问题
I need javascript regex that will match words that are NOT followed by space character and has @ before, like this:
@bug - finds "@bug", because no space afer it
@bug and me - finds nothing because there is space after "@bug"
@bug and @another - finds "@another" only
@bug and @another and something - finds nothing because both words are followed by space.
Help? Added: string is fetched from and FF puts it's own tags at end of it. Although I basically need only the last word starting with @, $ (end-of-string) can not be used.
回答1:
Try re = /@\w+\b(?! )/
. This looks for a word (making sure it captures the whole word) and uses a negative lookahead to make sure the word is not followed by a space.
Using the setup above:
var re = /@\w+\b(?! )/, // etc etc
for ( var i=0; i<cases.length; i++ ) {
print( re2.exec(cases[i]) )
}
//prints
@bug
null
@another
null
The only way this won't work is if your word ends in an underscore and you wanted that punctuation to be part of the word: For example '@bug and @another_ blahblah' will pick out @another since @another
wasn't followed by a space.
This doesn't seem very likely but if you wanted to deal with that case too, you could use /@\w+\b(?![\w ]/
and that would return null
for @bug and @another_
and @bug_
for @another and @bug_
.
回答2:
It sounds like you're really just looking for words at the end of the input:
/@\w+$/
Tests:
var re = /@\w+$/,
cases = ['@bug',
'@bug and me',
'@bug and @another',
'@bug and @another and something'];
for (var i=0; i<cases.length; i++)
{
console.log(cases[i], ':', re.test(cases[i]), re.exec(cases[i]));
}
// prints
@bug : true ["@bug"]
@bug and me : false null
@bug and @another : true ["@another"]
@bug and @another and something : false null
来源:https://stackoverflow.com/questions/8584981/javascript-regex-find-a-word-not-followed-by-space-character