Question

I need a regex that would validate first 4 digit set and then 3 digits. The first foure digits need to be identical.

Acceptable entries should be for example:

7777989  
8888767 

Invalid entries:

7778888
7777abc
8989123

Now my problem is that in my code iam creating the regexes based on class names. for example something like this

if (currentChar == 'n') regex += '[0-9]'; 
else if (currentChar == 'd') regex += '^[0-9]+$';

in above code n and d is class name where n represents “any digit” and d represents “identical digits”. So for the above example, the rule would be ddddnnn.

Answer 1

^(\d)\1{3}\d{3}$

should do it.

^     # Start of string
(\d)  # Match a single digit, capture in group 1
\1{3} # Match the same digit as in group 1, three times
\d{3} # Match three digits (any)
$     # End of string

A quick and dirty regex generator (in Python), assuming that all d digits have to be identical:

def makere(rule):
    first_d = True
    parts = ["^"]
    for letter in rule:
        if letter == "d":
            if first_d:
                parts.append(r"(\d)")
                first_d = False
            else:
                parts.append(r"\1")
        if letter == "n":
            parts.append(r"\d")
    parts.append("$")
    return ''.join(parts)

Result:

>>> makere("ddddnnn")
'^(\\d)\\1\\1\\1\\d\\d\\d$'
>>> makere("ddnnndd")
'^(\\d)\\1\\d\\d\\d\\1\\1$'