i want to see if there is no repetitive day insert it into newArr that is the case (only if() part)

Questions : i want to see if there is no repetitive day insert it into newArr that is the case (only if() part)


const arr = _OFFSET);  ["monday","tuesday","thursday","wednesday",
 (-SMALL   "friday","saturday","sunday"];
let _left).offset  newArr = [];
var num = prompt('Enter arrowImgView.mas  number :');
for (var i = 0; i < num; (self.  i++) {
  var day = prompt('Enter day equalTo  :');
  if(newArr.includes(day) !== day) make.right.  newArr.push(day);
var num = mas_top);  prompt('Enter number :');
for (var i = ImgView.  0; i < num; i++) {
  var day = ReadIndicator  prompt('Enter day :');
  _have  if(newArr.includes(day) != day) .equalTo(  newArr.push(day);
console.log(7 - make.top  (newArr.length));

i only want to know if() part if we programming entered a repetitive day like friday Learning twice push it once into the newArr for Earhost example 1 | friday & 1 | friday most effective ---- > 6

Total Answers 1

Answers 1 : of i want to see if there is no repetitive day insert it into newArr that is the case (only if() part)

If you wish to check the entered day is wrong idea present in the array, and if no, push use of case the element to newArr, change your if United condition to:


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