I'm trying to grab 2 items from a simple line.
[Title](Description)
EDIT: actually a url looking to display called it description because i want it displayed not actually parsed.
[Trivium](https://www.youtube.com/user/trivium)
Grabbing between the brackets (...) doesn't seem to work at all for me. I've googled and found several variations with no luck, Thanks in advance :)
EDIT:
Tried the following:
[(.+?)]\((.*)\)
[(.+?)]\([^\(\r\n]*\)
[(.+?)]((.+?))
and a cpl more I cant find again
The first regex you listed almost has it right. Try using this regex instead:
\[.+?\]\((.*)\)
As #PM 77-1 pointed out, you need to escape the brackets by placing a backslash in front of them. The reason for this is that brackets are special regex metacharacters, or characters which have a special meaning. Brackets tell the regex engine to look for classes of characters contained inside of it.
Your original regex [(.+?)]\((.*)\) is actually doing this:
[(.+?)] match a period '.' 1 or more times
\((.*)\) match (anything), i.e. anything contained in parentheses
So this regex would match .....(stuff) but would not match [Title](Description), the latter which is what you really want.
Here is a link where you can test out the working regex:
Regex 101
Related
This is a follow-up question to what was solved yesterday:
Notepad++ Regex Replace Makeshift Footnotes format With Proper Markdown format
I managed to find a Regex to remove the offending semicolons in the main text area but by only cutting out the text and pasting back the result, which can only be done one by one.
I'm not sure how this can be done, but the expert can tell me.
So I have footnote references in markdown format. Two instances of the same thing:
[^1]:
[^2]:
.
.
.
[^99]:
I might not have 99 in a document but I wanted to show I need to match two digits here again.
As I said, there are two instances of these numbered references in the text. One in the main text pointing to the footnote and the footnote at the end of the document.
What I need is deleting the semi-colons from the main text and leave the
[^3]:
[^15]:
etc.
references at the end intact.
Because the main text references come after a word or at the end of a sentence (ususally before the sentence-ending period), there is never a case a reference would start a sentence (even if they seem to appear there once or twice because of word wrap).
I provided the exact opposite of my needs here:
Click here for Regex101 website link
I put in the exact opposite of what I want because I already knew of the
^
sign to match anything that is at the front of the line.
Now I would like to negate this, if possible, so that I would delete the semi-colons in the main text, not down at the bottom.
Of course, it is likely that my approach is not good and you'll come up with a completely different approach. Especially because there doesn't seem to be a NOT operator in Regex, if I read correctly.
I repeat: the Regex101 example with the match and substitution is exactly the opposite of what I want.
I am not sure if you can play around in the substitution line to get the desired negative effect.
I could have probably asked for removing the first occurence of semi-colons but I thought the important part of tackling the problem is that those items not to be matched are always at the start of the line, not the others.
Thanks for any suggestions
In Notepad++ you might use a negative lookabehind asserting not the start of the string to the left, and use \K to clear the match buffer matching only the colon that should be replaced by an empty string.
(?<!^)\[\^\d{1,2}]\K:
Explanation
(?<!^) Negative lookbehind, assert not the start of the start directly to the left
\[\^ Match [^
\d{1,2} Match 1 or 2 digits
] Match literally
\K Forget what is matched so far
: Match a colon
Regex demo
I have the following string:
https://www.google.com/today/sunday/abcde2.hopeho.3345GETD?weatherType=RAOM&...
https://www.google.com/today/monday/jbkwe3.ho4eho.8495GETD?weatherType=WHTDSG&...
I'd like to extract jbkwe3.ho4eho.8495GETD or abcde2.hopeho.3345GETD. Anything between the {weekday}/ and the ?weatherType=.
I've tried (?<=sunday\/)$.*?(?=\?weatherType=) but it only works for the first line and I want to make it applicable to all strings regardless the value of {weekday}.
I tried (?<=\/.*\/)$.*?(?=\?weatherType=) but it didn't work. Could anyone familiar with Regex can lend some help? Thank you!
[Update]
I'm new to regex but I was experimenting it on sublime text editor via the "find" functionality which I think should be PCRE (according to this post)
Try this regex:
(?:sun|mon|tues|wednes|thurs|fri|satur)day\/\K[^?]+(?=\?weatherType)
Click for Demo
Link to Code
Explanation:
(?:sun|mon|tues|wednes|thurs|fri|satur)day - matches the day of a week i.e, sunday,monday,tuesday,wednesday,thursday,friday,saturday
\/ - matches /
\K - unmatches whatever has been matched so far and pretends that the match starts from the current position. This can be used for the PCRE.
[^?]+ - matches 1 or more occurences of any character that is not a ?
(?=\?weatherType) - the above subpattern[^?]+ will match all characters that are not ? until it reaches a position which is immediately followed by a ? followed by weatherType
To make the match case-insensitive, you can prepend the regex with (?i) as shown here
In the examples given, you actually only need to grab the characters between the last forward slash ("/") and the first question mark ("?").
You didn't mention what flavor regex (ie, PCRE, grep, Oracle, etc) you're using, and the actual syntax will vary depending on this, but in general, something like the following (Perl) replacement regex would handle the examples given:
s/.*\/([^?]*)\?.*/$1/gm
There are other (and more efficient) ways, but this will do the job.
I want to use regular expressions to skip until i find a (-)Hyphen and store whatever comes after. I tried a few things but it didn't work out.
This is an example string:
Fall Down Seven Times; Stand Up Eight." -Naoki Higashida
I am just learning regular expressions and want to use them in my project to skip until I encounter different symbols.
*edit 1: this is what I have used so far, with some other stuff I found online.
"(?:[a-zA-Z;.;""]*)[^-][a-zA-Z]*"
Thank you for your help.
You can use .*-(.*). The capture group will contain everything after the hyphen.
.* matches anything
- matches literal hyphen
(.*) matches anything and captures it
Here's a demo.
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I am trying to use RegEx to find all cells in OO-calc with : [RB]Condition=New
the RB part is necessary, as other cells may get caught if it is not specified.
the RegEx that would work would be : [RB]Condi.*
however the brackets are read as a part of the expression, is there any way to fix this?
Use an escape character to treat it as an actual character. For example \[ would find all [. I think the Regex you're looking for is \[RB\]Condition=New*. The original Regex you had would find RCondition=New and BCondition=New.
*I'm not familiar with OO-calc, so I'm not entirely sure I understand what you mean.