如何从结果集中跳过假期?

更新时间：2023-02-06 08:54:30

您是这个意思吗?

 function nth(d)  { if (d > 3 && d < 21) return 'th';  switch (d % 10) { case 1:  return "st"; case 2:  return "nd"; case 3:  return "rd"; default: return "th"; } }
function dateToYMD(date) { var strArray=['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']; var d = date.getDate(); var m = strArray[date.getMonth()]; var y = date.getFullYear(); return '' + (d <= 9 ? '0' + d : d) + '-' + m + '-' + y; }
Date.prototype.addDays = function(days) { var date = new Date(this.valueOf()); date.setDate(date.getDate() + days); return date; }
function pad(str) { return (" "+str).slice(-2) }
var cnt = 0, dataSet = [];

function printNextPeriod(startDate, endDate, periodInDays) {
  var numWorkDays = 0;
  var currentDate = new Date(startDate);
  while (numWorkDays < periodInDays && currentDate <= endDate) {
    currentDate = currentDate.addDays(1);
    // Skips friday
    if (currentDate.getDay() !== 5) {
      numWorkDays++;
    }
    if (numWorkDays == periodInDays) {
      numWorkDays = 0;
      cnt++;
      let date = dateToYMD(currentDate);
      let pos = holidays.indexOf(date);
      if (pos != -1) {
        console.log("replace",date,"with",instead[pos])
        date = instead[pos];
      }  
      let treatment = pad(cnt) + nth(cnt) + (cnt == 1 ? " Basic" : " Control") + " Treatment"
      dataSet.push([date, treatment])
    }
  }
}

var holidays = ["2019-02-07" // Thursday
  , "2019-10-08", "2019-12-17",
  "2019-02-05" // Friday
];
var instead = [];
holidays.forEach((hol,i) => {
  let d = new Date(hol);
  let date = d.getDate() + 1
  d.setDate(date); // next day (could be Weekend);
  while (d.getDay() === 5) {
    date = d.getDate()
    date++;
    d.setDate(date); // is any day not friday ok?
  }
  instead.push(dateToYMD(d))
  holidays[i] = dateToYMD(new Date(hol))
})
console.log(JSON.stringify(instead))
var start = new Date("2018-12-15");
var end = new Date("2019-12-31");
var period = 15;
printNextPeriod(start, end, period);

$(document).ready(function() {
  $('#example').DataTable({
    data: dataSet,
    columns: [{ title: "Date" },
              {title: "Frequency"}],
    order: [[1, "asc"]]
  });
});

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://cdn.datatables.net/1.10.19/css/jquery.dataTables.min.css" />
<script src="https://cdn.datatables.net/1.10.19/js/jquery.dataTables.min.js"></script>
<table id="example">
  </div>

Is it possible to skip Holidays also?? Assume i have set of holiday list in database..holidays will not comes to the result date.I want to skip the holiday date and skip to next date is fine

if holiday is 05-Feb-2019 then skip this date and need to show 06-Feb-2019(exclude Fridays)..Then next 30 days will be 13-Mar-2019 (30days calculating from 06 Feb)

for example

var holidays = array("05-Feb-2019", "08-Oct-2019", "17-Dec-2019"); 

function nth(d) {
  if (d > 3 && d < 21) return 'th'; 
  switch (d % 10) {
    case 1:  return "st";
    case 2:  return "nd";
    case 3:  return "rd";
    default: return "th";
  }
}

function dateToYMD(date) { var strArray=['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']; var d = date.getDate(); var m = strArray[date.getMonth()]; var y = date.getFullYear(); return '' + (d <= 9 ? '0' + d : d) + '-' + m + '-' + y; }

Date.prototype.addDays = function(days) {
  var date = new Date(this.valueOf());
  date.setDate(date.getDate() + days);
  return date;
}

var cnt = 0;
function printNextPeriod(startDate, endDate, periodInDays) {
  var numWorkDays = 0;
  var currentDate = new Date(startDate);
  while (numWorkDays < periodInDays && currentDate <= endDate) {
    currentDate = currentDate.addDays(1);
    // Skips friday
    if (currentDate.getDay() !== 5) {
      numWorkDays++;
    }
    if (numWorkDays == periodInDays) {
      numWorkDays = 0;
      cnt++; 
      document.getElementById("first").innerHTML += dateToYMD(currentDate)+"<br/>";
      document.getElementById("second").innerHTML += cnt+nth(cnt)+(cnt==1?" Basic":" Control")+ " Treatment"+"<br/>";

    }
  }
}

var start = new Date("2019-01-01");
var end = new Date("2019-12-31");
var period = 30;
printNextPeriod(start, end, period);

Now the code results like

**Date**
05-Feb-2019   
12-Mar-2019
-----------
12-Nov-2019
17-Dec-2019

Expecting output with skiping holidays

06-Feb-2019(exclude fridays)
13-Mar-2019
----------
so on

function nth(d) {
  if (d > 3 && d < 21) return 'th';
  switch (d % 10) {
    case 1:
      return "st";
    case 2:
      return "nd";
    case 3:
      return "rd";
    default:
      return "th";
  }
}

function dateToYMD(date) {
  var strArray = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec'];
  var d = date.getDate();
  var m = strArray[date.getMonth()];
  var y = date.getFullYear();
  return '' + (d <= 9 ? '0' + d : d) + '-' + m + '-' + y;
}

Date.prototype.addDays = function(days) {
  var date = new Date(this.valueOf());
  date.setDate(date.getDate() + days);
  return date;
}

var cnt = 0;

function printNextPeriod(startDate, endDate, periodInDays) {
  var numWorkDays = 0;
  var currentDate = new Date(startDate);
  while (numWorkDays < periodInDays && currentDate <= endDate) {
    currentDate = currentDate.addDays(1);
    // Skips friday
    if (currentDate.getDay() !== 5) {
      numWorkDays++;
    }
    if (numWorkDays == periodInDays) {
      numWorkDays = 0;
      cnt++;
      document.getElementById("first").innerHTML += dateToYMD(currentDate) + "<br/>";
      document.getElementById("second").innerHTML += cnt + nth(cnt) + (cnt == 1 ? " Basic" : " Control") + " Treatment" + "<br/>";

    }
  }
}

var start = new Date("2019-01-01");
var end = new Date("2020-01-01");
var period = 30;
printNextPeriod(start, end, period);

.period {
  float: left;
  padding: 5px;
  text-align: right;
  font-family: monospace;
}

<div class="period" id="first">**Date**
  <hr/>
</div>
<div class="period" id="second">**Frequency**
  <hr/>
</div>

You mean this?

function nth(d)  { if (d > 3 && d < 21) return 'th';  switch (d % 10) { case 1:  return "st"; case 2:  return "nd"; case 3:  return "rd"; default: return "th"; } }
function dateToYMD(date) { var strArray=['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']; var d = date.getDate(); var m = strArray[date.getMonth()]; var y = date.getFullYear(); return '' + (d <= 9 ? '0' + d : d) + '-' + m + '-' + y; }
Date.prototype.addDays = function(days) { var date = new Date(this.valueOf()); date.setDate(date.getDate() + days); return date; }
function pad(str) { return (" "+str).slice(-2) }
var cnt = 0, dataSet = [];

function printNextPeriod(startDate, endDate, periodInDays) {
  var numWorkDays = 0;
  var currentDate = new Date(startDate);
  while (numWorkDays < periodInDays && currentDate <= endDate) {
    currentDate = currentDate.addDays(1);
    // Skips friday
    if (currentDate.getDay() !== 5) {
      numWorkDays++;
    }
    if (numWorkDays == periodInDays) {
      numWorkDays = 0;
      cnt++;
      let date = dateToYMD(currentDate);
      let pos = holidays.indexOf(date);
      if (pos != -1) {
        console.log("replace",date,"with",instead[pos])
        date = instead[pos];
      }  
      let treatment = pad(cnt) + nth(cnt) + (cnt == 1 ? " Basic" : " Control") + " Treatment"
      dataSet.push([date, treatment])
    }
  }
}

var holidays = ["2019-02-07" // Thursday
  , "2019-10-08", "2019-12-17",
  "2019-02-05" // Friday
];
var instead = [];
holidays.forEach((hol,i) => {
  let d = new Date(hol);
  let date = d.getDate() + 1
  d.setDate(date); // next day (could be Weekend);
  while (d.getDay() === 5) {
    date = d.getDate()
    date++;
    d.setDate(date); // is any day not friday ok?
  }
  instead.push(dateToYMD(d))
  holidays[i] = dateToYMD(new Date(hol))
})
console.log(JSON.stringify(instead))
var start = new Date("2018-12-15");
var end = new Date("2019-12-31");
var period = 15;
printNextPeriod(start, end, period);

$(document).ready(function() {
  $('#example').DataTable({
    data: dataSet,
    columns: [{ title: "Date" },
              {title: "Frequency"}],
    order: [[1, "asc"]]
  });
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://cdn.datatables.net/1.10.19/css/jquery.dataTables.min.css" />
<script src="https://cdn.datatables.net/1.10.19/js/jquery.dataTables.min.js"></script>
<table id="example">
  </div>

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如何从结果集中跳过假期?

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