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Is there are a simple way to ensure with RegEx that input String will contain exactly, let's say one 'a', five 'b, seven 'z' characters - with no order checking?

Jonathan Hall
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Ares02
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    What have you tried? Include your code. What problems did you encounter? – Jonathan Hall Nov 28 '18 at 14:43
  • Yes, there is. What have you tried? What didn't work? What did you get? What did you expect? What doesn't work with your code and where is it? – Toto Nov 28 '18 at 14:43
  • https://stackoverflow.com/questions/22411445/regex-match-specific-characters-in-any-order-without-more-occurrences-of-each-c – Freiheit Nov 28 '18 at 14:46
  • https://regex101.com/ is a great tool to try and test regexes enter your regex and then several strings to see what matches. For example a regex that matches on one of each character is https://regex101.com/r/UtoK45/1 – Freiheit Nov 28 '18 at 14:47
  • let's say if you want to make sure that there is only one 'a' in your input then you can use regex: "[.&&[^a]]*[a][.&&[^a]]* It's quite simple but when you want to add more "a" and more characters it becomes much more complicated to check it via that type of regex style – Ares02 Nov 28 '18 at 14:48

2 Answers2

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In Java you can use StringUtils.countMatches as described here: How do I count the number of occurrences of a char in a String?

This is not a regex based solution but it is a Java solution. Several answers there show other techniques.

Freiheit
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The "ordered" way to do your task using regex is as follows:

  • Start with ^ anchor.
  • Then put 3 positive lookaheads for particular conditions (details below).
  • And finally put .*$ matching the whole string.

And now get down to particular conditions and respective lookaheads:

  1. A single a: (?=[^a]*a[^a]*$) - there should be:

    • [^a]* - A number of characters different than a, possibly 0.
    • a - Then just a single a.
    • [^a]* - Then again a number of characters different than a, possibly 0.
    • $ - And finally the end of string.

  2. Five times b: (?=[^b]*(?:b[^b]*){5}$). This time b[^b]* is the content of a non-capturing group, needed due to the quantifier after it.

  3. Seven times z: (?=[^z]*(?:z[^z]*){7}$). Like before, but this time the letter in question is z and the quantifier is 7.

So the whole regex can look like below:

^(?=[^a]*a[^a]*$)(?=[^b]*(?:b[^b]*){5}$)(?=[^z]*(?:z[^z]*){7}$).*$

For a working example see https://regex101.com/r/gUxC2C/1

Try to experiment with the content, adding or removing a, b or z and you will see appearing / disappearing match.

Valdi_Bo
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