Javascript 如何以 HH:MM:SS 格式比较两个时间字符串？
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How can I compare two time strings in the format HH:MM:SS?
提问by Null Pointer
I have two time strings in HH:MM:SSformat. For example, str1contains 10:20:45, str2contains 5:10:10.
我有两个HH:MM:SS格式的时间字符串。例如，str1包含10:20:45、 str2包含5:10:10。
How can I compare the above values?
如何比较上述值？
回答by Andy E
As Felix Klingsaid in the comments, provided your times are based on a 24 hour clock (and they should be if there's no AM/PM) and provided they are always in the format HH:MM:SSyou can do a direct string comparison:
正如Felix Kling在评论中所说，如果您的时间基于 24 小时制（如果没有 AM/PM 则应该是），并且如果它们始终采用HH:MM:SS您可以进行直接字符串比较的格式：
var str1 = "10:20:45",
str2 = "05:10:10";
if (str1 > str2)
alert("Time 1 is later than time 2");
else
alert("Time 2 is later than time 1");
回答by Evgeny Shadchnev
Date.parse('01/01/2011 10:20:45') > Date.parse('01/01/2011 5:10:10')
> true
The 1st January is an arbitrary date, doesn't mean anything.
1 月 1 日是一个任意日期，没有任何意义。
回答by Suresh Krishnamoorthi
Try this code for the 24 hrs format of time.
尝试使用此代码获取 24 小时格式的时间。
<script type="text/javascript">
var a="12:23:35";
var b="15:32:12";
var aa1=a.split(":");
var aa2=b.split(":");
var d1=new Date(parseInt("2001",10),(parseInt("01",10))-1,parseInt("01",10),parseInt(aa1[0],10),parseInt(aa1[1],10),parseInt(aa1[2],10));
var d2=new Date(parseInt("2001",10),(parseInt("01",10))-1,parseInt("01",10),parseInt(aa2[0],10),parseInt(aa2[1],10),parseInt(aa2[2],10));
var dd1=d1.valueOf();
var dd2=d2.valueOf();
if(dd1<dd2)
{alert("b is greater");}
else alert("a is greater");
}
</script>
回答by hien
Dateobject in js support comparison, set them same date for compare hh:mm:ss :
Datejs 中的对象支持比较，将它们设置为相同的日期进行比较 hh:mm:ss ：
new Date ('1/1/1999 ' + '10:20:45') > new Date ('1/1/1999 ' + '5:10:10')
> true
回答by Kishan Patel
You can easily do it with below code:
您可以使用以下代码轻松完成：
Note: The second argument in RegExp is 'g' which is the global search flag. The global search flag makes the RegExp search for a pattern throughout the string, creating an array of all occurrences it can find matching the given pattern. Below code only works if the time is in HH:MM:SS format i.e. 24 hour time format.
注意：RegExp 中的第二个参数是“g”，它是全局搜索标志。全局搜索标志使 RegExp 在整个字符串中搜索模式，创建一个数组，其中包含它可以找到的与给定模式匹配的所有匹配项。下面的代码仅适用于时间为 HH:MM:SS 格式即 24 小时时间格式的情况。
var regex = new RegExp(':', 'g'),
timeStr1 = '5:50:55',
timeStr2 = '6:17:05';
if(parseInt(timeStr1.replace(regex, ''), 10) < parseInt(timeStr2.replace(regex, ''), 10)){
console.log('timeStr1 is smaller then timeStr2');
} else {
console.log('timeStr2 is smaller then timeStr1');
}
回答by niksvp
Try this code.
试试这个代码。
var startTime = "05:01:20";
var endTime = "09:00:00";
var regExp = /(\d{1,2})\:(\d{1,2})\:(\d{1,2})/;
if(parseInt(endTime .replace(regExp, "")) > parseInt(startTime .replace(regExp, ""))){
alert("End time is greater");
}
回答by patapizza
You could compare the two values right after splitting them with ':'.
您可以在用 ':' 分割后立即比较这两个值。
回答by Ghasem Aladaghlou
I improved this function from @kamil-p solution. I ignored seconds compare . You can add seconds logic to this function by attention your using.
我从@kamil-p 解决方案改进了这个功能。我忽略了 seconds compare 。您可以通过注意您的使用向此功能添加秒逻辑。
Work only for "HH:mm" time format.
仅适用于“HH:mm”时间格式。
function compareTime(str1, str2){
if(str1 === str2){
return 0;
}
var time1 = str1.split(':');
var time2 = str2.split(':');
if(eval(time1[0]) > eval(time2[0])){
return 1;
} else if(eval(time1[0]) == eval(time2[0]) && eval(time1[1]) > eval(time2[1])) {
return 1;
} else {
return -1;
}
}
example
例子
alert(compareTime('8:30','11:20'));
Thanks to @kamil-p
感谢@kamil-p
回答by GameBoy
I think you can put it like this.
我想你可以这样说。
var a = "10:20:45";
var b = "5:10:10";
var timeA = new Date();
timeA.setHours(a.split(":")[0],a.split(":")[1],a.split(":")[2]);
timeB = new Date();
timeB.setHours(b.split(":")[0],b.split(":")[1],b.split(":")[2]);
var x= "B is later than A";
if(timeA>timeB) x = "A is later than B";
document.getElementById("demo").innerHTML = x;<p id="demo"></p>回答by Sublime Laravel Dev
Best way to compare times using convert into ms
使用转换为毫秒来比较时间的最佳方法
var minutesOfDay = function(m){
return m.minutes() + m.hours() * 60;
}
return minutesOfDay(now) > minutesOfDay(end);
