Regex match all strings except - regex

I have to create a regex that finds next strings:
stackoverflow //not found
stackexchange //found 'stackexchange'
stacksomething //found 'stacksomething '
stacksomething another words //find 'stacksomething '
Explanation: find a string that:
starts with 'stack'
ends with any word except 'overflow'
find only this word.
I created regex which corresponds to the first point, but can't implement the second and the third. I tried solutions like ^((?!overflow).)*$ and ^(?!.*\boverflow\b) but they don't work. That's what I have:
stack.*


You can use this negative lookahead regex:
\bstack(?!\w*overflow\b)\w*
RegEx Demo
Breakup:
\b # word boundary
stack # match literal text stack
(?!\w*overflow\b) # negative lookahead to fail the match if word ends with overflow
\w* # match 0 or more word characters to get full word match

Related

Create regular expression remove word

Hello good afternoon!!
I'm new to the world of regular expressions and would like some help creating the following expression!
I have a query that returns the following values:
caixa-pod
config-pod
consultas-pod
entregas-pod
monitoramento-pod
vendas-pod
I would like the results to be presented as follows:
caixa
config
consultas
entregas
monitoramento
vendas
In this case, it would exclude the word "-pod" from each value.

I would try (.*)-pod. It is not clear, where do you want to use that regexp (so regexp can be different). I guess it is dashboard variable.

You can try
\b[a-z]*(?=-pod)\b
This regex basically tells the regex engine to match
\b a word boundary
[a-z]* any number of lowercase characters in range a-z (feel free to extend to whatever is needed e.g. [a-zA-Z0-9] matches all alphanumeric characters)
(?=-pod) followed by -pod but exclude that from the result (positive lookahead)
\b another word boundary
\b matches a word boundary position between a word character and non-word character or position (start / end of string).

regex grab a word

i'm trying to grab a regex from source, but only name from this type.
"name":"HELP-PERP","posOnly":false,"price":40.3,"priceIncrement":0.01,"quote":null,"quoteV":73851918.483,"restricted":false,"sizeIncrement":0.01,"type":"future",
So i got here \b(\w*-PERP\w*)\b
This grabs the word HELP-PERP but duplicates it, so i'm trying to grab that word that matches the type =future.
Grab help-perp that is in the same line with type":"future".
Total nub at this, i've tried several things on regex101 and can't come up :(
Thank you

You can use
/\w*-PERP\w*\b(?=.*type":"future")/g
See the regex demo.
Details
\w*-PERP\w* - zero or more word chars, -PERP, and again zero or more chars
\b - a word boundary
(?=.*type":"future") - a positive lookahead that matches a location in string that is immediately followed with any zero or more chars other than line break chars as many as possible (.*) and then a type":"future" string.

Regex - Match any word but ignore specific word [duplicate]

This question already has answers here:
Regex: match everything but a specific pattern
(6 answers)
Closed 5 years ago.
I want to match any word that starts/ends or not contain with word "end" but ignore word "end", for example:
hello - would match
boyfriend - would match
endless - would match
endend - would match
but
end - would NOT match
I'm using ^(?!end)).*$ but its not what I want.
Sorry for my english

Try this:
^(?!(end)$).+$
This will match everything except end.

You can use this \b(?!(?:end\b))[\w]+
Components:
\b -> Start of the word boundary for each words.
(?! Negative lookahead to eliminate the word end.
(?:end\b) Non capturing parenthesis with the word end and word boundary.
) Closing tag for negative lookahead.
[\w]+ character class to capture words.
Explanation: The regex search will only look for locations starting with word boundaries, and will remove matches with end as only word. i.e [WORD BOUNDARY]end[END OF WORD BOUNDARY]. \w will capture rest of the word. You can keep incrementing this character class if you wish to capture some special characters like $ etc.

So you want to match any word, but not "end" ?
Unless I'm misunderstanding, a conditional statement is everything that is needed... In pseudocode:
if (word != "end") {
// Match
}
If you want to match all the words in a text that are not "end" you could just remove all the non-alpha characters, replace pattern (^end | end | end$) by an empty string, and then do a string split.
The other answers with a single regex might be better then, because regex matches are O(n), no matter of the pattern.

How to find a word left of a specific word if it exists else return the last word in regex

For example,
MR-L6LQN-LP MR-L6LQN-LQ Way
should return MR-L6LQN-LQ here the specific word is 'Way'
if the string is
MR-L6LQN-LP MR-L6LQN-LQ
output should be
MR-L6LQN-LQ
I think lookahead ideas would come in handy but am not able to work it out. Please suggest regex only solution.

You can use this regex with a positive lookahead:
\b[A-Z0-9-]+(?=\s+Way|$)
RegEx Demo
RegEx Breakup:
\b: Assert word boundary
[A-Z0-9-]+: Match one or more of uppercase letters, digits or hyphens
(?=\s+Way|$): Positive Lookahead to assert we have spaces and Way ahead or end of line.

Regex to match any words except words with given pattern

I want to match all words except following words :
1) any-random-word
2) any-random-word/
3) any-random-word/123
4) any-random-word/abcdef
so that following similar words can be matched.
1) any-random-word123
2) any-random-word(any non-word character other than '/')123
2) any-random-wordabcdef
4) any-random-word(any non-word character other than '/')abcdef
In fact any number or any word can be appened after 'any-random-word/'.
I tried with
^(?!any-random-word(\/?)(\w+)$|any-random-word$)
but its escaping all words having any-random-word in it.
Thanks.

You can change your current regex a little:
^(?!.*\bany-random-word\b)
And if you want to actually match something, add .+ at the end:
^(?!.*\bany-random-word\b).+
regex101 demo
\b (a word boundary) ensures that there's no other \w character around the word you don't want to match.
Edit: As per your further clarification, I would suggest this regex:
^(?!.*\bany-random-word(?:/|$)).+
The main part of the regex is the negative lookahead: (?!.*\bany-random-word(?:/|$)). It will cause the whole match to fail if what's inside matches.
.*\bany-random-word(?:/|$) will match any-random-word at the end of a string or followed by /, anywhere in the string that it is being tested against.
So, if you have any-random-word/, it will match, and cause the whole match to fail. If you have the string ending in any-random-word, again, the whole match will fail.