for 循环每 5 次迭代递增变量？

Question

我不是 100% 确定如何表达这个问题，所以请随意将问题标题更改为有意义的内容。

我有一个对象solution，其中包含一个days包含 10 个数组的属性名称，请参见下面的示例

{
    "sameShiftHolds": true,
    "sameStaffHolds": true,
    "sameRoomHolds": true,
    "days": [{
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": false
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": false
    }, {
        "availableStaff": [
            [0, 1, 4, 3, 5, 9, 22, 44],
            [0, 1, 4, 3, 5, 9, 22, 44],
            [4, 8, 7]
        ],
        "availableRooms": [
            [3, 6, 77, 89, 23],
            [3, 6, 77, 89, 23],
            [2, 7, 9]
        ],
        "suitableStaff": [
            [22, 44],
            [22, 44],
            [4]
        ],
        "suitableRooms": [
            [89, 23],
            [22, 44],
            [2]
        ],
        "ValidStartDate": true
    }]
}

在我的网站上，我有一个自定义的一周日历视图，它是根据模板创建的，从星期一到星期五。默认情况下，有 2 周，但用户可以更改他们想要查看的周数。每个星期都被一个具有唯一 id 的 div 分割"solCol0"，"solCol1"等等......

然后我循环选择选定的周数，在这种情况下，我们有2. 和然后通过长度环路solution这是10在这种情况下。我只想循环 5 次（对于一周中的每一天），然后在 5 次循环之后，将 columdId 增加 1 以将详细信息附加到下一周，例如附加前 5 个循环"solCol0"，然后附加接下来的 5个循环到“ solCol1”，如果用户选择了超过 2 周，到 3，solution长度将增加到 15，因此接下来的 5 个循环将附加到“'solCol2”` 等...

抱歉，如果这不是很清楚，通常我只需要在每 5 次循环后增加一个值。任何帮助将不胜感激。

loadSolutionStartRows: function(dates, solution) {
    var self = this;
    for (var i = 0; i < dates.length; i++) {
        var columnId = "#solCol" + i;
        var startDate = moment(dates[i], 'Do MMM');
        var rowDates = [];
        var iterate = 5;
        for (var d = 0; d < solution.days.length; d++) {
            //Every 5 loops - columnId = "#solCol" + i + 1;
            rowDates.push(moment(startDate).format('ddd (Do MMM)'));
            startDate.add(1, 'days');
            var selectedDate = rowDates[d];

            var statusClass;
            var statusIconClass;
            if (solution.days[d].ValidStartDate === true) {
                statusClass = "sxpTableHeaderIconGreenStatus";
                statusIconClass = "octicon " + "octicon-check";
            }
            if (solution.days[d].ValidStartDate === false) {
                statusClass = "sxpTableHeaderIconRedStatus";
                statusIconClass = "octicon " + "octicon-x";
            }


            $(columnId).append(self.solutionTableRow({
                rowId: i + 1,
                date: selectedDate,
                statusClass: statusClass,
                statusIconClass: statusIconClass,
                trainerCountEarly: 1,
                trainerListEarly: 1,
                roomCountEarly: 1,
                roomListEarly: 1,
                trainerCountLate: 1,
                trainerListLate: 1,
                roomCountLate: 1,
                roomListLate: 1
            }));
        }
    }
}

},

使用上面的代码我实现了下面的图像，它增加了 10 天而不是我想要的 5 天。

Answer 1

您将希望有一个每 5 次迭代递增的变量，因此您需要检查索引是否为 5 的倍数：

var foo = 0;

for (var i; i < someLength; ++i) {
  if (i % 5 === 0) {
    foo++;
  }
}

这使用模运算符来获取 的余数i / 5。如果它是 0，那么我们知道索引是 5 的倍数。

请注意，这完全没有必要，您也可以这样做

Math.floor(someLength / 5);

或者更简洁：

someLength / 5 | 0;

两者都会将除法的结果截断为整数，并且您将知道someLength包含多少个 5 。