Javascript 中 MM/DD/YYYY 的正则表达式
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Regular Expression for MM/DD/YYYY in Javascript
提问by Mark Sandman
I've just written this regular expression in javaScript however it doesn't seem to work, here's my function:
我刚刚在 javaScript 中编写了这个正则表达式，但它似乎不起作用，这是我的函数：
function isGoodDate(dt){
var reGoodDate = new RegExp("/^((0?[1-9]|1[012])[- /.](0?[1-9]|[12][0-9]|3[01])[- /.](19|20)?[0-9]{2})*$/");
return reGoodDate.test(dt);
}
and this is how I call it in my form validation
这就是我在表单验证中调用它的方式
if(!isGoodDate(userInput[1].value)){
alert("date not in correct format of MM/dd/YYYY");
return false;
}
now I want it to return MM/DD/YYYY however if I put a valid date in it raises the alert? Any ideas anyone?
现在我希望它返回 MM/DD/YYYY 但是如果我在其中输入有效日期会引发警报？任何人的想法？
回答by Jan
Attention, before you copy+paste: The question contains some syntactic errors in its regex. This answer is correcting the syntax. It is not claiming to be the best regex for date/time parsing.
注意，在复制+粘贴之前：该问题的正则表达式中包含一些语法错误。这个答案正在纠正语法。它并不声称是日期/时间解析的最佳正则表达式。
Try this:
尝试这个：
function isGoodDate(dt){
var reGoodDate = /^((0?[1-9]|1[012])[- /.](0?[1-9]|[12][0-9]|3[01])[- /.](19|20)?[0-9]{2})*$/;
return reGoodDate.test(dt);
}
You either declare a regular expression with:
您可以声明一个正则表达式：
new RegExp("^((0?[1-9]|1[012])[- /.](0?[1-9]|[12][0-9]|3[01])[- /.](19|20)?[0-9]{2})*$")
Or:
或者：
/^((0?[1-9]|1[012])[- /.](0?[1-9]|[12][0-9]|3[01])[- /.](19|20)?[0-9]{2})*$/
Notice the /
请注意 /
回答by Shef
Maybe because you are declaring the isGoodDate()function, and then you are calling the isCorrectDate()function?
也许是因为你声明了isGoodDate()函数，然后调用了isCorrectDate()函数？
Try:
尝试：
function isGoodDate(dt){
var reGoodDate = /^(?:(0[1-9]|1[012])[\/.](0[1-9]|[12][0-9]|3[01])[\/.](19|20)[0-9]{2})$/;
return reGoodDate.test(dt);
}
Works like a charm, test it here.
就像一个魅力，在这里测试一下。
Notice, this regex will validate dates from 01/01/1900through 31/12/2099. If you want to change the year boundaries, change these numbers (19|20)on the last regex block. E.g. If you want the year ranges to be from 01/01/1800through 31/12/2099, just change it to (18|20).
请注意，此正则表达式将验证从01/01/1900到 的日期31/12/2099。如果要更改年份边界，请(19|20)在最后一个正则表达式块上更改这些数字。例如，如果您希望年份范围从01/01/1800到31/12/2099，只需将其更改为(18|20)。
回答by KooiInc
I don't think you need a regular expression for this. Try this:
我认为您不需要为此使用正则表达式。尝试这个：
function isGoodDate(dt){
var dts = dt.split('/').reverse()
,dateTest = new Date(dts.join('/'));
return isNaN(dateTest) ? false : true;
}
//explained
var dts = dt.split('/').reverse()
// ^ split input and reverse the result
// ('01/11/2010' becomes [2010,11,01]
// this way you can make a 'universal'
// datestring out of it
,dateTest = new Date(dts.join('/'));
// ^ try converting to a date from the
// array just produced, joined by '/'
return isNaN(dateTest) ? false : true;
// ^ if the date is invalid, it returns NaN
// so, if that's the case, return false
回答by mplungjan
I agree with @KooiInc, but it is not enough to test for NaN
我同意@KooiInc，但它不足以测试 NaN
function isGoodDate(dt){
var dts = dt.split('/').reverse()
,dateTest = new Date(dts.join('/'));
return !isNaN(dateTest) &&
dateTest.getFullYear()===parseInt(dts[0],10) &&
dateTest.getMonth()===(parseInt(dts[1],10)-1) &&
dateTest.getDate()===parseInt(dts[2],10)
}
which will handle 29/2/2001 and 31/4/2011
这将处理 29/2/2001 和 31/4/2011
For this script to handle US dates do
对于这个脚本来处理美国日期做
function isGoodDate(dt){
var dts = dt.split('/')
,dateTest = new Date(dt);
return !isNaN(dateTest) &&
dateTest.getFullYear()===parseInt(dts[2],10) &&
dateTest.getMonth()===(parseInt(dts[0],10)-1) &&
dateTest.getDate()===parseInt(dts[1],10)
}
回答by Usman Ali
Add this in your code, it working perfectly fine it here. click here http://jsfiddle.net/Shef/5Sfq6/
将此添加到您的代码中，它在这里工作得很好。点击这里http://jsfiddle.net/Shef/5Sfq6/
function isGoodDate(dt){
var reGoodDate = /^(?:(0[1-9]|1[012])[\/.](0[1-9]|[12][0-9]|3[01])[\/.](19|20)[0-9]{2})$/;
return reGoodDate.test(dt);
}
}
回答by Mostafa Khedeer
Validate (DD-MM-YYYY) format :)
验证 (DD-MM-YYYY) 格式 :)
function isGoodDate(dt) {
var reGoodDate = /^(?:(0[1-9]|[12][0-9]|3[01])[\-.](0[1-9]|1[012])[\-.](19|20)[0-9]{2})$/;
return reGoodDate.test(dt);
}
