正则表达式不匹配字符串中的模式

Question

I am a newbie and have been struggling the last hour to figure this out. Let's say you have these strings:

baa cec haw heef baas bat jackaay

I want to match all the words which don't have two aa's consecutively, so in the above it will match cec , haw , heef , bat .

This is what i have done so far, but it's completely wrong i can sense :D

\w*[^\s]*[^a\s]{2}[^\s]*\w*

Answer 1

You maybe want to use negative lookahead:

/(^|\s)(?!\w*aa\w*)(\w+)/gi

You can check your string by paste this code on console on Chrome/Firefox (F12):

var pattern = /(^|\s)(?!\w*aa\w*)(\w+)/gi;
var str = 'baa cec haw heef baas bat jackaay';
while(match = pattern.exec(str))
    console.log(match[2]); // position 2 is (\w+) in regex

You can read more about lookahead here . See it on Regex101 to see how this regex work.

Answer 2

You need a regex that has 2 things: a word boundary \\b and a negative lookahead right after it (it will be sort of anchored that way) that will lay restrictions to the subpattern that follows.

\b(?!\w*aa)\w+

See the regex demo

Regex breakdown:

• \\b - word boundary
• (?!\\w*aa) - the negative lookahead that will cancel a match if the word has 0 or more word characters followed by two a s
• \\w+ - 1 or more word characters.

Code demo:

 var re = /\\b(?!\\w*aa)\\w+/gi; var str = 'baa cec haw heef bAas bat jackaay bar ha aa lar'; var res = str.match(re); document.write(JSON.stringify(res)); 

Answer 3

in javascript, you could use filter and regex invert ! a non-capturing group ?: .

var strings = ['baa','cec','haw','heef','baas','bat','jackaay'];
strings = $(strings).filter(function(index, element){
   return !/.*(?:aa).*/.test(element);                // regex => .*(?:aa).*
});