In this case we want a delimiter that’s a little more concrete, so “one or more” ('ended', () => tState( from 'react'Ĭonst currentIndex = players.findIndex(p => p. The most common forms of whitespace you will use with regular expressions are the space ( ), the tab ( \t ), the new line ( ) and the carriage return ( \r) (useful in Windows environments), and these special characters match each of their respective whitespaces. If you want to search for 'my cats', instead of: myString.
Regex whitespace plus#
The reason I used a + instead of a ‘*’ is because a plus is defined as one or more of the preceding element, where an asterisk is zero or more. The regex is good, but the explanation is a bit misleading. While the accepted answer is technically correct, a more practical approach, if possible, is to just strip whitespace out of both the regular expression and the search string. Furthermore, you can find the Troubleshooting Login Issues section which can answer your unresolved problems and equip you with a lot of relevant information. I also preserved the capture group around the whitespace, if you don’t want this then remove the brackets as already described above. Regex Whitespace Characters LoginAsk is here to help you access Regex Whitespace Characters quickly and handle each specific case you encounter. If you want to match multiple whitespaces then see my comments above.
Assuming that the spaces after the first and. Note as I have already pointed out that the negative lookahead ?! will not match when wordB is followed by a single whitespace and wordc. The problem with this regex is that the capture group is greedy and will match anything including whitespace. I replaced wordc with swordc since that is more explicit. The only difference is that not the regex matches whitespace OR. Preserving your original regex you can use:
Hello the problem is not the expression but the HTML out put that are not considered as whitespace. Matching discontinuous text with regex is not possible. Or you can match non-whitespace parts into capturing groups, and then join them. Are you capturing the whitespace to a group for a reason? If not you could just remove the brackets, i.e. 1 You need to match whitespace with \s+ and remove it using regex-based search and replace. (s*) – the brackets indicate a capturing group.
You may want to consider + if you want at least one space. * will match 0 or more spaces so it will match wordAwordB. Of course, if you do want to match lines with wordc after wordB then you shouldn’t use a negative lookahead. If you want to match against more than one space before wordc you can use (?!s*wordc) for 0 or more spaces or (?!s*+wordc) for 1 or more spaces depending on what your intention is. You may want to be more precise and use (?!swordc). Thus, it does more than just 'check if there is at least one non-whitespace character'. Your regex matches a string that consists entirely of non-whitespace characters and is at least one character in length. matches 13 + at least one of any character (including a space): 1133. Currently you are relying on the space after ?! to match the whitespace. The regex is good, but the explanation is a bit misleading. Regular Expressions are sets of characters and/or metacharacters that match (or. (?! wordc) is a negative lookahead, so you wont match lines wordA wordB wordc which is assume is intended (and is why the last line is not matched). Note that all matches are replaced no matter how many spaces. Here are some example matches and the associated replacement output: Note the single space between ?! and wordc which means that wordA wordB wordc will not match, but wordA wordB wordc will. Assuming that it is doing what you want it to.
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