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

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 1

Using (?<!hede) is a better answer. (?<!whateverYouDontWantToMatch) is a negative look behind as opposed to (?!whateverYouDontWantToMatch) which is a negative look ahead. which means that with (?<!) it will check right at the current position of the string instead of only looking after the match. So for example. You will run into issues using (?!) and it only works in this case because of the anchor.

Answer 2

While its true that you can use a look around, I read an article which uses another method which seems more elegant and less syntactically cumbersome.

The idea is counterintuitive: to actually match the stuff you don't want to match but only match what you want to match inside groups & reference them later.

E.g. Blacklisting words: pattern='\bTarzan\b|\bJane\b|(\w+)' Then just use group(1) e.g. '\1' to get your matches which exclude the prefixed words.

Great article describing this: https://www.rexegg.com/regex-best-trick.html#simplecase & Great SO Answer which also describes it: How do (*SKIP) or (*F) work on regex?

Answer 3

# 一个简单的方式
import re
skip_word = 'hede'
stranger_char = '虩'
content = '''hoho
hihi
haha
hede'''
print(
    '\n'.join(re.findall(
        '([^{}]*?)\n'.format(stranger_char), 
        content.replace(skip_word, stranger_char)
    )).replace(stranger_char, skip_word) 
)

# hoho
# hihi
# haha

Answer 4

Maybe you'll find this on Google while trying to write a regex that is able to match segments of a line (as opposed to entire lines) which do not contain a substring. Tooke me a while to figure out, so I'll share:

Given a string:

<span class="good">bar</span><span class="bad">foo</span><span class="ugly">baz</span>

I want to match <span> tags which do not contain the substring "bad".

/<span(?:(?!bad).)*?> will match <span class=\"good\"> and <span class=\"ugly\">.

Notice that there are two sets (layers) of parentheses:

• The innermost one is for the negative lookahead (it is not a capture group)
• The outermost was interpreted by Ruby as capture group but we don't want it to be a capture group, so I added ?: at it's beginning and it is no longer interpreted as a capture group.

Demo in Ruby:

s = '<span class="good">bar</span><span class="bad">foo</span><span class="ugly">baz</span>'
s.scan(/<span(?:(?!bad).)*?>/)
# => ["<span class=\"good\">", "<span class=\"ugly\">"]

Answer 5

A simpler solution is to use the not operator !

Your if statement will need to match "contains" and not match "excludes".

var contains = /abc/;
var excludes =/hede/;

if(string.match(contains) && !(string.match(excludes))){  //proceed...

I believe the designers of RegEx anticipated the use of not operators.

Answer 6

The below function will help you get your desired output

<?PHP
      function removePrepositions($text){
            
            $propositions=array('/\bfor\b/i','/\bthe\b/i'); 
        
            if( count($propositions) > 0 ) {
                foreach($propositions as $exceptionPhrase) {
                    $text = preg_replace($exceptionPhrase, '', trim($text));

                }
            $retval = trim($text);

            }
        return $retval;
    }
     
        
?>

Answer 7

The TXR Language supports regex negation.

$ txr -c '@(repeat)
@{nothede /~hede/}
@(do (put-line nothede))
@(end)'  Input

A more complicated example: match all lines that start with a and end with z, but do not contain the substring hede:

$ txr -c '@(repeat)
@{nothede /a.*z&~.*hede.*/}
@(do (put-line nothede))
@(end)' -
az         <- echoed
az
abcz       <- echoed
abcz
abhederz   <- not echoed; contains hede
ahedez     <- not echoed; contains hede
ace        <- not echoed; does not end in z
ahedz      <- echoed
ahedz

Regex negation is not particularly useful on its own but when you also have intersection, things get interesting, since you have a full set of boolean set operations: you can express "the set which matches this, except for things which match that".

Answer 8

If you want to match a character to negate a word similar to negate character class:

For example, a string:

<?
$str="aaa        bbb4      aaa     bbb7";
?>

Do not use:

<?
preg_match('/aaa[^bbb]+?bbb7/s', $str, $matches);
?>

Use:

<?
preg_match('/aaa(?:(?!bbb).)+?bbb7/s', $str, $matches);
?>

Notice "(?!bbb)." is neither lookbehind nor lookahead, it's lookcurrent, for example:

"(?=abc)abcde", "(?!abc)abcde"

Answer 9

Not regex, but I've found it logical and useful to use serial greps with pipe to eliminate noise.

eg. search an apache config file without all the comments-

grep -v '\#' /opt/lampp/etc/httpd.conf      # this gives all the non-comment lines

and

grep -v '\#' /opt/lampp/etc/httpd.conf |  grep -i dir

The logic of serial grep's is (not a comment) and (matches dir)

Answer 10

Since no one else has given a direct answer to the question that was asked, I'll do it.

The answer is that with POSIX grep, it's impossible to literally satisfy this request:

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

The reason is that with no flags, POSIX grep is only required to work with Basic Regular Expressions (BREs), which are simply not powerful enough for accomplishing that task, because of lack of alternation in subexpressions. The only kind of alternation it supports involves providing multiple regular expressions separated by newlines, and that doesn't cover all regular languages, e.g. there's no finite collection of BREs that matches the same regular language as the extended regular expression (ERE) ^(ab|cd)*$.

However, GNU grep implements extensions that allow it. In particular, \| is the alternation operator in GNU's implementation of BREs. If your regular expression engine supports alternation, parentheses and the Kleene star, and is able to anchor to the beginning and end of the string, that's all you need for this approach. Note however that negative sets [^ ... ] are very convenient in addition to those, because otherwise, you need to replace them with an expression of the form (a|b|c| ... ) that lists every character that is not in the set, which is extremely tedious and overly long, even more so if the whole character set is Unicode.

Thanks to formal language theory, we get to see how such an expression looks like. With GNU grep, the answer would be something like:

grep "^\([^h]\|h\(h\|eh\|edh\)*\([^eh]\|e[^dh]\|ed[^eh]\)\)*\(\|h\(h\|eh\|edh\)*\(\|e\|ed\)\)$" input

(found with Grail and some further optimizations made by hand).

You can also use a tool that implements EREs, like egrep, to get rid of the backslashes, or equivalently, pass the -E flag to POSIX grep (although I was under the impression that the question required avoiding any flags to grep whatsoever):

egrep "^([^h]|h(h|eh|edh)*([^eh]|e[^dh]|ed[^eh]))*(|h(h|eh|edh)*(|e|ed))$" input

Here's a script to test it (note it generates a file testinput.txt in the current directory). Several of the expressions presented in other answers fail this test.

#!/bin/bash
REGEX="^\([^h]\|h\(h\|eh\|edh\)*\([^eh]\|e[^dh]\|ed[^eh]\)\)*\(\|h\(h\|eh\|edh\)*\(\|e\|ed\)\)$"

# First four lines as in OP's testcase.
cat > testinput.txt <<EOF
hoho
hihi
haha
hede

h
he
ah
head
ahead
ahed
aheda
ahede
hhede
hehede
hedhede
hehehehehehedehehe
hedecidedthat
EOF
diff -s -u <(grep -v hede testinput.txt) <(grep "$REGEX" testinput.txt)

In my system it prints:

Files /dev/fd/63 and /dev/fd/62 are identical

as expected.

For those interested in the details, the technique employed is to convert the regular expression that matches the word into a finite automaton, then invert the automaton by changing every acceptance state to non-acceptance and vice versa, and then converting the resulting FA back to a regular expression.

As everyone has noted, if your regular expression engine supports negative lookahead, the regular expression is much simpler. For example, with GNU grep:

grep -P '^((?!hede).)*$' input

However, this approach has the disadvantage that it requires a backtracking regular expression engine. This makes it unsuitable in installations that are using secure regular expression engines like RE2, which is one reason to prefer the generated approach in some circumstances.

Using Kendall Hopkins' excellent FormalTheory library, written in PHP, which provides a functionality similar to Grail, and a simplifier written by myself, I've been able to write an online generator of negative regular expressions given an input phrase (only alphanumeric and space characters currently supported, and the length is limited): http://www.formauri.es/personal/pgimeno/misc/non-match-regex/

For hede it outputs:

^([^h]|h(h|e(h|dh))*([^eh]|e([^dh]|d[^eh])))*(h(h|e(h|dh))*(ed?)?)?$

which is equivalent to the above.

Answer 11

With negative lookahead, regular expression can match something not contains specific pattern. This is answered and explained by Bart Kiers. Great explanation!

However, with Bart Kiers' answer, the lookahead part will test 1 to 4 characters ahead while matching any single character. We can avoid this and let the lookahead part check out the whole text, ensure there is no 'hede', and then the normal part (.*) can eat the whole text all at one time.

Here is the improved regex:

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

Note the (*?) lazy quantifier in the negative lookahead part is optional, you can use (*) greedy quantifier instead, depending on your data: if 'hede' does present and in the beginning half of the text, the lazy quantifier can be faster; otherwise, the greedy quantifier be faster. However if 'hede' does not present, both would be equal slow.

Here is the demo code.

For more information about lookahead, please check out the great article: Mastering Lookahead and Lookbehind.

Also, please check out RegexGen.js, a JavaScript Regular Expression Generator that helps to construct complex regular expressions. With RegexGen.js, you can construct the regex in a more readable way:

var _ = regexGen;

var regex = _(
    _.startOfLine(),             
    _.anything().notContains(       // match anything that not contains:
        _.anything().lazy(), 'hede' //   zero or more chars that followed by 'hede',
                                    //   i.e., anything contains 'hede'
    ), 
    _.endOfLine()
);

Answer 12

FWIW, since regular languages (aka rational languages) are closed under complementation, it's always possible to find a regular expression (aka rational expression) that negates another expression. But not many tools implement this.

Vcsn supports this operator (which it denotes {c}, postfix).

You first define the type of your expressions: labels are letter (lal_char) to pick from a to z for instance (defining the alphabet when working with complementation is, of course, very important), and the "value" computed for each word is just a Boolean: true the word is accepted, false, rejected.

In Python:

In [5]: import vcsn
        c = vcsn.context('lal_char(a-z), b')
        c
Out[5]: {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z} → 𝔹

then you enter your expression:

In [6]: e = c.expression('(hede){c}'); e
Out[6]: (hede)^c

convert this expression to an automaton:

In [7]: a = e.automaton(); a

finally, convert this automaton back to a simple expression.

In [8]: print(a.expression())
        \e+h(\e+e(\e+d))+([^h]+h([^e]+e([^d]+d([^e]+e[^]))))[^]*

where + is usually denoted |, \e denotes the empty word, and [^] is usually written . (any character). So, with a bit of rewriting ()|h(ed?)?|([^h]|h([^e]|e([^d]|d([^e]|e.)))).*.

You can see this example here, and try Vcsn online there.

Answer 13

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 14

Simplest thing that I could find would be

[^(hede)]

Tested at https://regex101.com/

You can also add unit-test cases on that site

Answer 15

Benchmarks

I decided to evaluate some of the presented Options and compare their performance, as well as use some new Features. Benchmarking on .NET Regex Engine: http://regexhero.net/tester/

Benchmark Text:

The first 7 lines should not match, since they contain the searched Expression, while the lower 7 lines should match!

Regex Hero is a real-time online Silverlight Regular Expression Tester.
XRegex Hero is a real-time online Silverlight Regular Expression Tester.
Regex HeroRegex HeroRegex HeroRegex HeroRegex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her Regex Her Regex Her Regex Her Regex Her Regex Her Regex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her is a real-time online Silverlight Regular Expression Tester.Regex Hero
egex Hero egex Hero egex Hero egex Hero egex Hero egex Hero Regex Hero is a real-time online Silverlight Regular Expression Tester.
RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRegex Hero is a real-time online Silverlight Regular Expression Tester.

Regex Her
egex Hero
egex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her is a real-time online Silverlight Regular Expression Tester.
Regex Her Regex Her Regex Her Regex Her Regex Her Regex Her is a real-time online Silverlight Regular Expression Tester.
Nobody is a real-time online Silverlight Regular Expression Tester.
Regex Her o egex Hero Regex  Hero Reg ex Hero is a real-time online Silverlight Regular Expression Tester.

Results:

Results are Iterations per second as the median of 3 runs - Bigger Number = Better

01: ^((?!Regex Hero).)*$                    3.914   // Accepted Answer
02: ^(?:(?!Regex Hero).)*$                  5.034   // With Non-Capturing group
03: ^(?!.*?Regex Hero).*                   7.356   // Lookahead at the beginning, if not found match everything
04: ^(?>[^R]+|R(?!egex Hero))*$             6.137   // Lookahead only on the right first letter
05: ^(?>(?:.*?Regex Hero)?)^.*$             7.426   // Match the word and check if you're still at linestart
06: ^(?(?=.*?Regex Hero)(?#fail)|.*)$       7.371   // Logic Branch: Find Regex Hero? match nothing, else anything

P1: ^(?(?=.*?Regex Hero)(*FAIL)|(*ACCEPT))  ?????   // Logic Branch in Perl - Quick FAIL
P2: .*?Regex Hero(*COMMIT)(*FAIL)|(*ACCEPT) ?????   // Direct COMMIT & FAIL in Perl

Since .NET doesn't support action Verbs (*FAIL, etc.) I couldn't test the solutions P1 and P2.

Summary:

The overall most readable and performance-wise fastest solution seems to be 03 with a simple negative lookahead. This is also the fastest solution for JavaScript, since JS does not support the more advanced Regex Features for the other solutions.

Answer 16

Another option is that to add a positive look-ahead and check if hede is anywhere in the input line, then we would negate that, with an expression similar to:

^(?!(?=.*\bhede\b)).*$

with word boundaries.

---

The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.

---

RegEx Circuit

jex.im visualizes regular expressions:

Answer 17

As long as you are dealing with lines, simply mark the negative matches and target the rest.

In fact, I use this trick with sed because ^((?!hede).)*$ looks not supported by it.

For the desired output

1. Mark the negative match: (e.g. lines with hede), using a character not included in the whole text at all. An emoji could probably be a good choice for this purpose.

   s/(.*hede)/🔒\1/g
   

2. Target the rest (the unmarked strings: e.g. lines without hede). Suppose you want to keep only the target and delete the rest (as you want):

   s/^🔒.*//g
   

For a better understanding

Suppose you want to delete the target:

1. Mark the negative match: (e.g. lines with hede), using a character not included in the whole text at all. An emoji could probably be a good choice for this purpose.

   s/(.*hede)/🔒\1/g
   

2. Target the rest (the unmarked strings: e.g. lines without hede). Suppose you want to delete the target:

   s/^[^🔒].*//g
   

3. Remove the mark:

   s/🔒//g
   

Answer 18

^((?!hede).)*$ is an elegant solution, except since it consumes characters you won't be able to combine it with other criteria. For instance, say you wanted to check for the non-presence of "hede" and the presence of "haha." This solution would work because it won't consume characters:

^(?!.*\bhede\b)(?=.*\bhaha\b) 

Answer 19

I wanted to add another example for if you are trying to match an entire line that contains string X, but does not also contain string Y.

For example, let's say we want to check if our URL / string contains "tasty-treats", so long as it does not also contain "chocolate" anywhere.

This regex pattern would work (works in JavaScript too)

^(?=.*?tasty-treats)((?!chocolate).)*$

(global, multiline flags in example)

Interactive Example: https://regexr.com/53gv4

Matches

(These urls contain "tasty-treats" and also do not contain "chocolate")

• example.com/tasty-treats/strawberry-ice-cream
• example.com/desserts/tasty-treats/banana-pudding
• example.com/tasty-treats-overview

Does Not Match

(These urls contain "chocolate" somewhere - so they won't match even though they contain "tasty-treats")

• example.com/tasty-treats/chocolate-cake
• example.com/home-cooking/oven-roasted-chicken
• example.com/tasty-treats/banana-chocolate-fudge
• example.com/desserts/chocolate/tasty-treats
• example.com/chocolate/tasty-treats/desserts

Answer 20

An, in my opinon, more readable variant of the top answer:

^(?!.*hede)

Basically, "match at the beginning of the line if and only if it does not have 'hede' in it" - so the requirement translated almost directly into regex.

Of course, it's possible to have multiple failure requirements:

^(?!.*(hede|hodo|hada))

Details: The ^ anchor ensures the regex engine doesn't retry the match at every location in the string, which would match every string.

The ^ anchor in the beginning is meant to represent the beginning of the line. The grep tool matches each line one at a time, in contexts where you're working with a multiline string, you can use the "m" flag:

/^(?!.*hede)/m # JavaScript syntax

or

(?m)^(?!.*hede) # Inline flag

Answer 21

If you want the regex test to only fail if the entire string matches, the following will work:

^(?!hede$).*

e.g. -- If you want to allow all values except "foo" (i.e. "foofoo", "barfoo", and "foobar" will pass, but "foo" will fail), use: ^(?!foo$).*

Of course, if you're checking for exact equality, a better general solution in this case is to check for string equality, i.e.

myStr !== 'foo'

You could even put the negation outside the test if you need any regex features (here, case insensitivity and range matching):

!/^[a-f]oo$/i.test(myStr)

The regex solution at the top of this answer may be helpful, however, in situations where a positive regex test is required (perhaps by an API).

Answer 22

With ConyEdit, you can use the command line cc.gl !/hede/ to get lines that do not contain the regex matching, or use the command line cc.dl /hede/ to delete lines that contain the regex matching. They have the same result.

Answer 23

with this, you avoid to test a lookahead on each positions:

/^(?:[^h]+|h++(?!ede))*+$/

equivalent to (for .net):

^(?>(?:[^h]+|h+(?!ede))*)$

Old answer:

/^(?>[^h]+|h+(?!ede))*$/

Answer 24

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 25

How to use PCRE's backtracking control verbs to match a line not containing a word

Here's a method that I haven't seen used before:

/.*hede(*COMMIT)^|/

How it works

First, it tries to find "hede" somewhere in the line. If successful, at this point, (*COMMIT) tells the engine to, not only not backtrack in the event of a failure, but also not to attempt any further matching in that case. Then, we try to match something that cannot possibly match (in this case, ^).

If a line does not contain "hede" then the second alternative, an empty subpattern, successfully matches the subject string.

This method is no more efficient than a negative lookahead, but I figured I'd just throw it on here in case someone finds it nifty and finds a use for it for other, more interesting applications.

Answer 26

Since the introduction of ruby-2.4.1, we can use the new Absent Operator in Ruby’s Regular Expressions

from the official doc

(?~abc) matches: "", "ab", "aab", "cccc", etc.
It doesn't match: "abc", "aabc", "ccccabc", etc.

Thus, in your case ^(?~hede)$ does the job for you

2.4.1 :016 > ["hoho", "hihi", "haha", "hede"].select{|s| /^(?~hede)$/.match(s)}
 => ["hoho", "hihi", "haha"]

Answer 27

It may be more maintainable to two regexes in your code, one to do the first match, and then if it matches run the second regex to check for outlier cases you wish to block for example ^.*(hede).* then have appropriate logic in your code.

OK, I admit this is not really an answer to the posted question posted and it may also use slightly more processing than a single regex. But for developers who came here looking for a fast emergency fix for an outlier case then this solution should not be overlooked.

Answer 28

Aforementioned (?:(?!hede).)* is great because it can be anchored.

^(?:(?!hede).)*$               # A line without hede

foo(?:(?!hede).)*bar           # foo followed by bar, without hede between them

But the following would suffice in this case:

^(?!.*hede)                    # A line without hede

This simplification is ready to have "AND" clauses added:

^(?!.*hede)(?=.*foo)(?=.*bar)   # A line with foo and bar, but without hede
^(?!.*hede)(?=.*foo).*bar       # Same

Answer 29

Through PCRE verb (*SKIP)(*F)

^hede$(*SKIP)(*F)|^.*$

This would completely skips the line which contains the exact string hede and matches all the remaining lines.

DEMO

Execution of the parts:

Let us consider the above regex by splitting it into two parts.

1. Part before the | symbol. Part shouldn't be matched.

   ^hede$(*SKIP)(*F)
   

2. Part after the | symbol. Part should be matched.

   ^.*$
   

PART 1

Regex engine will start its execution from the first part.

^hede$(*SKIP)(*F)

Explanation:

• ^ Asserts that we are at the start.
• hede Matches the string hede
• $ Asserts that we are at the line end.

So the line which contains the string hede would be matched. Once the regex engine sees the following (*SKIP)(*F) (Note: You could write (*F) as (*FAIL)) verb, it skips and make the match to fail. | called alteration or logical OR operator added next to the PCRE verb which inturn matches all the boundaries exists between each and every character on all the lines except the line contains the exact string hede. See the demo here. That is, it tries to match the characters from the remaining string. Now the regex in the second part would be executed.

PART 2

^.*$

Explanation:

• ^ Asserts that we are at the start. ie, it matches all the line starts except the one in the hede line. See the demo here.

• .* In the Multiline mode, . would match any character except newline or carriage return characters. And * would repeat the previous character zero or more times. So .* would match the whole line. See the demo here.

  Hey why you added .* instead of .+ ?

  Because .* would match a blank line but .+ won't match a blank. We want to match all the lines except hede , there may be a possibility of blank lines also in the input . so you must use .* instead of .+ . .+ would repeat the previous character one or more times. See .* matches a blank line here.

• $ End of the line anchor is not necessary here.

Answer 30

Here's how I'd do it:

^[^h]*(h(?!ede)[^h]*)*$

Accurate and more efficient than the other answers. It implements Friedl's "unrolling-the-loop" efficiency technique and requires much less backtracking.