Date Regular Expression Match 2 or 1 Digit Date or Month

Question

I Have this regex a perfect to to match mm/dd/yyyy on all condition time format but i need to modify it a little bit

if someone insert 09/04/1980 its work perfect, but this 5/9/1990 (1 digit) will error :(

can someone fix it for me ?

this is my current regex :

(?:(?:0[1-9]|1[0-2])/(?:0[1-9]|1[0-9]|2[0-9])|(?:(?!02)(?:0[1-9]|1[0-2])-(?:30))|(?:(?:0[13578]|1[02])-31))/(?:19|20)[0-9]{2}

Why, just why developers attempt to solve every problem in the world using regex — anubhava, Oct 17 '14 at 12:12
I'm just trying learn regex, but when it become this long it's so hard to figure it out — himawan_r, Oct 17 '14 at 12:16

score 1 · Accepted Answer · answered Oct 17 '14 at 12:14

1

(?:(?:0?[1-9]|1[0-2])\/(?:0?[1-9]|1[0-9]|2[0-9])|(?:(?!02)(?:0?[1-9]|1[0-2])-(?:30))|(?:(?:0?[13578]|1[02])-31))\/(?:19|20)[0-9]{2}

Try this.Made 0 optional.

See demo.

http://regex101.com/r/dZ1vT6/65

answered Oct 17 '14 at 12:14

vks

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wow, i've been placing those here and there to no avail. how could it could be so easy. anyway nice website it is – himawan_r Oct 17 '14 at 12:24
@machun yes it is.Leaning regex is fun there. – vks Oct 17 '14 at 12:24

score 0 · Answer 2 · answered Oct 17 '14 at 12:16

0

DON'T use regex for this. I'll provide pesudocode since you've not specified language. You just need to split on \ and convert the array elements to numbers and then do simple if condition.

if(arr[0] < 0 && arr[0] > 32){
   // validated the month
}

answered Oct 17 '14 at 12:16

Amit Joki

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Date Regular Expression Match 2 or 1 Digit Date or Month

2 Answers2