Nie*_*and 1 javascript compare date range
我有多个日期范围.我想检查它们是否在javascript中重叠.只有两个很容易,我使用:
if(start_times1 <= end_times2 && end_times1 >= start_times2) {}
Run Code Online (Sandbox Code Playgroud)
但是当有超过2个日期范围时,公式是什么?
Pau*_* S. 17
您可以使用嵌套for循环arguements
function dateRangeOverlaps(a_start, a_end, b_start, b_end) {
if (a_start <= b_start && b_start <= a_end) return true; // b starts in a
if (a_start <= b_end && b_end <= a_end) return true; // b ends in a
if (b_start < a_start && a_end < b_end) return true; // a in b
return false;
}
function multipleDateRangeOverlaps() {
var i, j;
if (arguments.length % 2 !== 0)
throw new TypeError('Arguments length must be a multiple of 2');
for (i = 0; i < arguments.length - 2; i += 2) {
for (j = i + 2; j < arguments.length; j += 2) {
if (
dateRangeOverlaps(
arguments[i], arguments[i+1],
arguments[j], arguments[j+1]
)
) return true;
}
}
return false;
}
Run Code Online (Sandbox Code Playgroud)
小智 6
这是保罗发布的内容的精炼版本:
脚本:
function dateRangeOverlaps(a_start, a_end, b_start, b_end) {
if (a_start < b_start && b_start < a_end) return true; // b starts in a
if (a_start < b_end && b_end < a_end) return true; // b ends in a
if (b_start < a_start && a_end < b_end) return true; // a in b
return false;
}
function multipleDateRangeOverlaps(timeEntries) {
let i = 0, j = 0;
let timeIntervals = timeEntries.filter(entry => entry.from != null && entry.to != null && entry.from.length === 8 && entry.to.length === 8);
if (timeIntervals != null && timeIntervals.length > 1)
for (i = 0; i < timeIntervals.length - 1; i += 1) {
for (j = i + 1; j < timeIntervals.length; j += 1) {
if (
dateRangeOverlaps(
timeIntervals[i].from.getTime(), timeIntervals[i].to.getTime(),
timeIntervals[j].from.getTime(), timeIntervals[j].to.getTime()
)
) return true;
}
}
return false;
}
Run Code Online (Sandbox Code Playgroud)
小智 5
下面的代码来自我的项目,也许对你有帮助:
function dateRangeOverlaps(startDateA, endDateA, startDateB, endDateB) {
if ((endDateA < startDateB) || (startDateA > endDateB)) {
return null
}
var obj = {};
obj.startDate = startDateA <= startDateB ? startDateB : startDateA;
obj.endDate = endDateA <= endDateB ? endDateA : endDateB;
return obj;
}Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
11861 次 |
| 最近记录: |