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I am using this regex:
((?:[a-z][a-z]+))_(\d+)_((?:[a-z][a-z]+)\d+)_(\d{13}) to match strings like this:
SH_6208069141055_BC000388_20110412101855 separating into 4 groups:
SH 6208069141055 BC000388 20110412101855 Question: How do I make the first group optional, so that the resulting group is a empty string?
I want to get 4 groups in every case, when possible.
Input string for this case: (no underline after the first group)
6208069141055_BC000388_20110412101855 
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For example, the pattern ab? c will match either the strings "abc" or "ac" because the b is considered optional. Similar to the dot metacharacter, the question mark is a special character and you will have to escape it using a slash \? to match a plain question mark character in a string.
So to make any group optional, we need to have to put a “?” after the pattern or group. This question mark makes the preceding group or pattern optional. This question mark is also known as a quantifier.
What you want to do is to grab the first group of the match result, surrounded by () in the regex and the way to do this is to use the non-capturing group syntax, i.e. ?: . So the regex (\p{Alpha}*[a-z])(?:@example.com) will return just the id part of the email.
Making a non-capturing, zero to more matching group, you must append ?.
(?: ..... )? ^ ^____ optional |____ group You can easily simplify your regex to be this:
(?:([a-z]{2,})_)?(\d+)_([a-z]{2,}\d+)_(\d+)$ ^ ^^ |--------------|| | first group ||- quantifier for 0 or 1 time (essentially making it optional) I'm not sure whether the input string without the first group will have the underscore or not, but you can use the above regex if it's the whole string.
regex101 demo
As you can see, the matched group 1 in the second match is empty and starts at matched group 2.
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