If you look at this text:
FIRST TEXT (IF CAPS AND IF IT ENDS WITH A PERIOD) SHOULD BE EXCLUDED. Here comes all the text we want to grab. And the ONLY problem with our current regular expression is that it also includes the period and space in front of this text. Does anyone know how to fix it so we grab from "Here comes..." and not ". Here comes..."? Thank you.
My current regular expression looks like this: (?![A-ZÆØÅ!´'/0-9\s()]+[.])[^=]*
But I simply can't figure out how to exclude the first ". " from the selection. Can anyone please help? You can try it out here:
https://regex101.com/r/UpRlOV/3
The dot and spaces are matched because your lookahead pattern does not match only up to the dot. To make sure your match does not start with . + space(s), you may consume them if they are present. An optional non-capturing group is quite handy in such situations:
(?![A-ZÆØÅ!´'\/0-9\s()]+[.])(?:\.\s*)?\K[^=]+
^^^ ^^
or, if your regex engine does not support \K match reset operator, use a capturing group:
(?![A-ZÆØÅ!´'\/0-9\s()]+[.])(?:\.\s*)?([^=]+)
^ ^
See the regex demo.
.+?\.(.+)
Is something like this what you're looking for?
this way you can just grab group 1 from the result
https://regex101.com/r/1Eo38B/1
Related
I'd like to match every dot or comma but not in href attribute. So I have this regular expression:
^(?!.*?href=)(.*?)([.,])(\S+)
But it matches only the first occurrence. I think it because of non-greedy .*? But I can't come up with anything else. Can you help me, please?
What you might do to match every dot or comma and assuming that the attribute value is between single or double quotes is to match what you don't want and to capture in a group what you want to keep.
If you don't want to match a dot in the href you could match it with href=" followed by [^"]*" or '[^']*'. Then you could use an alternation | to capture in a group a dot or a comma using ([.,])
href=(?:"[^"]*"|'[^']*')|([.,])
If you want to match every occurrence, you will need to run the regex with the global (g) flag:
e.g.
/^(?!.*?href=)(.*?)([.,])(\S+)/g
I suggest you use a tool such as https://regex101.com/ to test and debug your regular expressions, it's super handy!
I need help with RegEx I just can't figure it out I need to search for broken Hashtags which have an space.
So the strings are for Example:
#ThisIsaHashtagWith Space
But there could also be the Words "With Space" which I don't want to replace.
So important is that the String starts with "#" then any character and then the words "With Space" which I want to replace to "WithSpace" to repair the Hashtags.
I have a Document with 10k of this broken Hashtags and I'm kind of trying the whole day without success.
I have tried on regex101.com
with following RegEx:
^#+(?:.*?)+(With Space)
Even I think it works on regex101.com it doesn't in Notepad++
Any help is appreciated.
Thanks a lot.
BR
In your current regex you match a # and then any character and in a capturing group match (With Space).
You could change the capturing group to capture the first part of the match.
(#+.*?)With Space
Then you could use that group in the replacement:
$1WithSpace
As an alternative you could first match a single # followed by zero or more times any character non greedy .*? and then use \K to reset the starting point of the reported match.
Then match With Space.
#+(?:.*?)\KWith Space
In the replacement use WithSpace
If you want to match one or more times # you could use a quantifier +. If the match should start at the beginning of string you could use an anchor ^ at the start of the regex.
Try using ^(#.+?)(With\s+Space) for your regex as it also matches multiple spaces and tab characters - if you have multiple rows that you want to affect do gmi for the flags. I just tried it with the following two strings, each on a separate line in Notepad++
#blablaWith Space
#hello###$aWith Space
The replace with value is set to $1WithSpace and I've tried both replaceAll and replace one by one - seems to result in the following.
#blablaWithSpace
#hello###$aWithSpace
Feel free to comment with other strings you want replaced. Also be sure that you have selected the Regular Extension search mode in NPP.
Try this? (#.*)( ).
I tried this in Notepad++ and you should be able to just replace all with $1. Make sure you set the find mode to regular expressions first.
const str = "#ThisIsAHashtagWith Space";
console.log(str.replace(/(#.*)( )/g, "$1"));
I'm trying to find all lines without ending period (dot) but without finding blank (empty) lines. And after that I want to add ending period to that sentence.
Example:
The good is whatever stops such things from happening.
Meaning as the Higher Good
It was from this that I drew my fundamental moral conclusions.
I have tried few regex but they also find empty lines as well.
Is there a regex for Notepad++ that can achieve that?
Enable Regular Expression match, then search for:
\S(?<!\.)\K\s*$
and replace with:
.$0
Breakdown:
\S Match a non-whitespace character
(?<!\.) It shouldn't be a period
\K Reset match
\s* Match optional whitespace characters
$ End of line
You could use something like this to find the lines that you are interested in adding capture group to it and appending you needed chars.
(?<!\.)\r\n
This works by using negative look behind (?<!\.) to check that there is no . before \r
There is a group or regex operators that can be used to accomplish this type of tasks.
Look ahead positive (?=)
Look ahead negative (?!)
Look behind positive (?<=)
Look behind negative (?
Try this short and effective solution too.
Search: \w$
Replace: $0.
I'd like to know how can I ignore characters that follows a particular pattern in a Regex.
I tried with positive lookaheads but they do not work as they preserves those character for other matches, while I want them to be just... discarded.
For example, a part of my regex is: (?<DoubleQ>\"\".*?\"\")|(?<SingleQ>\".*?\")
in order to match some "key-parts" of this string:
This is a ""sample text"" just for "testing purposes": not to be used anywhere else.
I want to capture the entire ""sample text"", but then I want to "extract" only sample text and the same with testing purposes. That is, I want the group to match to be ""sample text"", but then I want the full match to be sample text. I partially achieved that with the use of the \K option:
(?<DoubleQ>\"\"\K.*?\"\")|(?<SingleQ>\"\K.*?\")
Which ignores the first "" (or ") from the full match but takes it into account when matching the group. How can I ignore the following "" (")?
Note: positive lookahead does not work: it does not ignore characters from the following matches, it just does not include them in the current match.
Thanks a lot.
I hope I got your questions right. So you want to match the whole string including the quotes, but you want to replace/extract it only the expression without the quotes, right?
You typically can use the regex replace functionality to extract just a part of the match.
This is the regex expression:
""?(.*?)""?
And this the replace expression:
$1
I have the following data:
SOMEDATA .test 01/45/12 2.50 THIS IS DATA
and I want to extract the number 2.50 out of this. I have managed to do this with the following RegEx:
(?<=\d{2}\/\d{2}\/\d{2} )\d+.\d+
However that doesn't work for input like this:
SOMEDATA .test 01/45/12 2500 THIS IS DATA
In this case, I want to extract the number 2500.
I can't seem to figure out a regex rule for that. Is there a way to extract something between two spaces ? So extract the text/number after the date until the next whitespace ? All I know is that the date will always have the same format and there will always be a space after the text and then a space after the number I want to extract.
Can someone help me out on this ?
Capture number between two whitespaces
A whitespace is matched with \s, and non-whitespace with \S.
So, what you can use is:
\d{2}\/\d{2}\/\d{2} +(\S+)
^^^
See the regex demo
The 1+ non-whitespace symbols are captured into Group 1.
If - for some reason - you need to only get the value as a whole match, use your lookbehind approach:
(?<=\d{2}\/\d{2}\/\d{2} )\S+
Or - if you are using PCRE - you may leverage the match reset operator \K:
\d{2}\/\d{2}\/\d{2} +\K\S+
^^
See another demo
NOTE: the \K and a capture group approaches allow 1 or more spaces after the date and are thus more flexible.
I see some people helped you already, but if you would want an alternative working one for some reason, here's what works too :)
.+ \d+\/\d+\/\d+ (\d+[\.\d]*)
So the .+ matches anything plus the first space
then the \d+/\d+/\d+ is the date parsing plus a space
the capturing group is the number, as you can see I made the last part optional, so both floating point values and normal values can be matched. Hope this helped!
Proof: https://regex101.com/r/fY3nJ2/1
Just make the fractal part optional:
(?<=\d{2}\/\d{2}\/\d{2} )\d+(?:\.\d+)?
Demo: https://regex101.com/r/jH3pU7/1
Update following clarifications in comments:
To match anything (but space) surrounded by spaces and prepended by date use:
(?<=\d{2}\/\d{2}\/\d{2} )\S+
Demo: https://regex101.com/r/jH3pU7/3
Rather than capture, you can make your entire match be the target text by using a look behind:
(?<=\d\d(\/\d\d){2} )\S+
This matches the first series of non-whitespace that follows a "date like" part.
Note also the reduction in the length of the "date like" pattern. You may consider using this part of the regex in whatever solution you use.