Question

下面是我的函数，dotkoy变量：iterator.format('DD-MM-YYYY') == props.activity[0] || iterator.format('DD-MM-YYYY') == props.activity[2]我不喜欢，props.activity是数组，我想要props.activity自动增加到数组长度，我想检查iterator.format('DD-MM-YYYY') == props.activity[0]...props.activity[props.acticity.length] 总结props.activity数组中有日期，我想用iterator.format（'DD-MM-YYYY'）加上时间和常数函数（如loadsh时间和常数方法）逐一检查日期。感谢您的任何建议

Calculate = (props) => {

  const constant = x => () => x;
  const times = (n, iterator) => {
    let accum = Array(Math.max(0, n));
    for (let i = 0; i < n; i++) accum[i] = iterator.call();
    return accum;
  };
  let result = [];
  const currentMonth = props.focus.month();
  let iterator = moment(props.focus);
  let i = 0;
  let h = 0;
  let control = []
  while (iterator.month() === currentMonth) {
    if (iterator.weekday() === 0 || result.length === 0) {
      result.push(times(7, constant({})));
    }
    let week = result[result.length - 1];
    week[iterator.weekday()] = {
      date: iterator.date(),
      dotkoy: iterator.format('DD-MM-YYYY') == props.activity[0] || iterator.format('DD-MM-YYYY') == props.activity[2],
      selected: props.selected && iterator.isSame(props.selected, 'day'),
      today: iterator.isSame(moment(), 'day'),
      haftasonumu: iterator.format('ddd') == "Cts" || iterator.format('ddd') == "Paz",
    };
    iterator.add(1, 'day');
  }
  return result;
};

Answer 1

只需使用Array#some。这将对照您提供的回调检查数组的每个成员，如果至少一个成员通过了检查，则返回true；如果没有成员通过，则返回false：

dotkoy: props.activity.some(activity => iterator.format('DD-MM-YYYY') == activity)

如果要避免每次迭代都进行线性扫描，则可以为prop.activity的所有成员预先计算查找并进行检查。在这种情况下，最简单的选择是Set：

const lookupDates = new Set(props.activity);

/* ... */

dotkoy: lookupDates.has(iterator.format('DD-MM-YYYY')),

JavaScript函数递增自动化问题

1 个答案: