We Keep Coding

Regex for matching string in parentheses including when opening or closing parenthesis is missing

I want to match strings in parentheses (including the parens themselves) and also match strings when a closing or opening parenthesis is missing.
From looking around my ideal solution would involve conditional regex however I need to work within the limitations of javascript's regex engine.
My current solution that almost works: /\(?[^()]+\)|\([^()]+/g. I could split this up (might be better for readability) but am curious to know if there is a way to achieve it without being overly verbose with multiple |'s.
Examples
Might help to understand what I'm trying to achieve through examples (highlighted sections are the parts I want to match):
(paren without closing
(paren in start) of string
paren (in middle) of string
paren (at end of string)
paren without opening)
string without any parens
(string with only paren)
string (with multiple) parens (in a row)
Here's a link to the tests I set up in regexr.com.

You can match the following regular expression.
^\([^()]*$|^[^()]*\)$|\([^()]*\)
Javascript Demo
Javascript's regex engine performs the following operations.
^ # match the beginning of the string
\( # match '('
[^()]*. # match zero or more chars other than parentheses,
# as many as possible
$ # match the end of the string
| # or
^ # match the beginning of the string
[^()]*. # match zero or more chars other than parentheses,
# as many as possible
\) # match ')'
$ # match the end of the string
| # or
\( # match '('
[^()]*. # match zero or more chars other than parentheses,
# as many as possible
\) # match ')'

As of question date (May 14th 2020) the Regexr's test mechanism was not working as expected (it matches (with multiple) but not (in a row)) Seems to be a bug in the test mechanism. If you copy and paste the 8 items in the "text' mode of Regexr and test your expression you'll see it matches (in a row). The expression also works as expected in Regex101.

I think you've done alright. The issue is that you need to match [()] in two places and only one of them needs to be true but both can't be false and regex isn't so smart as to keep state like that. So you need to check if there is 0 or 1 opening or 0 or 1 closing in alternatives like you have.
Update:
I stand corrected since all you seem to care about is where there is an open or closing parenthesis you could just do something like this:
.*[\(?\)]+.*
In English: any number of characters with eith an ( or ) followed by any number of characters. This will match them in any order though, so if you need ( to be before closed even though you don't seem to care if both are present, this won't work.

Optional regex string/pattern sections with and without non-capturing groups

Here's what I'm trying to do:
http://i.imgur.com/Xqrf8Wn.png
Simply take a URL with 3 groups, $1 not so important, $2 & $3 are but $2 is totally optional including (obviously) the corresponding backslash when present, which is all I am trying to make optional. I get that it can/should? be in a non-cap group, but does it HAVE to be? I've seen enough now seems to indicate it does not HAVE to be. If possible, I'd really like to have someone explain it so I can try to fully understand it, and not just get one possible working answer handed to me to simply copy, like some come here seeking.
Here's my regex string(s) tried and at best only currently matching second URL string with optional present:
^https:\/\/([a-z]{0,2})\.?blah\.com(?:\/)(.*)\/required\/B([A-Z0-9]{9}).*
^https:\/\/([a-z]{0,2})\.?blah\.com(\/)?(.*)\/required\/B([A-Z0-9]{9}).*
^https:\/\/([a-z]{0,2})\.?blah\.com(?:\/)?(.*)?\/required\/B([A-Z0-9]{9}).*
Here are the two URLs that I want to capture group 2 & 3, with 1 and 2 being optional, but $2 being the problem. I've tried all the strings above and have yet to get it to match the string when the optional is NOT present and I believe it must be due to the backslashes?
https://blah.com/required/B7BG0Z0GU1A
https://blah.com/optional/required/B7BG0Z0GU1A

Making a part of the pattern optional is as simple as adding ?, and your last two attempts both work: https://regex101.com/r/RIKvYY/1
Your mistake is that your test is wrong - you are using ^ which matches the beginning of the string. You need to add the /m flag (multiline) to make it match the beginning of each line. This is the reason your patterns never match the second line...
Note that you're allowing two slashes (//required, for example). You can solve it by joining the first slash and the optional part to the same capturing group (of course, as long as you are using .* you can still match multiple slashes):
https:\/\/([a-z]{0,2})\.?blah\.com(?:\/(.*))?\/required\/B([A-Z0-9]{9}).*

regex - how to select all double slashes except followed by colon

I need some help with RegEx, it may be a basic stuff but I cannot find a correct way how to do it. Please help!
So, here's my question:
I have a list of URLs, that are invalid because of double slash, like this:
http://website.com//wp-content/folder/file.jpg, to fix it I need to remove all double slashes except the first one followed by colon (http://), so fixed URL is this: http://website.com/wp-content/folder/file.jpg.
I need to do it with RegExp.
Variant 1
url.replace(/\/\//g,'/'); // => http:/website.com/wp-content/folder/file.jpg
will replace all double slashed (//), including the first one, which is not correct.
example here:
https://regex101.com/r/NhCVMz/2

You may use
url = url.replace(/(https?:\/\/)|(\/){2,}/g, "$1$2")
See the regex demo
Note: a ^ anchor at the beginning of the pattern might be used if the strings are entire URLs.
This pattern will match and capture http:// or https:// and will restore it in the resulting string with the $1 backreference and all other cases of 2 or more / will be matched by (\/){2,} and only 1 occurrence will be put back into the resulting string since the capturing group does not include the quantifier.

Find (^|[^:])/{2,}
Replace $1/
delimited: /(^|[^:])\/{2,}/

Capture multiline content between two words in javascript with RegExp [duplicate]

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

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!).

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.

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".

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

The given answers are perfectly fine, just an academic point:
Regular Expressions in the meaning of theoretical computer sciences ARE NOT ABLE do it like this. For them it had to look something like this:
^([^h].*$)|(h([^e].*$|$))|(he([^h].*$|$))|(heh([^e].*$|$))|(hehe.+$)
This only does a FULL match. Doing it for sub-matches would even be more awkward.

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).

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.

Here's a good explanation of why it's not easy to negate an arbitrary regex. I have to agree with the other answers, though: if this is anything other than a hypothetical question, then a regex is not the right choice here.

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()
);

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.

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)

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.

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))*$/

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

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

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.

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:

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"

The OP did not specify or Tag the post to indicate the context (programming language, editor, tool) the Regex will be used within.
For me, I sometimes need to do this while editing a file using Textpad.
Textpad supports some Regex, but does not support lookahead or lookbehind, so it takes a few steps.
If I am looking to retain all lines that Do NOT contain the string hede, I would do it like this:
1. Search/replace the entire file to add a unique "Tag" to the beginning of each line containing any text.
Search string:^(.)
Replace string:<##-unique-##>\1
Replace-all
2. Delete all lines that contain the string hede (replacement string is empty):
Search string:<##-unique-##>.*hede.*\n
Replace string:<nothing>
Replace-all
3. At this point, all remaining lines Do NOT contain the string hede. Remove the unique "Tag" from all lines (replacement string is empty):
Search string:<##-unique-##>
Replace string:<nothing>
Replace-all
Now you have the original text with all lines containing the string hede removed.
If I am looking to Do Something Else to only lines that Do NOT contain the string hede, I would do it like this:
1. Search/replace the entire file to add a unique "Tag" to the beginning of each line containing any text.
Search string:^(.)
Replace string:<##-unique-##>\1
Replace-all
2. For all lines that contain the string hede, remove the unique "Tag":
Search string:<##-unique-##>(.*hede)
Replace string:\1
Replace-all
3. At this point, all lines that begin with the unique "Tag", Do NOT contain the string hede. I can now do my Something Else to only those lines.
4. When I am done, I remove the unique "Tag" from all lines (replacement string is empty):
Search string:<##-unique-##>
Replace string:<nothing>
Replace-all

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"]

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.
Part before the | symbol. Part shouldn't be matched.
^hede$(*SKIP)(*F)
Part after the | symbol. Part should be matched.
^.*$
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.

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".

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.

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;
}
?>

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

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
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
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:
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
Target the rest (the unmarked strings: e.g. lines without hede). Suppose you want to delete the target:
s/^[^🔒].*//g
Remove the mark:
s/🔒//g

^((?!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)

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.

Simplest thing that I could find would be
[^(hede)]
Tested at https://regex101.com/
You can also add unit-test cases on that site

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.

Regex explanation

I am looking at the code in the tumblr bookmarklet and was curious what the code below did.
try{
if(!/^(.*\.)?tumblr[^.]*$/.test(l.host))
throw(0);
tstbklt();
}
Can anyone tell me what the if line is testing? I have tried to decode the regex but have been unable to do so.

Initially excluding the specifics of the regex, this code is:
if ( ! /.../.test(l.host) )
"if not regex.matches(l.host)" or "if l.host does not match this regex"
So, the regex must correctly describe the contents of l.host text for the conditional to fail and thus avoid throwing the error.
On to the regex itself:
^(.*\.)?tumblr[^.]*$
This is checking for the existence of tumblr but only after any string ending in . that might exist:
^ # start of line
( # begin capturing group 1
.* # match any (non-newline) character, as many times as possible, but zero allowed
\. # match a literal .
) # end capturing group 1
? # make whole preceeding item optional
tumblr # match literal text tumblr
[^.]* # match any non . character, as many times as possible, but zero allowed
$ # match end of line
I thought it was testing to see if the host was tumblr
Yeah, it looked like it might be intended to check that, but if so it's the wrong way to do it.
For that, the first bit should be something like ^(?:[\w-]+\.)? to capture an alphanumeric subdomain (the ?: is a non-capturing group, the [\w-]+ is at least 1 alphanumeric, underscore or hyphen) and the last bit should be either \.(?:com|net|org)$ or perhaps like (?:\.[a-zA-Z]+)+$ depending on how flexible the tld section might need to be.

My attempt to break it down. I'm no expert with regex however:
if(!/^(..)?tumblr[^.]$/.test(l.host))
This part isn't really regex but tells us to only execute the if() if this test does not work.
if(!/^(.*\.)?tumblr[^.]*$/.test(l.host))
This part allows for any characters before the tumblr word as long as they are followed by a . But it is all optional (See the ? at the end)
if(!/^(.*.)?tumblr**[^.]*$/**.test(l.host))
Next, it matches any character except the . and it the *$ extends that to match any character afterwards (so it doesn't break after 1) and it works until the end of the string.
Finally, the .test() looks to test it against the current hostname or whatever l.host contains (I'm not familiar with the tumblr bookmarklet)
So basically, it looks like that part is checking to see that if the host is not part of tumblr, then throw that exception.
Looking forward to see how wrong I am :)