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I need help putting together a regex that will match word that ends with "Id" with case sensitive match.
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To run a “whole words only” search using a regular expression, simply place the word between two word boundaries, as we did with ‹ \bcat\b ›. The first ‹ \b › requires the ‹ c › to occur at the very start of the string, or after a nonword character.
$ means "Match the end of the string" (the position after the last character in the string).
Basically (0+1)* mathes any sequence of ones and zeroes. So, in your example (0+1)*1(0+1)* should match any sequence that has 1. It would not match 000 , but it would match 010 , 1 , 111 etc. (0+1) means 0 OR 1.
In regex, the uppercase metacharacter is always the inverse of the lowercase counterpart. \d (digit) matches any single digit (same as [0-9] ). The uppercase counterpart \D (non-digit) matches any single character that is not a digit (same as [^0-9] ).
Try this regular expression:
\w*Id\b \w* allows word characters in front of Id and the \b ensures that Id is at the end of the word (\b is word boundary assertion).
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Gumbo gets my vote, however, the OP doesn't specify whether just "Id" is an allowable word, which means I'd make a minor modification:
\w+Id\b 1 or more word characters followed by "Id" and a breaking space. The [a-zA-Z] variants don't take into account non-English alphabetic characters. I might also use \s instead of \b as a space rather than a breaking space. It would depend if you need to wrap over multiple lines.
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