Using regular expression, I want to select only the words which:

• are alphanumeric
• do not contain only numbers
• do not contain only alphabets
• have unique numbers(1 or more)

I am not really good with the regex but so far, I have tried [^\d\s]*(\d+)(?!.*\1) which takes me nowhere close to the desired output :(

Here are the input strings:

I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3

Expected Matches:

abc123
ab12s
hel1lo2haha3hoho4

Using regular expression, I want to select only the words which:

• are alphanumeric
• do not contain only numbers
• do not contain only alphabets
• have unique numbers(1 or more)

I am not really good with the regex but so far, I have tried [^\d\s]*(\d+)(?!.*\1) which takes me nowhere close to the desired output :(

Here are the input strings:

I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3

Expected Matches:

abc123
ab12s
hel1lo2haha3hoho4

• javascript
• regex

Share Improve this question Follow edited Feb 2, 2019 at 6:58 CertainPerformance 372k5555 gold badges352352 silver badges357357 bronze badges asked Feb 2, 2019 at 6:44 ManJoeyManJoey 21333 silver badges77 bronze badges

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 8

You can use

\b(?=\d*[a-z])(?=[a-z]*\d)(?:[a-z]|(\d)(?!\w*\1))+\b

https://regex101./r/TimjdW/3

Anchor the start and end of the pattern at word boundaries with \b, then:

• (?=\d*[a-z]) - Lookahead for an alphabetical character somewhere in the word
• (?=[a-z]*\d) - Lookahead for a digit somewhere in the word
• (?:[a-z]|(\d)(?!\w*\1))+ Repeatedly match either:
  • [a-z] - Any alphabetical character, or
  • (\d)(?!\w*\1) - A digit which does not occur again in the same word

Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:

/\b(?=[a-z]*\d)(?=\d*[a-z])(?!\w*(\d)\w*\1)[a-z\d]+\b/ig

RegEx Demo

RegEx Details:

• \b: Word boundary
• (?=[a-z]*\d): Make sure we have at least a digit
• (?=\d*[a-z]): Make sure we have at least a letter
• (?!\w*(\d)\w*\1): Make sure digits are not repeated anywhere in the word
• [a-z\d]+: Match 1+ alphanumericals
• \b: Word boundary

You could assert all the conditions using one negative lookahead:

\b(?![a-z]+\b|\d+\b|\w*(\d)\w*\1)[a-z\d]+\b

See live demo here

The important parts are starting match from \b and immediately looking for the conditions:

• [a-z]+\b Only alphabetic

• \d+\b Only numeric

• \w*(\d)\w*\1 Has a repeating digit

You can use this

\b(?!\w*(\d)\w*\1)(?=(?:[a-z]+\d+)|(?:\d+[a-z]+))[a-z0-9]+\b

• \b - Word boundary.
• (?!\w*(\d)\w*\1) - Condition to check unique digits.
• (?=(?:[a-z]+\d+)|(?:\d+[a-z]+)) - Condition to check alphanumeric words.
• [a-z0-9]+ - Matches a to z and 0 to 9

Demo