Regex - Expression for 6 digits or 9 digits not less or between or more

Question

I have to write a regex that detects either 6 digits or 9 digits - meaning:

ID 123 - it should not meet the regext as it is 3 digits
ID 1234 - it should not meet the regext as it is 4 digits
ID 12345 - it should not meet the regext as it is 5 digits
ID 123456 - it should meet the regex it is 6 digits <==========
ID 1234567 - it should not meet the regext as it is 7 digits
ID 12345678 - it should not meet the regext as it is 8 digits
ID 123456789 - it should meet the regext as it is 9 digits <==========
ID 1234567890 - it should not meet the regext as it is 10 digits
ID 12345678901 - it should not meet the regext as it is 11 digits

and so on... you got the idea?

Can anyone help me with the regex for this one?

Try `^ID (\d{6}|\d{9})$` – The fourth bird Jun 12 '19 at 15:24 — The fourth bird, Jun 12 '19 at 15:24

score 3 · Answer 1 · answered Jun 12 '19 at 15:23

3

You may try to use this pattern:

\b(\d{6}|\d{9})\b

answered Jun 12 '19 at 15:23

cn007b

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score 2 · Answer 2 · answered Jun 12 '19 at 15:46

2

Or another one:

\b\d{6}(?:\d{3})?\b

See demo at regex101

\b matches a word boundary
\d{6} six digits (\d is a short for digit)
(?:\d{3})? an optional non capturing group containing another three digits

answered Jun 12 '19 at 15:46

bobble bubble

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The fourth bird · Answer 3 · 2019-06-12T17:37:05.450

1

If the ID part should be matched, you could use an alternation to match either 6 or 9 digits.

Use anchors to assert the start ^ and the end $ of the string or add word boundaries \b to the beginning and the end of the pattern.

The digits are in the first capturing group.

^ID (\d{6}|\d{9})$

edited Jun 12 '19 at 17:37

answered Jun 12 '19 at 17:04

The fourth bird

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Regex - Expression for 6 digits or 9 digits not less or between or more

3 Answers3