Regex: matching up to the first occurrence of word with character 'a' in it

Question

I need a regular expression to match the first word with character 'a' in it for each line. For example my test string is this:

bbsc abcd aaaagdhskss
dsaa asdd aaaagdfhdghd
wwer wwww awww wwwd

Only the ones in BOLD fonts should be matched. How can I do that? I can match all the words with 'a' in it, but can't figure out how to only match the first occurrence.

Answer 1

Under the assumption that the only characters being used are word characters, ie \\w characters, and white space then use:

/^(?:[^a ]+ +)*([^a ]*a\w*)\b/gm

1. ^ Matches the start of the line
2. (?:[^a ]+ +)* Matches 0 or more occurrences of words composed of any character other than an a followed by one or more spaces in a non-capturing group.
3. ([^a ]*a\\w*)\\b Matches a word ending on a word boundary (it is already guaranteed to begin on a word boundary) that contains an a . The word-boundary constraint allows for the word to be at the end of the line.

The first word with an a in it will be in group #1.

See demo

If we cannot assume that only word ( \\w ) and white space characters are present, then use:

^(?:[^a ]+ +)*(\w*a\w*)\b

The difference is in scanning the first word with an a in it, (\\w*a\\w*) , where we are guaranteed that we are scanning a string composed of only word characters.

Answer 2

What are you using? In many programs you can set limit. If possible: \\b[bz]*a[az]* with 1 limit.

If it is not possible, use group to capture and match latter: ([bz]*a[az]*).*

Answer 3

Try:

^(?:[^a ]+ )*(\w*a\w*) .*$

Basically what it says is: capture a bunch of words that are composed of anything but the letter a (or <space> ) then capture a word that must include the letter a .

should hold the first word with a .应该保持的第一个字a 。