Regular expression with two non-repeating symbols in any order

Question

I need to create the regex that will match such string:

AA+1.01*2.01,BB*2.01+1.01,CC

Order of * and + should be any I've created the following regex:

^(([A-Z][A-Z](([*+][0-9]+(\.[0-9])?[0-9]?){0,2}),)*[A-Z]{2}([*+][0-9]+(\.[0-9])?[0-9]?){0,2})$

But the problem is that with this regex + or * could be used twice but I only need any of them once so the following strings matches should be:

AA+1*2,CC - true
AA+1+2,CC - false (now is true with my regex)
AA*1+2,CC - true
AA*1*2,CC - false (now is true with my regex)

Answer 1

Either of the [+*] should be captured first and then use negative lookahead to match the other one.

Regex: [AZ]{2}([+*])(?:\\d+(?:\\.\\d+)?)(?!\\1)[+*](?:\\d+(?:\\.\\d+)?),[AZ]{2}

Explanation:

• [AZ]{2} Matches two upper case letters.

• ([+*]) captures either of + or * .

• (?:\\d+(?:\\.\\d+)?) matches number with optional decimal part.

• (?!\\1)[+*] looks ahead for symbol captured and matched the other one. So if + is captured previously then * will be matched.

• (?:\\d+(?:\\.\\d+)?) matches number with optional decimal part.

• ,[AZ]{2} matches , followed by two upper case letters.

Regex101 Demo

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To match the first case AA+1.01*2.01,BB*2.01+1.01,CC which is just a little advancement over previous pattern, use following regex.

Regex: (?:[AZ]{2}([+*])(?:\\d+(?:\\.\\d+)?)(?!\\1)[+*](?:\\d+(?:\\.\\d+)?),)+[AZ]{2}

Explanation: Added whole pattern except ,CC in first group and made it greedy by using + to match one or more such patterns.

Regex101 Demo

Answer 2

To get a regex to match your given example, extended to an arbitrary number of commas, you could use:

^(?:[A-Z]{2}([+*])?\d*\.?\d*(?!\1)[+*]?\d*\.?\d*,?)*$

Note that this example will also allow a trailing comma. I'm not sure if there is much you can do about that.

Regex 101 Example

If the trailing comma is an issue:

^(?:[A-Z]{2}([+*])?\d*\.?\d*(?!\1)[+*]?\d*\.?\d*,?)*?(?:[A-Z]{2}([+*])?\d*\.?\d*(?!\2)[+*]?\d*\.?\d*?)$

Regex 101 Example