Regex to find 2 identical characters at the end of a string
I need to find a regular expression with the following rules.
- contains 8 to 20 characters (regular or normal).
- contains no whitespace characters.
- cannot start with a number (0-9) or an underscore (_).
- at the end of the string, which should be 2 of the same char.
- must contain at least 1 number.
OK
+234567899
a_1de*Gg
xy1Me*__
!41deF_hij2lMnopq3ss
C234567890123$^67800
*5555555
sDF564zer""
!!!!!!!!!4!!!!!!!!!!
abcdefghijklmnopq9ss
Out of order:
has more or less 8-20 characters:
a_1+Eff B41def_hIJ2lmnopq3stt abCDefghijklmnopqrss5
has whitespace characters:
A_4 e*gg
starts with a number or underscore:
__1+Eff 841DEf_hij2lmnopq3stt
ends with two different characters:
a_1+eFg b41DEf_hij2lmnopq3st
does not contain numbers:
abCDefghijklmnopqrss
abcdef+++dF
!!!!!!!!!!!!!!!!!!!!
As long as I have it
((?m:[^0-9_]^(?=.*[0-9])\S{8,20}$))
But I can't figure out the two same characters at the end?
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2 answers
For most regex variants (PCRE, Python, PHP, JavaScript) the following will work:
/^(?=\S{8,20}$)(?=\D*\d)(?![0-9_]).{6,18}?(.)\1$/i
Demonstration with unit tests versus your example cases
Explanation:
-
/
-
^
start of line -
(?=\S{8,20}$)
followed by 8-20 characters without spaces. -
(?=\D*\d)
contains a digit -
(?![0-9_])
cannot start with a number or underscore -
.{6,18}?
Unwanted character match (moves us from the beginning of the line to the end) -
(.)\1
match any character followed by the same character again -
$
end of line -
/
-
i
flag: case insensitive (required to seeGg
, for example, the same character twice)
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