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Regular expression to find exact words

I want to write a regex that will only give me words containing the letters that I specify, and the length of the matched word must be the same as the number of characters specified.

Therefore, if I give letters OMHE

, it should only match words that contain these and only those letters, and also only the number of times the letter appears in a sequence of letters

In regex, I am a concrete example so far (I create regex dynamically otherwise) ...

.*?[O{1}M{1}H{1}E{1}]{4}

This works to some extent, but I get multiple words that shouldn't be the same. Words that should correspond to the example, HOME

but MEMO

and HOMEE

have not. I feel really bad when it comes to regexes :(

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4 answers


You can use this lookahead based regex:

^(?=.*?O)(?=.*?M)(?=.*?H)(?=.*?E)[OMHE]{4}$


Demo version of RegEx

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you can use ^(([OMHE])(?!.*\2)){4}$

It uses negative prediction, saying that after each match [OMHE]

, no other occurrence of the captured text is allowed. Then four reps are required. Since the outer group only exists to define repetition, it can be optimized as a non-capturing group:



^(?:([OMHE])(?!.*\1)){4}$

It's easy to expand to more symbols ...

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You can use this template

(O|M|H|E){4}
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This is the way to do it if you want to make the match mobile inside
String.

 # (?<!\S)([OMHE])(?!\1)([OMHE])(?!\1|\2)([OMHE])(?!\1|\2|\3)([OMHE])(?!\S)

 (?<! \S )
 ( [OMHE] )                    # (1)
 (?! \1 )
 ( [OMHE] )                    # (2)
 (?! \1 | \2 )
 ( [OMHE] )                    # (3)
 (?! \1 | \2 | \3 )
 ( [OMHE] )                    # (4)
 (?! \S )
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