How to check if a string is a legal “dd/mm/yyyy” d

2019-01-09 04:23发布

Given a string str, how could I check if it is in the dd/mm/yyyy format and contains a legal date ?

Some examples:

bla bla      // false
14/09/2011   //         true
09/14/2011   // false
14/9/2011    // false
1/09/2011    // false
14/09/11     // false
14.09.2011   // false
14/00/2011   // false
29/02/2011   // false
14/09/9999   //         true

7条回答
forever°为你锁心
2楼-- · 2019-01-09 04:31

you can use regular exp to validate date . try like this :

  re = /^\d{1,2}\/\d{1,2}\/\d{4}$/; 
if(form.mydate.value != '' && !form.mydate.value.match(re))
  //do something here

note: this will only work for dd/mm/yyyy

for exact match of your requirement use

 re = /^\d{2}\/\d{2}\/\d{4}$/; 
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Summer. ? 凉城
3楼-- · 2019-01-09 04:35

I'm going to answer a different question, as Misha Moroshko's has already been well-answered: use HTML5. That is, on the assumption that the strings in question arise as user inputs through a Web browser, I propose that the entries be received as

   <input type = "date" ...

I recognize that not all browsers likely to be in use will interpret "date" in a way that rigorously enforces validity. It's the right thing to do, though, will certainly improve as time goes on, and might well be good enough in a particular context even now simply to eliminate the need to validate the date-string after the fact.

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我想做一个坏孩纸
4楼-- · 2019-01-09 04:36

for dd/mm/yyyy format only

^(0?[1-9]|[12][0-9]|3[01])[\/](0?[1-9]|1[012])[\/]\d{4}$
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男人必须洒脱
5楼-- · 2019-01-09 04:40

Edit: exact solution below

You could do something like this, but with a more accurate algorithm for day validation:

function testDate(str) {
  var t = str.match(/^(\d{2})\/(\d{2})\/(\d{4})$/);
  if(t === null)
    return false;
  var d = +t[1], m = +t[2], y = +t[3];

  // Below should be a more acurate algorithm
  if(m >= 1 && m <= 12 && d >= 1 && d <= 31) {
    return true;  
  }

  return false;
}

http://jsfiddle.net/aMWtj/

Date validation alg.: http://www.eee.hiflyers.co.uk/ProgPrac/DateValidation-algorithm.pdf

Exact solution: function that returns a parsed date or null, depending exactly on your requirements.

function parseDate(str) {
  var t = str.match(/^(\d{2})\/(\d{2})\/(\d{4})$/);
  if(t !== null){
    var d = +t[1], m = +t[2], y = +t[3];
    var date = new Date(y, m - 1, d);
    if(date.getFullYear() === y && date.getMonth() === m - 1) {
      return date;   
    }
  }

  return null;
}

http://jsfiddle.net/aMWtj/2/

In case you need the function to return true/false and for a yyyy/mm/dd format

function IsValidDate(pText) {
    var isValid = false ;
    var t = pText.match(/^(\d{4})\/(\d{2})\/(\d{2})$/);

    if (t !== null) {
        var y = +t[1], m = +t[2], d = +t[3];
        var date = new Date(y, m - 1, d);

        isValid = (date.getFullYear() === y && date.getMonth() === m - 1) ;
    }

    return isValid ;
}
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你好瞎i
6楼-- · 2019-01-09 04:40

Personally, I think the best solution would be to modify the UI to use dropdowns for the month and possibly day selections.

Trying to figure out if 1/2/2001 is January 2nd or February 1st based solely on that input string is impossible.

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闹够了就滚
7楼-- · 2019-01-09 04:49

Try -

   var strDate = '12/03/2011';
   var dateParts = strDate.split("/");
   var date = new Date(dateParts[2], (dateParts[1] - 1) ,dateParts[0]);

There's more info in this question - Parse DateTime string in JavaScript (the code in my answer is heavily influenced by linked question)

Demo - http://jsfiddle.net/xW2p8/

EDIT

Updated answer, try -

function isValidDate(strDate) {
    if (strDate.length != 10) return false;
    var dateParts = strDate.split("/");
    var date = new Date(dateParts[2], (dateParts[1] - 1), dateParts[0]);
    if (date.getDate() == dateParts[0] && date.getMonth() == (dateParts[1] - 1) && date.getFullYear() == dateParts[2]) {
        return true;
    }
    else return false;
}

This function passes all the test cases. As far as I'm aware, Adam Jurczyk had posted an accurate answer well before I corrected my original wrong answer. He deserves credit for this.

Demo - http://jsfiddle.net/2r6eX/1/

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