Grep regular expression for digits in character st

Question

I need some way to find words that contain any combination of characters and digits but exactly 4 digits only, and at least one character.

EXAMPLE:

a1a1a1a1        // Match
1234            // NO match (no characters)
a1a1a1a1a1      // NO match
ab2b2           // NO match
cd12            // NO match
z9989           // Match
1ab26a9         // Match
1ab1c1          // NO match
12345           // NO match
24              // NO match
a2b2c2d2        // Match
ab11cd22dd33    // NO match

Answer 1

Assuming you only need ASCII, and you can only access the (fairly primitive) regexp constructs of grep, the following should be pretty close:

grep ^[a-zA-Z]*[0-9][a-zA-Z]*[a-zA-Z]*[0-9][a-zA-Z]*[a-zA-Z]*[0-9][a-zA-Z]*[a-zA-Z]*[0-9][a-zA-Z]*$ | grep [a-zA-Z]

Answer 2

The regex for that is:

([A-Za-z]\d){4}

• [A-Za-z] - for character class
• \d - for number
• you wrapp them in () to group them indicating the format character follow by number
• {4} - indicating that it must be 4 repetitions

Answer 3

With grep:

grep -iE '^([a-z]*[0-9]){4}[a-z]*$' | grep -vE '^[0-9]{4}$'

Do it in one pattern with Perl:

perl -ne 'print if /^(?!\d{4}$)([^\W\d_]*\d){4}[^\W\d_]*$/'

The funky [^\W\d_] character class is a cosmopolitan way to spell [A-Za-z]: it catches all letters rather than only the English ones.