How to make OR operator in regular expression?

Question

What i want is something like this:

(
[integer OR (ANY but not integer or white-space)]
[(ONE white-space OR NONE)]
[integer OR (ANY but not integer or white-space)]
)

Example of strings that will match: 99 $ 99$ $ 99 $99

what i have now is two regular expression :

^[^\d\s](\s{0,1})\d+ AND ^\d+(\s{0,1})[^\d\s]

any ideas of how to replace those two with only one regular expression?

Answer 1

The standard way to make a or in a regular expression is the pipe | (ref)

If you need to match one or the other of your regexp, you need to match (A|B) with A=^[^\d\s](\s{0,1})\d+ and B=^\d+(\s{0,1})[^\d\s]

Result is ^(([^\d\s](\s{0,1})\d+)|(\d+(\s{0,1})[^\d\s]))

Answer 2

"|" is the "or" equivalent in regex. Group them and put | in between.

[^\d\s](\s{0,1})\d+|^\d+(\s{0,1})[^\d\s]

works just fine for your examples.

Answer 3

^((?:\d+\s?[^\d\s]+)|(?:[^\d\s]+\s?\d+))$

Try this.See demo.

https://regex101.com/r/hI0qP0/2