例如，在输入框中给定两个日期:
<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>
<script>
alert(datediff("day", first, second)); // what goes here?
</script>
如何在JavaScript中获得两个日期之间的天数?
当前回答
使用毫秒时要小心。
date.getTime()返回毫秒，用毫秒做数学运算需要包含
日光节约时间(DST) 检查两个日期的时间是否相同(小时，分钟，秒，毫秒) 请确定需要哪些天数差异:2016年9月19日- 2016年9月29日= 1天或2天的差异?
上面评论中的例子是我迄今为止找到的最好的解决方案 https://stackoverflow.com/a/11252167/2091095。但是，如果你想计算所有涉及的天数，则对其结果使用+1。
function treatAsUTC(date) {
var result = new Date(date);
result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
return result;
}
function daysBetween(startDate, endDate) {
var millisecondsPerDay = 24 * 60 * 60 * 1000;
return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}
var diff = daysBetween($('#first').val(), $('#second').val()) + 1;
其他回答
您可以使用UnderscoreJS来格式化和计算差异。
演示https://jsfiddle.net/sumitridhal/8sv94msp/
var startDate = moment(“206 -08- 29t23:35:01”); var endDate = moment(“206 -08- 30t23:35:01”); 游戏机。log (startDate); 游戏机。log (endDate); var结果= endDate。diff(startDate, hours, true); 文件。 文档全身appendChild(文档。createTextNode (resultHours)); 身体(白色空间:pre;font-family: monospace;) “https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.5.1/moment.min.js”< script src = > / < script >
我来找这个小工具在里面你会找到这个的函数。这里有一个简短的例子:
<script type="text/javascript" src="date.js"></script>
<script type="text/javascript">
var minutes = 1000*60;
var hours = minutes*60;
var days = hours*24;
var foo_date1 = getDateFromFormat("02/10/2009", "M/d/y");
var foo_date2 = getDateFromFormat("02/12/2009", "M/d/y");
var diff_date = Math.round((foo_date2 - foo_date1)/days);
alert("Diff date is: " + diff_date );
</script>
function formatDate(seconds, dictionary) {
var foo = new Date;
var unixtime_ms = foo.getTime();
var unixtime = parseInt(unixtime_ms / 1000);
var diff = unixtime - seconds;
var display_date;
if (diff <= 0) {
display_date = dictionary.now;
} else if (diff < 60) {
if (diff == 1) {
display_date = diff + ' ' + dictionary.second;
} else {
display_date = diff + ' ' + dictionary.seconds;
}
} else if (diff < 3540) {
diff = Math.round(diff / 60);
if (diff == 1) {
display_date = diff + ' ' + dictionary.minute;
} else {
display_date = diff + ' ' + dictionary.minutes;
}
} else if (diff < 82800) {
diff = Math.round(diff / 3600);
if (diff == 1) {
display_date = diff + ' ' + dictionary.hour;
} else {
display_date = diff + ' ' + dictionary.hours;
}
} else {
diff = Math.round(diff / 86400);
if (diff == 1) {
display_date = diff + ' ' + dictionary.day;
} else {
display_date = diff + ' ' + dictionary.days;
}
}
return display_date;
}
我也有同样的问题，但如果你在SQL查询上做的话会更好:
DateDiff(DAY, StartValue,GETDATE()) AS CountDays
查询将自动生成一个列CountDays
从DatePicker小部件使用formatDate怎么样?您可以使用它来转换时间戳格式的日期(从01/01/1970开始的毫秒)，然后做一个简单的减法。
