Regular expression to match a line that doesn't contain a word – Dev

The best answers to the question “Regular expression to match a line that doesn't contain a word” in the category Dev.

QUESTION:

I know it’s possible to match a word and then reverse the matches using other tools (e.g. grep -v). However, is it possible to match lines that do not contain a specific word, e.g. hede, using a regular expression?

Input:

hoho
hihi
haha
hede

Code:

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

Desired output:

hoho
hihi
haha

ANSWER:

Note that the solution to does not start with “hede”:

^(?!hede).*$

is generally much more efficient than the solution to does not contain “hede”:

^((?!hede).)*$

The former checks for “hede” only at the input string’s first position, rather than at every position.

ANSWER:

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-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 "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!).

ANSWER:

Answer:

^((?!hede).)*$

Explanation:

^the beginning of the string,
( group and capture to \1 (0 or more times (matching the most amount possible)),
(?! look ahead to see if there is not,

hede your string,

) end of look-ahead,
. any character except \n,
)* end of \1 (Note: because you are using a quantifier on this capture, only the LAST repetition of the captured pattern will be stored in \1)
$ before an optional \n, and the end of the string

ANSWER:

If you’re just using it for grep, you can use grep -v hede to get all lines which do not contain hede.

ETA Oh, rereading the question, grep -v is probably what you meant by “tools options”.