Regular expression to match text that "doesn't" contain a word?

Question

I would like to exclude certain words using a regular expression.

Input Text:

aaa1234 cc bbb1234 c1234 cc dd aacc cccc ccadf cc

Output Text:

aaa1234 bbb1234 c1234 dd aacc cccc ccadf

exclude word: cc

I used (^|\s)[^(cc)]+(\s|$)

How to make the regular expression work?

Using [^(cc)] is same as [^c)(] I have changed the position to make it clear — Rakholiya Jenish, Jul 08 '15 at 05:35
A possible duplicate to http://stackoverflow.com/questions/2078915/a-regular-expression-to-exclude-a-word-string — menakas, Jul 08 '15 at 05:43
It's easier to just match what you want to exclude and replace it with the empty string. — Uphill_ What '1, Jul 08 '15 at 06:20

score 1 · Accepted Answer · answered Jul 08 '15 at 05:35

1

\s+\bcc\b|\bcc\b\s+

Try this.Replace by empty string.See demo.

answered Jul 08 '15 at 05:35

vks

Pranav C Balan · Answer 2 · 2015-07-08T06:20:11.560

0

You can use something like this (^|\s+)cc(\s+|$) to filter word cc, then replace it by empty string

var word='aaa1234 cc bbb1234 c1234 cc dd aacc cccc ccadf cc';

console.log(word.replace(/(^|\s+)cc(\s+|$)/g,' ').trim());

edited Jul 08 '15 at 06:20

answered Jul 08 '15 at 05:38

Pranav C Balan

Hmm ok, I see that now. It doesn't give the required output `aaa1234 bbb1234 c1234 dd aacc cccc ccadf` though but gives `aaa1234bbb1234 c1234dd aacc cccc ccadf` – Jerry Jul 08 '15 at 05:49
The first leaves double spaces, the second leaves a trailing space (at the end). – Jerry Jul 08 '15 at 05:58

2 Answers2