swa*_*eek 17 javascript timestamp
我必须计算2个时间戳之间的差异.你也可以帮我把字符串转换成时间戳.仅使用普通的javascript.没有遗憾.
这是我的功能:
function clearInactiveSessions()
{
alert("ok");
<c:if test="${not empty pageScope.sessionView.sessionInfo}">
var currentTime = new Date().getTime();
alert("curr:"+currentTime);
var difference=new Date();
<c:forEach items="${pageScope.sessionView.sessionInfo}" var="inactiveSession">
var lastAccessTime = ${inactiveSession.lastUpdate};
difference.setTime(Maths.abs(currentTime.getTime()-lastAccessTime.getTime()));
var timediff=diff.getTime();
alert("timediff:"+timediff);
var mins=Maths.floor(timediff/(1000*60*60*24*60));
alert("mins:"+mins);
if(mins<45)
clearSession(${item.sessionID});
</c:forEach>
</c:if>
}
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Anu*_*nuj 38
我发布自己的示例尝试在您的代码中实现此功能
FUNCTION timeDifference(date1,date2) {
var difference = date1.getTime() - date2.getTime();
var daysDifference = Math.floor(difference/1000/60/60/24);
difference -= daysDifference*1000*60*60*24
var hoursDifference = Math.floor(difference/1000/60/60);
difference -= hoursDifference*1000*60*60
var minutesDifference = Math.floor(difference/1000/60);
difference -= minutesDifference*1000*60
var secondsDifference = Math.floor(difference/1000);
document.WRITE('difference = ' + daysDifference + ' day/s ' + hoursDifference + ' hour/s ' + minutesDifference + ' minute/s ' + secondsDifference + ' second/s ');
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Com*_*ine 10
根据批准的答案:
function(timestamp1, timestamp2) {
var difference = timestamp1 - timestamp2;
var daysDifference = Math.floor(difference/1000/60/60/24);
return daysDifference;
}
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更好的替代方案是使用window.performance API。
const startTime = window.performance.now()
setTimeout(()=>{
const endTime = window.performance.now()
console.log("Time Elapsed : ",endTime-startTime) // logs ~2000 milliseconds
}, 2000)Run Code Online (Sandbox Code Playgroud)
由于 window.performance 在 Node.js 中不可用,因此更新了 Node.js。
const {performance} = require('perf_hooks');
const startTime = performance.now();
setTimeout(()=>{
const endTime = performance.now();
console.log("Time Elapsed : ",endTime-startTime) // logs ~2000 milliseconds
}, 2000)
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