Javascript正则表达式:找到一个单词NOT，后跟空格字符 [英] Javascript regex: find a word NOT followed by space character

问题描述

我需要javascript正则表达式，以匹配不带空格字符且之前带有@的单词，例如:

I need javascript regex that will match words that are NOT followed by space character and has @ before, like this:

@bug-查找"@bug"，因为之后没有空格

@bug - finds "@bug", because no space afer it

@bug和我-什么也找不到，因为"@bug"后有空格

@bug and me - finds nothing because there is space after "@bug"

@bug和@another-仅找到"@another"

@bug and @another - finds "@another" only

@bug和@another之类的东西-找不到任何东西，因为两个词都跟着空格.

@bug and @another and something - finds nothing because both words are followed by space.

有帮助吗? 添加: 从中获取字符串，并且FF在其末尾放置自己的标签.尽管我基本上只需要以@开头的最后一个单词，但是不能使用$(字符串的结尾).

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.

推荐答案

尝试re = /@\w+\b(?! )/.这将查找一个单词(确保它捕获了整个单词)，并使用否定的超前查询来确保单词后没有空格.

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.

使用上面的设置:

var re = /@\w+\b(?! )/, // etc etc

for ( var i=0; i<cases.length; i++ ) {
    print( re2.exec(cases[i]) )
}

//prints
@bug
null
@another
null

这唯一不起作用的方法是，如果您的单词以下划线结尾，并且您希望标点符号成为单词的一部分:例如，由于@another，'@ bug和@another_ blahblah'将选择@another后面没有空格. 似乎不太可能，但是如果您也想处理这种情况，则可以使用/@\w+\b(?![\w ]/，对于@bug and @another_来说返回null，对于@another and @bug_则返回@bug_.

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_.

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