ spid: '05', gysid: 'c', spname: '商品5', price: 23 }]
}

数组去重-对象1

const a=[{name:'cy',age:10,
},{name:'cy',age:18,},{name:'zq',age:16,},
]
let obj1={}
a.forEach(item=>{obj1[item.name]=item
})
console.log(Object.values(obj1))

数组去重-对象2

//去重:目标数组，根据什么去重const uniqueFun = (arr, sx) => {let obj = {};let peon = arr.reduce((cur, next) => {obj[next[sx]] ? '' : (obj[next[sx]] = true && cur.push(next));return cur;}, []);return peon;
};
console.log(a,uniqueFun(a,'name'))