查找大于或等于 2014 的正则表达式

Question

I want to find the years (4 digit nos) which are greater than or equal to 2014 using regular expression. [2014 - 9999]

I've tried this, but its not matching if the last digit is less than 4.

^[2-9]\d{1,}[1-9]0*[4-9]0*$

Eg: 3013

Please help.

Answer 1

Try the below regex to match the years from 2014 to 9999,

^(?:20(?:1[4-9]|[2-9][0-9])|2[1-9][0-9][0-9]|[3-9][0-9][0-9][0-9])$

DEMO

Explanation:

• 20(?:1[4-9]|[2-9][0-9]) Matches the years from 2014 to 2099.
• 2[1-9][0-9][0-9] Matches the years from 2100 to 2999.
• [3-9][0-9][0-9][0-9] Matches the years from 3000 to 9999.

Answer 2

Obviously, regular expressions are a terrible way to do this, so I'm assuming you're doing it this way as a mental exercise, or as a homework assignment. Either way - like most programming problems, you can solve this by breaking it down into smaller, manageable chunks:

(This may not be the coolest solution, but it's easy to understand)

1. Match any numbers with 5 or more digits (>=10000) (not strictly needed for your problem description, but…)
2. Match any 4-digit number of the form [3-9]xxx (3000-9999)
3. Match any 4-digit number of the form 2[1-9]xx (2100-2999)
4. Match any 4-digit number of the form 20[2-9]x (2020-2099)
5. Match any 4-digit number of the form 201[4-9] (2015-2019)
6. Take the regexes you formed for steps 0-4 and OR them together so you match on any of these patterns.

(See Avinash Raj's answer, for actual regexes that do this)

Answer 3

^[3-9]\d{3}|(2(0(1[4-9]|[2-9]\d)|[1-9]\d{2}))$

pass

2014 
2015
3014
3000
9999
5014

fail

2013
0014
2012
1013

Answer 4

((?:20(?:(?:1[4-9])|(?:[2-9]\\d)))|(?:2[1-9]\\d\\d)|(?:[3-9]\\d\\d\\d))

is what you are looking for. Extra surrounding parantheses is to enable un-ambiguous capturing or back referencing.