I'm trying to use a regular expression to match on a string. Brackets are special characters within regex, am I'm unsure of how'd i'd go about including them in my regex.
To provide more context, I want to find a string such as test[test]
My regex currently looks like this: ^*test[test]. My expression is built out more much than this, but this example is enough to understand the problem.
How can i search for brackets in my string without triggering a character class. I need to use a regex, please don't recommend switching to something else.
You can escape a character with a backslash so \[
I can highly recommend https://regex101.com/ to test your regex without having to code it.
Try: ^.*test\[test\] - This mean {start of line}, {anything}, "test[test]".
Related
I like the %r<…> delimiters because it makes it really easy to spot the beginning and end of the regex, and I don't have to escape any /. But it seems that they have an insurmountable limitation that other delimiters don't have?
Every other delimiter imaginable works fine:
/(?<!foo)/
%r{(?<!foo)}
%r[(?<!foo)]
%r|(?<!foo)|
%r/(?<!foo)/
But when I try to do this:
%r<(?<!foo)>
it gives this syntax error:
unterminated regexp meets end of file
Okay, it probably doesn't like that it's not a balanced pair, but how do you escape it such that it does like it?
Does something need to be escaped?
According to wikibooks.org:
Any single non-alpha-numeric character can be used as the delimiter,
%[including these], %?or these?, %~or even these things~.
By using this notation, the usual string delimiters " and ' can appear
in the string unescaped, but of course the new delimiter you've chosen
does need to be escaped.
Indeed, escaping is needed in these examples:
%r!(?<\!foo)!
%r?(\?<!foo)?
But if that were the only problem, then I should be able to escape it like this and have it work:
%r<(?\<!foo)>
But that yields this error:
undefined group option: /(?\<!foo)/
So maybe escaping is not needed/allowed? wikibooks.org does list %<pointy brackets> as one of the exceptions:
However, if you use
%(parentheses), %[square brackets], %{curly brackets} or
%<pointy brackets> as delimiters then those same delimiters
can appear unescaped in the string as long as they are in balanced
pairs
Is it a problem with balanced pairs?
Balanced pairs are no problem as long as you are doing something in the Regexp that requires them, like...
%r{(?<!foo{1})} # repetition quantifier
%r[(?<![foo])] # character class
%r<(?<name>foo)> # named capture group
But what if you need to insert a left-side delimiter ({, [, or <) inside the regex? Just escape it, right? Ruby seems to have no problem with escaped unbalanced delimiters most of the time...
%r{(?<!foo\{)}
%r[(?<!\[foo)]
%r<\<foo>
It's just when you try to do it in the middle of the "group options" (which I guess is what the <! characters are classified as here) following a (? that it doesn't like it:
%r<(?\<!foo)>
# undefined group option: /(?\<!foo)/
So how do you do that then and make Ruby happy? (without changing the delimiters)
Conclusion
The workaround is easy. I'll just change this particular regex to just use something else instead like %r{…} instead.
But the questions remain...
Is there really no way to escape the < here?
Are there really some regular expression that are simply impossible to write using certain delimiters like %r<…>?
Is %r<…> the only regular expression delimiter pair that has this problem (where some regular expressions are impossible to write when using it). If you know of a similar example with %r{…}/%r[…], do share!
Version info
Not that it probably matters since this syntax probably hasn't changed, but I'm using:
⟫ ruby -v
ruby 2.6.0p0 (2018-12-25 revision 66547) [x86_64-linux]
Reference:
https://ruby-doc.org/core-2.6.3/Regexp.html
% Notation
As others have mentioned, seems like an oversight based on how this character differs from other paired boundaries.
As far as "Is there really no way to escape the < here?" there is a way... but you're not going to like it:
%r<(?#{'<'}!foo)> == %r((?<!foo))
Using interpolation to insert the < character seems to work. But given that there are much better options, I would avoid it unless you were planning on splitting the regex into sections anyway...
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
I have a string with multiline as below.
rawMessage=sysUpTimeInstance-->0:0:00:05.00
snmpTrapOID.0-->linkDown.0.0
In the drools when portion i have written the condition as below.
rawMessage matches "(?i).*linkDown(.|\n|\r)*"
but it is not working.Please provide me some pointers to handle multiline.
Its not clear to me what you want to do/achieve. Your regex looks not wrong (I don't know the drools flavour and what you want to match).
In general (.|\n|\r)* is able to match any character including newlines. In your example there is no newline after "linkDown", so what should it match there?
Maybe you need to double escape (I don't know for drools) like this: (.|\\n|\\r)*.
Another possibility is to use the singleline modifer s (Again, I don't know if drools supports this modifier). This makes the . match also newline characters, could then look something like this
rawMessage matches "(?i)(?s).*linkDown.*"
or if it should only match multiline from "linkdown" on
rawMessage matches "(?i).*linkDown(?s).*"
Drools uses standard java regular expressions. As the previous answer mention, your expression looks wrong. And yes, you need to double escape special chars like you would do in java. Just check the javadoc for the Pattern class in the java API.
I'm sure this is a simple question for someone at ease with regular expressions:
I need to match everything up until the character #
I don't want the string following the # character, just the stuff before it, and the character itself should not be matched. This is the most important part, and what I'm mainly asking. As a second question, I would also like to know how to match the rest, after the # character. But not in the same expression, because I will need that in another context.
Here's an example string:
topics/install.xml#id_install
I want only topics/install.xml. And for the second question (separate expression) I want id_install
First expression:
^([^#]*)
Second expression:
#(.*)$
[a-zA-Z0-9]*[\#]
If your string contains any other special characters you need to add them into the first square bracket escaped.
I don't use C#, but i will assume that it uses pcre... if so,
"([^#]*)#.*"
with a call to 'match'. A call to 'search' does not need the trailing ".*"
The parens define the 'keep group'; the [^#] means any character that is not a '#'
You probably tried something like
"(.*)#.*"
and found that it fails when multiple '#' signs are present (keeping the leading '#'s)?
That is because ".*" is greedy, and will match as much as it can.
Your matcher should have a method that looks something like 'group(...)'. Most matchers
return the entire matched sequence as group(0), the first paren-matched group as group(1),
and so forth.
PCRE is so important i strongly encourage you to search for it on google, learn it, and always have it in your programming toolkit.
Use look ahead and look behind:
To get all characters up to, but not including the pound (#): .*?(?=\#)
To get all characters following, but not including the pound (#): (?<=\#).*
If you don't mind using groups, you can do it all in one shot:
(.*?)\#(.*) Your answers will be in group(1) and group(2). Notice the non-greedy construct, *?, which will attempt to match as little as possible instead of as much as possible.
If you want to allow for missing # section, use ([^\#]*)(?:\#(.*))?. It uses a non-collecting group to test the second half, and if it finds it, returns everything after the pound.
Honestly though, for you situation, it is probably easier to use the Split method provided in String.
More on lookahead and lookbehind
first:
/[^\#]*(?=\#)/ edit: is faster than /.*?(?=\#)/
second:
/(?<=\#).*/
For something like this in C# I would usually skip the regular expressions stuff altogether and do something like:
string[] split = exampleString.Split('#');
string firstString = split[0];
string secondString = split[1];
I just stumbled on a case where I had to remove quotes surrounding a specific regex pattern in a file, and the immediate conclusion I came to was to use vim's search and replace util and just escape each special character in the original and replacement patterns.
This worked (after a little tinkering), but it left me wondering if there is a better way to do these sorts of things.
The original regex (quoted): '/^\//' to be replaced with /^\//
And the search/replace pattern I used:
s/'\/\^\\\/\/'/\/\^\\\/\//g
Thanks!
You can use almost any character as the regex delimiter. This will save you from having to escape forward slashes. You can also use groups to extract the regex and avoid re-typing it. For example, try this:
:s#'\(\\^\\//\)'#\1#
I do not know if this will work for your case, because the example you listed and the regex you gave do not match up. (The regex you listed will match '/^\//', not '\^\//'. Mine will match the latter. Adjust as necessary.)
Could you avoid using regex entirely by using a nice simple string search and replace?
Please check whether this works for you - define the line number before this substitute-expression or place the cursor onto it:
:s:'\(.*\)':\1:
I used vim 7.1 for this. Of course, you can visually mark an area before (onto which this expression shall be executed (use "v" or "V" and move the cursor accordingly)).