Regex that will match 6 characters that only allows digits, leading, and trailing spaces

Question

The regex that I'm trying to implement should match the following data:

1. 123456
2. 12345 
3.  23456
4.      5
5. 1     
6.       
7.   2   
8.  2345 

It should not match the following:

1. 12 456
2. 1234 6
3.  1 6
4. 1 6

It should be 6 characters in total including the digits, leading, and trailing spaces. It could also be 6 characters of just spaces. If digits are used, there should be no space between them.

I have tried the following expressions to no avail:

^\s*[0-9]{6}$ 

\s*[0-9]\s* 

Answer 1

You can also do:

^(?!.*?\d +\d)[ \d]{6}$ 

• The zero width negative lookahead (?!.*?\d +\d) ensures that the lines having space(s) in between digits are not selected

• [ \d]{6} matches the desired lines that have six characters having just space and/or digits.

Answer 2

You can just use a *\d* * pattern with a restrictive (?=.{6}$) lookahead:

^(?=.{6}$) *\d* *$ 

See the regex demo

Explanation:

• ^ - start of string
• (?=.{6}$) - the string should only have 6 any characters other than a newline
• * - 0+ regular spaces (NOTE to match horizontal space - use [^\S\r\n])
• \d* - 0+ digits
• * - 0+ regular spaces
• $ - end of string.

Java demo (last 4 are the test cases that should fail):

List<String> strs = Arrays.asList("123456", "12345 ", " 23456", "     5", // good "1     ", "      ", "  2   ", " 2345 ",  // good "12 456", "1234 6", " 1   6", "1    6"); // bad for (String str : strs)     System.out.println(str.matches("(?=.{6}$) *\\d* *")); 

Note that when used in String#matches(), you do not need the intial ^ and final $ anchors as the method requires a full string match by anchoring the pattern by default.