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  • 给定一个时间,获取该时间所在周的周一及周日

    近期在做数据报表,时间处理部分非常繁琐,其中一个需求是给定一个时间需要获取该时间所在周的周一及周末,下面是代码:

    //调用,不传时间就默认为今天
    console.log(getWeek('2017-11-08'));
    //给定一个时间获取本周的星期一及星期日
        function getWeek(time){
            var now='';
            time?(now = new Date(time)):(now = new Date());
            var day = now.getDay();
            if(day == 0){
                day = 7;
            }
            var nowTime = now.getTime(),
                MondayTime = nowTime - (day-1)*24*60*60*1000,    // 周一
                SundayTime =  nowTime + (7-day)*24*60*60*1000;   // 周日var monday = new Date(MondayTime),                   // 格式化周一
                start_Year =monday.getFullYear(),
                start_Month = monday.getMonth()+ 1,
                start_Day = monday.getDate();
            var sunday = new Date(SundayTime),                   // 格式化周末
                end_Year =sunday.getFullYear(),
                end_Month = sunday.getMonth()+ 1,
                end_Day = sunday.getDate();
            if(start_Month<10) {
                start_Month="0"+start_Month;
            }
            if(start_Day < 10) {
                start_Day="0"+start_Day;
            }
            if(end_Month<10) {
                end_Month="0"+end_Month;
            }
            if(end_Day < 10) {
                end_Day="0"+end_Day;
            }
            var start_time = start_Year+'-'+start_Month+'-'+start_Day;
            var end_time = end_Year+'-'+end_Month+'-'+end_Day;
            if(new Date()<new Date(end_time)){
                var today = new Date().toLocaleDateString().replace(/\//g, "-");
                var today_year = today.substring(0,4);
                var today_month = today.substring(5,7);
                var today_day = today.substring(8,10);
                if(today_day<10) {
                    today_day="0"+today_day;
                }
                end_time = today_year+'-'+today_month+'-'+today_day;
            }
            return start_time+' -- '+end_time;
        }
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  • 原文地址:https://www.cnblogs.com/huangtao5921/p/7802540.html
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