Regular expression to match string not containing a word.

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

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^((?!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.

Explanation

A string is just a list of

characters. Before, and after each character, there’s an empty string. So a list of

characters will have

empty strings. Consider the string

:

1
2
3
4
    +--+---+--+---+--+---+--+---+--+---+--+---+--+---+--+---+--+

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

‘s are the empty strings. The regex

looks ahead to see if there’s no substring

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-assertions because 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

up ahead, before a character is consumed by the

(dot). The regex

will do that only once, so it is wrapped in a group, and repeated zero or more times:

. Finally, the start- and end-of-input are anchored to make sure the entire input is consumed:

As you can see, the input

will fail because on

, the regex

fails (there is

up ahead!).

Jan 18 at 15:49 community-wiki Thanks to Bart Kiers