How to negate specific word in regex? [duplicate]

Answer 1

Extracted from this comment by bkDJ:

^(?!bar$).*

The nice property of this solution is that it's possible to clearly negate (exclude) multiple words:

^(?!bar$|foo$|banana$).*

Answer 2

A great way to do this is to use negative lookahead:

^(?!.*bar).*$

The negative lookahead construct is the pair of parentheses, with the opening parenthesis followed by a question mark and an exclamation point. Inside the lookahead [is any regex pattern].

Answer 3

I wish to complement the accepted answer and contribute to the discussion with my late answer.

@ChrisVanOpstal shared this regex tutorial which is a great resource for learning regex.

However, it was really time consuming to read through.

I made a cheatsheet for mnemonic convenience.

This reference is based on the braces [], (), and {} leading each class, and I find it easy to recall.

Regex = {
 'single_character': ['[]', '.', {'negate':'^'}],
 'capturing_group' : ['()', '|', '\\', 'backreferences and named group'],
 'repetition'      : ['{}', '*', '+', '?', 'greedy v.s. lazy'],
 'anchor'          : ['^', '\b', '$'],
 'non_printable'   : ['\n', '\t', '\r', '\f', '\v'],
 'shorthand'       : ['\d', '\w', '\s'],
 }

Answer 4

If it's truly a word, bar that you don't want to match, then:

^(?!.*\bbar\b).*$

The above will match any string that does not contain bar that is on a word boundary, that is to say, separated from non-word characters. However, the period/dot (.) used in the above pattern will not match newline characters unless the correct regex flag is used:

^(?s)(?!.*\bbar\b).*$

Alternatively:

^(?!.*\bbar\b)[\s\S]*$

Instead of using any special flag, we are looking for any character that is either white space or non-white space. That should cover every character.

But what if we would like to match words that might contain bar, but just not the specific word bar?

(?!\bbar\b)\b\[A-Za-z-]*bar[a-z-]*\b

1. (?!\bbar\b) Assert that the next input is not bar on a word boundary.
2. \b\[A-Za-z-]*bar[a-z-]*\b Matches any word on a word boundary that contains bar.

See Regex Demo

Answer 5

I came across this forum thread while trying to identify a regex for the following English statement:

Given an input string, match everything unless this input string is exactly 'bar'; for example I want to match 'barrier' and 'disbar' as well as 'foo'.

Here's the regex I came up with

^(bar.+|(?!bar).*)$

My English translation of the regex is "match the string if it starts with 'bar' and it has at least one other character, or if the string does not start with 'bar'.

Answer 6

The following regex will do what you want (as long as negative lookbehinds and lookaheads are supported), matching things properly; the only problem is that it matches individual characters (i.e. each match is a single character rather than all characters between two consecutive "bar"s), possibly resulting in a potential for high overhead if you're working with very long strings.

b(?!ar)|(?<!b)a|a(?!r)|(?<!ba)r|[^bar]