Regex – How to match everything except a particular pattern?

Match everything except for specified strings

 

You could use a look-ahead assertion:

(?!999)\d{3}

This example matches three digits other than 999.

But if you happen not to have a regular expression implementation with this feature (see Comparison of Regular Expression Flavors), you probably have to build a regular expression with the basic features on your own.

A compatible regular expression with basic syntax only would be:

[0-8]\d\d|\d[0-8]\d|\d\d[0-8]

This does also match any three digits sequence that is not 999.

Example 1

If you want to match a word A in a string and not to match a word B. For example: If you have a text:

1. I have a two pets - dog and a cat
2. I have a pet - dog

If you want to search for lines of text that HAVE a dog for a pet and DOESN’T have cat you can use this regular expression:

^(?=.*?\bdog\b)((?!cat).)*$

It will find only second line:

2. I have a pet - dog

Example 2

Regular expression to match a line that doesn’t contain a word?

Input:

hoho
hihi
haha
hede

Code:

grep "<Regex for 'doesn't contain hede'>" input

Desired output:

hoho
hihi
haha

Solution:

The notion that regex doesn’t support inverse matching is not entirely true. You can mimic this behavior by using negative look-arounds:

^((?!hede).)*$

The regex above will match any string, or line without a line break, not containing the (sub)string ‘hede’. As mentioned, this is not something regex is “good” at (or should do), but still, it is possible.

And if you need to match line break chars as well, use the DOT-ALL modifier (the trailing s in the following pattern):

/^((?!hede).)*$/s

or use it inline:

/(?s)^((?!hede).)*$/

(where the /.../ are the regex delimiters, i.e., not part of the pattern)

If the DOT-ALL modifier is not available, you can mimic the same behavior with the character class [\s\S]:

/^((?!hede)[\s\S])*$/

Explanation:

A string is just a list of n characters. Before, and after each character, there’s an empty string. So a list of n characters will have n+1 empty strings. Consider the string "ABhedeCD":

    ┌────┬───┬────┬───┬────┬───┬────┬───┬────┬───┬────┬───┬────┬───┬────┬───┬────┐ 
S = │ e1 │ A │ e2 │ B │ e3 │ h │ e4 │ e │ e5 │ d │ e6 │ e │ e7 │ C │ e8 │ D │ e9 │ 
    └────┴───┴────┴───┴────┴───┴────┴───┴────┴───┴────┴───┴────┴───┴────┴───┴────┘ 

index      0        1        2        3        4        5        6        7

where the e‘s are the empty strings. The regex (?!hede). looks ahead to see if there’s no substring "hede" to be seen, and if that is the case (so something else is seen), then the . (dot) will match any character except a line break. Look-arounds are also called zero-width-assertionsbecause they don’t consume any characters. They only assert/validate something.

So, in my example, every empty string is first validated to see if there’s no "hede" up ahead, before a character is consumed by the . (dot). The regex (?!hede). will do that only once, so it is wrapped in a group, and repeated zero or more times: ((?!hede).)*. Finally, the start- and end-of-input are anchored to make sure the entire input is consumed: ^((?!hede).)*$

As you can see, the input "ABhedeCD" will fail because on e3, the regex (?!hede) fails (there is "hede" up ahead!).

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