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，它包含一个属性名称 days，它包含 10 个数组，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，我有一个定制的一周日历 View ，它是从一个模板创建的，从周一到周五。默认情况下，有 2 周，但用户可以更改他们希望看到的周数。每周由具有唯一 ID “solCol0”、“solCol1” 等的 div 分割....
然后我遍历选定的周数，在本例中我们有 2。
然后遍历 solution 的长度，在本例中为 10。
我只想循环 5 次(一周中的每一天)，然后在 5 次循环后，将 columdId 增加 1 以将详细信息附加到下一周，例如前 5 次循环附加到 "solCol0" 然后接下来的 5 个附加到“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 个循环后递增一个值。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图，它增加了 10 天而不是我想要的 5 天。
最佳答案
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++;
}
}
这使用取模运算符来获取 i/5 的余数。如果它是 0，那么我们知道索引是 5 的倍数。
请注意，这是完全没有必要的，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 chop 为整数，你就会知道someLength包含多少个5。
关于javascript - for循环每5次迭代递增变量？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39793967/