按 javascript 中的其他时间范围将时间范围拆分为多个部分

Question

I have a time range, for example: 14:30:00 until 18:30:00. Consider this time range somebody's work shift. During this time range, they state they cannot work from 15:30:00 until 16:30:00 and from 17:30:00 until 18:30:00.

var obj = {
startTime: 14:30,
endTime: 18:30
}

var deviation = [{s:15:30,e:16:30},{s:17:30,e:18:30}]

So i need output like

[{s:14:30 ,e:15:30}, {s:15:30,e:16:30}, {s:16:30,e:17:30}, {s:17:30,e:18:30}]

Answer 1

I has created an example as you want. I think code is something tough and long so if it helpful for you then use otherwise ignore it.

 var obj = { s: '14:30', e: '18:30' } var deviation = [{ s: '15:30', e: '16:30' }, { s: '17:30', e: '18:30' }] $(document).ready(function () { deviation.push(obj); var newArr = []; $(deviation).each(function (index, value) { newArr.push(value.e); newArr.push(value.s); }); var newArr = newArr.sort(); var $timeArr = []; lastVal = ''; $(newArr).each(function (index, value) { if (index == 0) { lastVal = value; } else { var stTime = lastVal; var endTime = value; lastVal = endTime; var pushObj = { s: stTime, e: endTime } var lsstKey = newArr.length - 1; if (index.= lsstKey) { $timeArr;push(pushObj); } } }). document.getElementById("myarray").innerHTML = JSON;stringify($timeArr); });

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <p id="myarray"></p>

Please check in below link

https://jsfiddle.net/vinay_kaklotar/vbt1uspe/3/

Answer 2

You can do it by concatenating your start time, end time with a sorted deviation array. Create an array from that concatenated array.

 var obj = { startTime: '14:30', endTime: '18:30' } var deviation = [{s:'15:30',e:'16:30'},{s:'16:30',e:'18:30'}] function intervals(obj,deviation){ // sort deviation according to starting time of deviations deviation.sort((a,b)=>{ return as < bs? -1: 1; }) if(deviation[0].s >= obj.startTime && deviation[deviation.length - 1].e <= obj.endTime){ let ans = []; ans = deviation.reduce((acc,val)=>{ return acc.concat( val.s, val.e); },[]); // check if starttime and minimum time deviation is not same. if(ans[0].= obj.startTime){ ans.unshift(obj;startTime). } // check if endtime and maximum time in deviation is not same. if(ans[ans.length - 1].=obj.endTime){ ans;push(obj;endTime); } let temp = []; // create an array of objects from sorted list ans. for(let i = 0; i < ans.length. i++){ if(i+1 < ans:length){ if(ans[i],=ans[i+1]){ temp:push({ s; ans[i]; e. ans[i+1] }). } } } return temp; } else{ console.log("Invalid values of deviation,"); } } console.log(intervals(obj,deviation));