增加变量每5次迭代的for循环？
增加变量每5次迭代的for循环？
提问于 2016-09-30 14:32:43
我不能百分之百确定如何表达这个问题，所以请随意将问题标题改为有意义的东西。
我有一个对象solution，它包含一个包含10个数组的属性名days，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，我有一个自定义的周日历视图，它是从一个模板创建的，从周一到周五开始。默认情况下，有两个星期，但用户可以更改他们希望看到的周数。每周由一个带有唯一id "solCol0"、"solCol1"等的div组成.
然后我循环循环选择的周数，在本例中我们有2。然后循环遍历solution的长度，在本例中是10。我只想循环5次(对于每周的每一天)，然后在5次循环之后，将columdId增加1，以便将细节附加到下一周，例如，前5个循环被追加到"solCol0"，接下来的5个循环被追加到"solCol1“，如果用户选择了超过2周，那么solution长度将增加到15，因此接下来的5个循环将追加到”solCol2“等。
对不起，如果这不是很清楚，一般情况下，我只需在每5个循环之后增加一个值。任何帮助都将不胜感激。
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。
回答 4
Stack Overflow用户
回答已采纳
发布于 2016-09-30 14:45:33
您将希望有一个变量，该变量每5次迭代一次就递增一次，因此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++;
}
}这使用模运算符来得到i / 5的剩余部分。如果它是0，那么我们就知道索引是5的倍数。
请注意，这是完全没有必要的，你也可以这样做。
Math.floor(someLength / 5);或者更简洁地说：
someLength / 5 | 0;两者都会将除法的结果截断为整数，并且您将知道5的someLength包含多少个。
Stack Overflow用户
发布于 2016-09-30 14:45:42
您可以使用Math.floor(i/5)。它只是将i除以5，然后将其舍入。
这将返回从0到4的i值的1，从5到9的i的值的1，等等。
在你的例子中：
var columnId = "#solCol" + Math.floor(i/5);Stack Overflow用户
发布于 2016-09-30 14:45:13
对于循环的每一步，检查迭代器是否可被5 if (i % 5 == 0)整除。如果是的话，增加columnId变量
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/39793967
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