blocks|key|2344896|text|perl样式:+12\d{3}/01\d/0-3\d+0-2\d(？：0-5\d){2}|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2344897|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

perl style: [12]\d{3}/[01]\d/[0-3]\d [0-2]\d(?::[0-5]\d){2}

blocks|key|940985|text|如果您只想验证语法，下面是POSIX正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|940986|[0-9]{1,4}/[0-9]{1,2}/[0-9]{1,2}+[0-9]{1,2}:[0-9]{1,2}:[0-9]{1,2}|code-block|syntax|javascript|940987|但是，如果您想检查语义，我将使用您选择的语言处理字符串，有太多的情况您无法用正则表达式(如闰年/秒、夏时制等)覆盖。|940988|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you just want to validate the syntax, here is the POSIX regex:

<pre><code>[0-9]{1,4}/[0-9]{1,2}/[0-9]{1,2} [0-9]{1,2}:[0-9]{1,2}:[0-9]{1,2}
</code></pre>

But if you want to check the semantics, I would process the string using your language of choice, there are too many cases you cannot cover with regular expressions (like leap years/seconds, daylight savings, etc)

blocks|key|2344989|text|Regular-Expressions.info是关于RegEx的一个很好的信息来源，它也有一个好的教程。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2344990|entityMap|0|LINK|mutability|MUTABLE|url|http://www.regular-expressions.info/|1|http://www.regular-expressions.info/tutorial.html^0|0|O|0|1E|2|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

<a href="http://www.regular-expressions.info/" rel="nofollow noreferrer">Regular-Expressions.info</a> is a nice source of information about RegEx, it also has a good <a href="http://www.regular-expressions.info/tutorial.html" rel="nofollow noreferrer">tutorial</a>.

blocks|key|2345009|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2345010|(\d\d)?\d\d\/\d\d?/\d\d?+\d\d?:\d\d:\d\d|code-block|syntax|javascript|2345011|可以验证语法，但正如balpha指出的那样，这并不意味着它是一个有效的日期。|2345012|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Using

<pre><code>(\d\d)?\d\d\/\d\d?/\d\d? \d\d?:\d\d:\d\d
</code></pre>

could validate the syntax, but as balpha points out, that doesn't make it a valid date.

blocks|key|818700|text|您应该考虑不使用正则表达式来执行此操作，而是使用正确的格式字符串在DateTime中运行字符串。这样你就可以确保它确实是一个有效的时间戳，而不仅仅是看起来像它的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|818701|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You should consider not doing this with regular expressions, but rather just run the string through DateTime with the proper format string. That way you can ensure that it is indeed a valid timestamp and not just something that looks like it.

blocks|key|818727|text|为此，我建议使用达特杰。自己解析日期是不必要的，Regex也不足以验证时间戳。使用datejs，您可以在一个日期中解析字符串，如果它无效，您将获得null：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|818728|Date.parse("2009/06/29+13:30:10",+"yyyy/MM/dd+HH:mm:ss");|code-block|syntax|javascript|818729|entityMap|0|LINK|mutability|MUTABLE|url|http://www.datejs.com/^0|8|3|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I would recommend to use <a href="http://www.datejs.com/" rel="noreferrer">Datejs</a> for this. Parsing the date yourself is not necessary and a Regex is not enough to validate a timestamp. With datejs you could parse the string in a date and you'll get null if its invalid:

<pre><code>Date.parse("2009/06/29 13:30:10", "yyyy/MM/dd HH:mm:ss");
</code></pre>

blocks|key|940522|text|function+validateTimestamp(timestamp)+{

++++if+(!/\d{4}\/\d{2}\/\d{2}+\d{2}:\d{2}:\d{2}/.test(timestamp))+{
++++++++return+false;
++++}

++++var+split+=+timestamp.split(/[%5E\d]%2B/);

++++var+year+=+parseFloat(split[0]);
++++var+month+=+parseFloat(split[1]);
++++var+day+=+parseFloat(split[2]);

++++var+hour+=+parseFloat(split[3]);
++++var+minute+=+parseFloat(split[4]);
++++var+second+=+parseFloat(split[5]);

++++return+hour+<+25+&&+minute+<+61+&&+second+<+61+&&+month+<+13+&&+day+<+32;

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|940523|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>function validateTimestamp(timestamp) {

 if (!/\d{4}\/\d{2}\/\d{2} \d{2}:\d{2}:\d{2}/.test(timestamp)) {
 return false;
 }

 var split = timestamp.split(/[^\d]+/);

 var year = parseFloat(split[0]);
 var month = parseFloat(split[1]);
 var day = parseFloat(split[2]);

 var hour = parseFloat(split[3]);
 var minute = parseFloat(split[4]);
 var second = parseFloat(split[5]);

 return hour &lt; 25 &amp;&amp; minute &lt; 61 &amp;&amp; second &lt; 61 &amp;&amp; month &lt; 13 &amp;&amp; day &lt; 32;

}
</code></pre>

blocks|key|818764|text|下面是我今天早些时候编写的用于验证字符串的正则表达式，格式与您提到的类似：YYYY-MM-DD+hh:mm:ss。它没有指出一些糟糕的日期(例如2月30日)，但可能比在每个位置使用简单的\d略好一些。要注意的事情：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|818765|您可以只指定一个日期，一个时间，或者同时指定日期%2B时间。|ordered-list-item|818766|时间可以采用12小时或24小时的格式。|818767|秒是可选的|818768|上午/下午是任选的
字符串dateAndTimeRegex+=+"%5E\s*“//忽略空格"(”//开始日期"20123“/年:2010年、2011年.到2039年"\W”//分隔符；通常为"-“(0-9\1012)+//月份:0至9，或00至09，或10至12+"\W”//分隔符之间的月份和日；通常为"-“(0120-9区)+//日:0至9，或00至29，或30，或31+")?”//结束可选日期"\s?“//可选空格"(”//+start+of+time“(010-9+x+20-3)”//小时:0至9，或00至19，或20至23+"\W“//分隔符在小时和分钟之间；通常为":“(0-5)”//分钟:+00到59+"(“//开始秒(可选)”)+"\W“//分隔符在分钟和秒之间；通常为":“(0-5)”/秒:+00至59“？”//结束可选秒"(\s*AaPp)?“//可选AM、am、PM、pm+")?”//可选时间的结束“”\*$“；//尾空白|818769|@kyrias的一条评论暗示，一旦我们进入2020年，这个regex将在几个月内失效。根据您使用它的方式，您需要将"2010-9“更改为其他内容。|818770|例如，如果要验证当前的日期%2B/-几年，可以将其更改为"2012“。若要验证2000到2099年，请将其更改为"200-9{2}“。|818771|我已经改变了原来的正则表达式以上寻找2010-2039。如果有必要，其他人可以在20年内编辑这个答案。|818772|entityMap^0|11|J|2L|2|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$]]|$1|I|3|J|5|H|7|12|8|@]|D|@]|E|$]]|$1|K|3|L|5|H|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|15|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|16|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|17|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|V|$]]

Here is a regex I wrote earlier today for validating strings in a format similar to what you mentioned: <code>YYYY-MM-DD hh:mm:ss</code>. It does not identify some bad dates (for example, February 30th) but may be slightly better than using the simplistic <code>\d</code> at each position. Things to note:

<ol>
<li>you can specify just a date, just a time, or both date + time</li>
<li>time can be in 12-hour or 24-hour format</li>
<li>seconds are optional</li>
<li>am/pm is optional

<pre><code>const std::string dateAndTimeRegex =
 "^\\s*" // ignore whitespace
 "(" // start of date
 "20[123][0-9]" // year: 2010, 2011, ..., through 2039
 "\\W" // delimiter between year and month; typically will be "-"
 "([0]?[0-9]|1[012])" // month: 0 through 9, or 00 through 09, or 10 through 12
 "\\W" // delimiter between month and day; typically will be "-"
 "([012]?[0-9]|3[01])" // day: 0 through 9, or 00 through 29, or 30, or 31
 ")?" // end of optional date
 "\\s?" // optional whitespace
 "(" // start of time
 "([01]?[0-9]|2[0-3])" // hour: 0 through 9, or 00 through 19, or 20 through 23
 "\\W" // delimiter between hours and minutes; typically will be ":"
 "([0-5][0-9])" // minute: 00 through 59
 "(" // start of seconds (optional)
 "\\W" // delimiter between minutes and seconds; typically will be ":"
 "([0-5][0-9])" // seconds: 00 through 59
 ")?" // end of optional seconds
 "(\\s*[AaPp][Mm])?" // optional AM, am, PM, pm
 ")?" // end of optional time
 "\\s*$"; // trailing whitespace
</code></pre></li>
</ol>

<hr>

A comment from @kyrias hints that this regex will fail in a few months once we hit the year 2020. Depending on how you use it, you'll need to change "201[0-9]" to something else.

For example, if you are looking to validate the current date +/- a few years, you could change it to "20[12][0-9]". To validate 2000 through 2099, change it to "20[0-9]{2}".

I've changed the original regex above to look for 2010-2039. Someone else can edit this answer in 20 years if necessary.

blocks|key|940579|text|我只是编写了一个Regex来控制MySQL日期时间(我认为这对于Oracle或其他数据库服务器并没有太大的不同)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|940580|此ReGex非常安全并控制日期的有效性(+28、30和31天中的月份，以及29/02中的偶数年)。然后，它控制时间。时间戳格式是这样的：|940581|YYYY+MM-DD+HH:MM:SS|blockquote|940582|这里是Regex+(相当长)：|940583|((((19%7C20)([2468][048]%7C[13579][26]%7C0[48])%7C2000)-02-29%7C((19%7C20)[0-9]{2}-(0[4678]%7C1[02])-(0[1-9]%7C[12][0-9]%7C30)%7C(19%7C20)[0-9]{2}-(0[1359]%7C11)-(0[1-9]%7C[12][0-9]%7C3[01])%7C(19%7C20)[0-9]{2}-02-(0[1-9]%7C1[0-9]%7C2[0-8])))\s([01][0-9]%7C2[0-3]):([012345][0-9]):([012345][0-9]))|code-block|syntax|javascript|940584|火柴|940585|2013-04-05+17:59:59+:59+2013-07-30+01:22:42+2099-12-30+23:59|940586|非匹配|940587|2016-02-29+-01:01:02+_|940588|我想我不能做一个更安全的了。非常功能和测试。请，如果你使用它，不要犹豫给我反馈;)|940589|FYI+:如果您只是在没有时间的情况下查找date+Regex控件，请到那里查找它：正则表达式以匹配有效日期+(查看我的帖子)。|offset|length|940590|干杯|940591|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/51224/regular-expression-to-match-valid-dates/15968019#15968019^0|0|0|0|0|0|0|0|0|0|0|16|C|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1C|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1E|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1F|8|@]|9|@]|A|$L|M]]|$1|N|3|O|5|6|7|1G|8|@]|9|@]|A|$]]|$1|P|3|Q|5|F|7|1H|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1I|8|@]|9|@]|A|$]]|$1|T|3|U|5|F|7|1J|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1K|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1L|8|@]|9|@$Z|1M|10|1N|1|1O]]|A|$]]|$1|11|3|12|5|6|7|1P|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A]]]]

I just wrote a Regex to control a MySQL datetime (and I think that it's not so different for Oracle or other database server).

This ReGex is very secure and controls the validity of the date (months in 28, 30 &amp; 31 days, and even years with 29/02). Then, it controls the time.
The timestamp format is this one : 

<blockquote>
 YYYY-MM-DD HH:MM:SS
</blockquote>

Here is the Regex (quite long) :

<pre><code>((((19|20)([2468][048]|[13579][26]|0[48])|2000)-02-29|((19|20)[0-9]{2}-(0[4678]|1[02])-(0[1-9]|[12][0-9]|30)|(19|20)[0-9]{2}-(0[1359]|11)-(0[1-9]|[12][0-9]|3[01])|(19|20)[0-9]{2}-02-(0[1-9]|1[0-9]|2[0-8])))\s([01][0-9]|2[0-3]):([012345][0-9]):([012345][0-9]))
</code></pre>

Matches

<blockquote>
 2013-04-05 17:59:59 | 2013-07-30 01:22:42 | 2099-12-30 23:59:59 | 2016-02-28 00:00:00
</blockquote>

Non-Matches

<blockquote>
 2016-02-29 -01:01:02 | 3000-04-24 17:42:21 | 2012-03-03 24:24:02 | 2012-03-03 21:60:45
</blockquote>

I don't think I can make a more secured one. Very functionnel and tested.
Please, if you use it, don't hesitate to give me feedback ;)

FYI : If you just look for the date Regex control without time, please go there to find it : <a href="https://stackoverflow.com/questions/51224/regular-expression-to-match-valid-dates/15968019#15968019">Regular Expression to match valid dates</a>
(check my post).

Cheers

blocks|key|818861|text|对于1000年至2999年之间的有效日期，请使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|818862|String+pattern+=+"((1\\d{3})%7C(20\\d{2}))-((0\\d)%7C(1[0-2]))-(([1-2]\\d)%7C(3[0-1]))+(([0-1]\\d)%7C(2[0-3])):([0-5]\\d):([0-5]\\d)";|code-block|syntax|javascript|818863|月份可以在01到12之间，白天在01到31之间，时间从00:00:00到23:59:59。|818864|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For a valid date between year 1000 and year 2999 use : 

<pre><code>String pattern = "((1\\d{3})|(20\\d{2}))-((0\\d)|(1[0-2]))-(([1-2]\\d)|(3[0-1])) (([0-1]\\d)|(2[0-3])):([0-5]\\d):([0-5]\\d)";
</code></pre>

The months can vary between 01 and 12, the days between 01 and 31 and the time between 00:00:00 and 23:59:59.

blocks|key|2345191|text|function+isTimestamp($input){
++++++++++++var+$regex+=+/%5E\d\d\d\d-(0?[1-9]%7C1[0-2])-(0?[1-9]%7C[12][0-9]%7C3[01])+(00%7C[0-9]%7C1[0-9]%7C2[0-3]):([0-9]%7C[0-5][0-9]):([0-9]%7C[0-5][0-9])$/;
++++++++++++return+$regex.test($input);
++++++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2345192|请检查一下|unstyled|2345193|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>function isTimestamp($input){
 var $regex = /^\d\d\d\d-(0?[1-9]|1[0-2])-(0?[1-9]|[12][0-9]|3[01]) (00|[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$/;
 return $regex.test($input);
 }
</code></pre>

check please

blocks|key|818875|text|下面是我如何在范围yyyy-mm-dd+hh:mm:ss+(+1970-01-01+00:00:00，2037-12-31+23:59:59+)中建立一个与SQL时间戳匹配的正则表达式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|818876|$leapYear+=+"(19[79][26]%7C198[048]%7C20[02][048]%7C20[13][26])";
$leapDate+=+"$leapYear-02-29";

$regularYear+=+"(19[789][0-9]%7C20[012][0-9]%7C203[0-7])";
$regularFebruaryDate+=+"02-([01][1-9]%7C2[0-8])";
$month31Date+=+"(0[13578]%7C10%7C12)-([012][1-9]%7C3[01])";
$month30Date+=+"(0[469]%7C11)-([012][1-9]%7C30)";
$regularDate+=+"$regularYear-($regularFebruaryDate%7C$month30Date%7C$month31Date)";

$hours+=+"(2[0-3]%7C[01][0-9])";
$minutes+=+"[0-5][0-9]";
$seconds+=+"[0-5][0-9]";

$sqlTimestamp+=+"($leapDate%7C$regularDate)+$hours:$minutes:$seconds";|code-block|syntax|javascript|818877|注:我避免了2038年，因为少了一个边缘箱。最大可能的SQL时间戳是2038-01-19+03:14:07。|818878|entityMap^0|9|J|V|J|1F|J|0|0|Y|J|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Here is how I build up a regex matching exactly SQL Timestamp <code>yyyy-mm-dd hh:mm:ss</code> in the range 
[<code>1970-01-01 00:00:00</code>, <code>2037-12-31 23:59:59</code>]

<pre class="lang-php prettyprint-override"><code>$leapYear = "(19[79][26]|198[048]|20[02][048]|20[13][26])";
$leapDate = "$leapYear-02-29";

$regularYear = "(19[789][0-9]|20[012][0-9]|203[0-7])";
$regularFebruaryDate = "02-([01][1-9]|2[0-8])";
$month31Date = "(0[13578]|10|12)-([012][1-9]|3[01])";
$month30Date = "(0[469]|11)-([012][1-9]|30)";
$regularDate = "$regularYear-($regularFebruaryDate|$month30Date|$month31Date)";

$hours = "(2[0-3]|[01][0-9])";
$minutes = "[0-5][0-9]";
$seconds = "[0-5][0-9]";

$sqlTimestamp = "($leapDate|$regularDate) $hours:$minutes:$seconds";
</code></pre>

Note: I avoided the year 2038 to have one less edge-case. The max possible SQL timestamp is <code>2038-01-19 03:14:07</code>.

blocks|key|940619|text|这是您正在寻找的正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|940620|[0-9]{4}/(0[1-9]%7C1[0-2])/(0[1-9]%7C[1-2][0-9]%7C3[0-1])+(2[0-3]%7C[01][0-9]):[0-5][0-9]:[0-5][0-9]|code-block|syntax|javascript|940621|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is the regex you are looking for: 

<pre><code>[0-9]{4}/(0[1-9]|1[0-2])/(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]:[0-5][0-9]
</code></pre>

I need a regular expression to validate a timestamp of the format, using Javascript:

<blockquote>
 YYYY/MM/DD HH:MI:SS
</blockquote>

I tried cooking up a few, but seems my regex skills fail to cover something or other.

Please give me a reference or way to do it.

P.S. : I mention regex, only as a suggestion. Im using Javascript and welcome any alternative.

Regular Expression to validate a timestamp

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要一个正则表达式来验证格式的时间戳，使用JavascriptYYYY/MM/DD HH:MI:SS我试着做了几个，但似乎我的判断力没能涵盖什么东西。请给我一个推荐信或做它的方法。P.S.：我提到regex只是作为一个建议。我正在使用Javascript和欢迎任何替代。

问验证时间戳的正则表达式
EN

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