I have the following reg ex:
"^((?!([\w!#$%&'*+\-/=?\^_`{|}~]+(\.[\w!#$%&'*+\-/=?\^_`{|}~]+)*#((([\-\w]+\.)+[a-zA-Z]{2,4})|(([0-9]{1,3}\.){3}[0-9]{1,3})))).)*$"
But it's not matching new lines as well:
https://regex101.com/r/nT6wK0/1
Any ideas how to make it match when there is a new line?
The . at the and actually means
All but a line break character. (source)
By replacing it with [\S\s], it means
All spacing characters and all non-spacing characters; so all characters.
Then it seems to work. You could have used other variants like [\W\w], [\D\d],...
So the "correct" regex (please don't take my word for it, first test this) is:
^((?!([\w!#$%&'*+\-/=?\^_`{|}~]+(\.[\w!#$%&'*+\-/=?\^_`{|}~]+)*#((([\-\w]+\.)+[a-zA-Z]{2,4})|(([0-9]{1,3}\.){3}[0-9]{1,3}))))[\S\s])*$
regex101 demo.
Assuming that you only want to match the first line, you can add the multiline option (/m) to include the newline.
If you want the second line to be included you'll need to read ahead an extra line. How you do that depends on the regex engine: N in sed; getline in awk; -n in perl; ...
Related
I'm using Java. So I have a comma separated list of strings in this form:
aa,aab,aac
aab,aa,aac
aab,aac,aa
I want to use regex to remove aa and the trailing ',' if it is not the last string in the list. I need to end up with the following result in all 3 cases:
aab,aac
Currently I am using the following pattern:
"aa[,]?"
However it is returning:
b,c
If lookarounds are available, you can write:
,aa(?![^,])|(?<![^,])aa,
with an empty string as replacement.
demo
Otherwise, with a POSIX ERE syntax you can do it with a capture:
^(aa(,|$))+|(,aa)+(,|$)
with the 4th group as replacement (so $4 or \4)
demo
Without knowing your flavor, I propose this solution for the case that it does know the \b.
I use perl as demo environment and do a replace with "_" for demonstration.
perl -pe "s/\baa,|,aa\b/_/"
\b is the "word border" anchor. I.e. any start or end of something looking like a word. It allows to handle line end, line start, blank, comma.
Using it, two alternatives suffice to cover all the cases in your sample input.
Output (with interleaved input, with both, line ending in newline and line ending in blank):
aa,aab,aac
_aab,aac
aab,aa,aac
aab_,aac
aab,aac,aa
aab,aac_
aa,aab,aac
_aab,aac
aab,aa,aac
aab_,aac
aab,aac,aa
aab,aac_
If the \b is unknown in your regex engine, then please state which one you are using, i.e. which tool (e.g. perl, awk, notepad++, sed, ...). Also in that case it might be necessary to do replacing instead of deleting, i.e. to fine tune a "," or "" as replacement. For supporting that, please show the context of your regex, i.e. the replacing mechanism you are using. If you are deleting, then please switch to replacing beforehand.
(I picked up an input from comment by gisek, that the cpaturing groups are not needed. I usually use () generously, including in other syntaxes. In my opinion not having to think or look up evaluation orders is a benefit in total time and risks taken. But after testing, I use this terser/eleganter way.)
If your regex engine supports positive lookaheads and positive lookbehinds, this should work:
,aa(?=,)|(?<=,)aa,|(,|^)aa(,|$)
You could probably use the following and replace it by nothing :
(aa,|,aa$)
Either aa, when it's in the begin or the middle of a string
,aa$ when it's at the end of the string
Demo
As you want to delete aa followed by a coma or the end of the line, this should do the trick: ,aa(?=,|$)|^aa,
see online demo
In a code file, I want to remove any (one or more) consecutive white lines (lines that may include only zero or more spaces/tabs and then a newline) that go between a code text and the concluding } of a block. This concluding } may have spaces for indentation before it, so I want to keep them.
Here is what I try to do:
perl -i -0777 -pe 's/\s+\n([ ]*)\}/\n($1)\}/g' file
For example, if my code file looks like (□ is the space character):
□□□□while (true) {\n
□□□□□□□□print("Yay!");□□□□□□\n
□□□□□□□□□□□□□□□□\n
□□□□}\n
Then I want it to become:
□□□□while (true) {\n
□□□□□□□□print("Yay!");\n
□□□□}\n
However it does not do the change I expected. Any idea what I am doing wrong here?
The only issues I can see with your regex are
you don't need the parenthesis around the matching variable,
and
the use of a character class when extracting the match is
redundant (unless you want to match tabs as well as spaces).
So, you could try
s/\s+\n( *)\}/\n$1\}/g
instead.
This works as expected when run on your test input.
To tidy it up even more, you could try the following.
s/\s+(\n *\})/$1/g
If there might be tabs as well as spaces, you can use a character class. (You do not need to include '|' inside the character class).
s/\s+(\n[ \t]*\})/$1/g
perl -pi -0777 -e's/^\s*\n(?=\s*})//mg' yourfile
(Remove whitespace from the beginning of a line through a newline that precedes a line with } as the first non-whitespace.)
Try using this regex instead, which uses a positive look-ahead assertion. This way you only capture the part that you want to remove, and then replace it with nothing:
s/\s+(?=\n[ ]*\})//g
You can try the following one liner
perl -0777 -pe 's/\s*\n*(\s*\n)/$1/g' test
I want to match two lines like the following using a Regular Expression:-
abcmnoxyz
=========
The first line is essentially random, the second line will be all the same character of a limited number of possibles (=, - and maybe a couple more). The lines can probably be required to be the same length but it would be nice if they didn't have to be. It would be OK to have multiple REs, one for each possible 'underline' character.
Can anyone come up with a way to do this?
This regex should do what you're trying to do :
regex = "(.*)\n(.)\2{2,}$"
group 1 will give you the line before the repeated linet
Live demo here
EXPLANATION
(.*)\n: match anything followed by a new line
(.)\2{2,} : capture something then check if its followed by same character 2+ more no. of times. You don't need to worry about which character is repeated.
In case you've a set of characters that can be repeated you can put a character set like this : [=-] instead of dot (.)
Use Grep's -B Flag
Matching with Alternation
Given your example, you can use extended regular expressions with alternations and a range operator. The -B flag tells grep how many lines before the match to include in the output.
$ grep -E -B1 '^(={5,}|-{5,})$' sample.txt
abcmnoxyz
=========
You can add alternations for additional characters if you want, although boundary markers ought to be as consistent as you can make them. You can also adjust the minimum number of sequential characters required for a match to suit your needs. I used a five-character range in the example because that's what was posted as the criterion in your original topic sentence, and because a shorter boundary marker is more likely to accidentally match truly random text.
Matching with a Character Class
Also, note that the following does the same job, but is a bit more concise. It uses a character class and a backreference to avoid alternations, which can get messy if you add many more boundary characters. Both versions are equally effective at matching your example.
$ grep -E -B1 '^([=-])\1{4,}$'
abcmnoxyz
========
A regex like this
^([^=\v]+)\v=+$
will do. Check it out at example 1
Explanation:
^([^=\v]+) # 1 or more matches of anything that is not a '=' or vertical space \v
\v=+$ # match a vertical space followed by 1 or more '='
If you want to extend this to more characters like '-' you could do this:
^([^=\-\v]+)\v(-|=)\2+$
Look at example 2
And, thanks to Ashish Ranjan, suppose you wanted to have = and/or - on the first line, use something like this:
^(.+)\v(-|=)\2+$
which would even allow you to have a first line like "=====". Having my doubts if OP had this in mind, though. Look at example 3
Hope this works
^([a-z]{1,})\n([=-]{1,})
\n and \r you have try both based on file format (unix or dos)
\1 will give you first line
\2 will give you second line
If the file contains same pattern over the text, then it might give you lot occurrence.
This answer is irrespective of number of characters in one line.
Ex: Tester
I have some string with multiple possible values:
e
(space)Exact
Exact
exact
phase
I want to get only the first four values, the regular expression I came up with is:
^\s*e
it means at the beginning of the string it has 0 or more white space followed by e(or E, case insensitive), howevever it always filters out the case
(space)Exact
my guess is it take ^ as not instead of beginning of string. How can i correct that? I use Perl Compatible Regular Expressions(PCRE) as the matching engine.
Try the using the mode modifiers in your regex to turn on ^$ match at linebreaks; and also, if necessary case insensitive
(?mi)^\s*e
The ^ character means only the beginning of a string. The beginning of a new line does not count as the beginning of a string. So this would not work if more than one are inside the same "string" object. Not sure how pcre works, but if you want to be able to match the begging of a line also you have to have the multi-line flag enabled.
Edit: If you want to pick up the beginnning of a new line go this route instead: \r\n at the beginning of the expression and remove the "^"
Edit #2 (because I feel like doing regex): here's what you're looking for:
(\b)[eE]+\w*
I have a css file and I want to add an empty line after every }.
How can I do this in Vim?
A substitution would work nicely.
:%s/}/\0\r/g
Replace } with the whole match \0 and a new line character \r.
or
:%s/}/&\r/g
Where & also is an alternative for the whole match, looks a bit funny though in my opinion. Vim golfers like it because it saves them a keystroke :)
\0 or & in the replacement part of the substitution acts as a special character. During the substitution the whole string that was matched replaces the \0 or the & character in the substitution.
We can demonstrate this with a more complex search and replace -
Which witch is which?
Apply a substitution -
:s/[wW][ih][ti]ch/The \0/g
Gives -
The Which The witch is The which?
The answer is :%s/}/}\r/ I guess.
:%s/pre/cur\r/g
%: operate on the entire buffer.
pre(previous pattern): which pattern will be to changed.
cur(current pattern): by which the previous pattern will be changed.
\r: new line.
g: repeat for every match on a line (default is to just replace the first).