I am using the following regex to locate in a document any series of characters that begins with characters dash dash -- and ends with a line feed character /n.
return #"(^--).*?(?=\r|\n)";
Almost works but only when there is a space between the -- and the next character.
return #"(?:--\s).*?(?=\r|\n)
Almost works but only when there is no space between the -- and the next character.
How do I get my return whether a space is following the -- or not?
I know nothing of regex other than what it's capable of. I found both of these sample patterns online. Thanks for your assistance.
You need to use \s? to capture either 0 or 1 spaces.
One use of the question mark in regex is to indicate that 0 or one matches of the previous character (or group of characters) will be matched, but not more than one.
Also, if you ever have the desire to learn regex for yourself, visit http://www.regular-expressions.info to learn and http://www.regexpal.com to practice.
Assuming that you are searching for substrings in a larger string and want to capture the the substring between -- and \n you could use an expression like:
--(.*)\r?\n
Which can be quoted in C# like this:
#"--(.*)\r?\n"
If you just want to make sure that a string starts with -- and ends with \n you could use:
(?s)^--.*\n\z
Related
What seemed to be a simple task ended up to not work as expected...
I'm trying to match \$\w+\b, unless it's preceded by an uneven number of backslashes.
Examples (only $result should be in the match):
This $result should be matched
This \$result should not be matched
This \\$result should be matched
This \\\$result should not be matched
etc...
The following pattern works:
(?<!\\)(\\\\)*\$\w+\b
However, even repeats of backslashes are included in the match, which is unwanted, so I'm trying to achieve this purely with a variable-length lookbehind, but nothing I tried so far seems to work.
Any regex virtuoso here can lend a hand?
You may use the following pattern:
(?<!(?:^|[^\\])\\(?:\\\\)*)\$\w+\b
Demo.
Breakdown of the Lookbehind; i.e., not preceded by:
(?:^|[^\\]) - Beginning of string/line or any character other than backslash.
\\ - Then, one backslash character.
(?:\\\\)* Then, any even number of backslash characters (including zero).
Looks like asking the question helped me answer my own question.
The part I don't want to be matched has to be wrapped with a positive lookbehind.
(?<=(?<!\\)(\\\\)*)\$\w+\b
Also works if the $result is at the start of the line.
If anyone has more optimal solutions, shoot!
This regular expression gets the wanted text in the third capture group:
(^| )(\\\\)*(\$\w+\b)
Explanation:
(^| ) Either beginning of line or a space
(\\\\)* An even number of backslash characters, including none
( Start of capture group 3
\$\w+\b The wanted text
) End of capture group 3
I'm trying to create a large regex expression where the plan is to capture 6 groups.
Is gonna be used to parse some Android log that have the following format:
2020-03-10T14:09:13.3250000 VERB CallingClass 17503 20870 Whatever content: this log line had (etc)
The expression I've created so far is the following:
(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}.\d{7})\t([A-Za-z]{4})\t(\w{+})\t(\d{5})\t(\d{5})\t(.*$)
The lines in this case are Tab separated, although the application that I'm developing will be dynamic to the point where this is not always the case, so regex I feel is still the best option even if heavier then performing a split.
Breaking down the groups in more detail from my though process:
Matches the date (I'm considering changing this to a x number of characters instead)
(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}.\d{7})
Match a block of 4 characters
([A-Za-z]{4})
Match any number of characters until the next tab
(\w{+})
Match a block of 5 numbers 2 times
\t(\d{5})
At last, match everything else until the end of the line.
\t(.*$)
If I use a reduced expression to the following it works:
(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}.\d{7})\t([A-Za-z]{4})\t(.*$)
This doesn't include 3 of the groups, the word and the 2 numbers blocks.
Any idea why is this?
Thank you.
The problem is \w{+} is going to match a word character followed by one or more { characters and then a final } character. If you want one or more word characters then just use plus without the curly braces (which are meant for specifying a specific number or number range, but will match literal curly braces if they do not adhere to that format).
(\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}.\d{7})\t([A-Za-z]{4})\t(\w+)\t(\d{5})\t(\d{5})\t(.*$)
I highly recommend using https://regex101.com/ for the explanation to see if your expression matches up with what you want spelled out in words. However for testing for use in C# you should use something else like http://regexstorm.net/tester
I have a problem with a regex command,
I have a file with a tons of lines and with a lot of sensitive characters,
this is an Example with all sensitive case 0123456789/*-+.&é"'(-è_çà)=~#{[|`\^#]}²$*ù^%µ£¨¤,;:!?./§<>AZERTYUIOPMLKJHGFDSQWXCVBNazertyuiopmlkjhgfdsqwxcvbn
I tried many regex commands but never get the expected result,
I have to select everything from Example to the end
I tried this command on https://www.regextester.com/ :
\sExample(.*?)+
Image of the result here
And when I tried it in C# the only result I get was : Example
I don't understand why --'
Here's a quick chat about greedy and pessimistic:
Here is test data:
Example word followed by another word and then more
Here are two regex:
Example.*word
Example.*?word
The first is greedy. Regex will match Example then it will take .* which consumes everything all the way to the END of the string and the works backwards spitting a character at a time back out, trying to make the match succeed. It will succeed when Example word followed by another word is matched, the .* having matched word followed by another (and the spaces at either end)
The second is pessimistic; it nibbled forwards along the string one character at a time, trying to match. Regex will match Example then it'll take one more character into the .*? wildcard, then check if it found word - which it did. So pessimistic matching will only find a single space and the full match in pessimistic mode is Example word
Because you say you want the whole string after Example I recommend use of a greedy quantifier so it just immediately takes the whole string that remains and declares a match, rather than nibbling forwards one at a time (slow)
This, then, will match (and capture) everything after Example:
\sExample(.*)
The brackets make a capture group. In c# we can name the group using ?<namehere> at the start of the brackets and then everything that .* matches can be retrieved with:
Regex r = new Regex("\sExample(?<x>.*)");
Match m = r.Match("Exampleblahblah");
Console.WriteLine(m.Groups["x"].Value); //prints: blahblah
Note that if your data contains newlines you should note that . doesn't match a newline, unless you enable RegexOptions.SingleLine when you create the regex
I need to find all the words that have between 15 and 20 characters in a big string. And I want to avoid getting a long words with something else at the end (for ex 1234567890abcdef#asdf.com). I don't want that to be a result, only words. Right now I'm spliting the string using white space as token and for each word I'm applying the following regular expression:
^[a-zA-Z0-9]{15,20}$
Is there any chance to do both things using one regular expression?
I'm using C#.
Good examples to catch:
1234567890abcdeg
qwertyuiopasdfgh
1234567890abcdeg, (catch it but remove ",")
Examples to avoid: 1234567890abcdeg#gmail.com
Don't use start/end anchors (^/$), but word delimiters (\b):
\b[a-zA-Z0-9]{15,20}(?=[\s,]|$)
I used (?=[\s,]|$) instead of the end delimiter to force a space character or a comma or the end of the string. Expand it as needed.
You may want to do likewise for the first \b if you need to, for instance: (?<=\s|^).
Normally, you would use word boundaries (\b) before and after the alphanumerics:
\b[a-zA-Z0-9]{15,20}\b
However, there's a small detail to take into account: uderscores ("_") are also considered a word character. The previous regex won't match the following text:
12345678901234567_
In order to avoid it, you can check if it's preceded and followed by either a \b or a "_", with lookarounds.
Regex:
(?<=\b|_)[a-zA-Z0-9]{15,20}(?=\b|_)
I am looking for a way to get words out of a sentence. I am pretty far with the following expression:
\b([a-zA-Z]+?)\b
but there are some occurrences that it counts a word when I want it not to. E.g a word followed by more than one period like "text..". So, in my regex I want to have the period to be at the end of a word zero or one time. Inserting \.? did not do the trick, and variations on this have not yielded anything fruitful either.
Hope someone can help!
A single dot means any character. You must escape it as
\.?
Maybe you want an expression like this:
\w+\.?
or
\p{L}+\.?
You need to add \.? (and not .?) because the period has special meaning in regexes.
to avoid a match on your example "test.." you ask for you not only need to put the \.? for checking first character after the word to be a dot but also look one character further to check the second character after the word.
I did end up with something like this
\w{2,}\.?[^.]
You should also consider that a sentence not always ends with a . but also ! or ? and alike.
I usually use rubulator.com to quick test a regexp