can any body tell me how to use regex for negation of string?
I wanna find all line that start with public class and then any thing except first,second and finally any thing else.
for example in the result i expect to see public class base but not public class myfirst:base
can any body help me please??
Use a negative lookahead:
public\s+class\s+(?!first|second).+
If Peter is correct and you're using Visual Studio's Find feature, this should work:
^:b*public:b+class:b+~(first|second):i.*$
:b matches a space or tab
~(...) is how VS does a negative lookahead
:i matches a C/C++ identifier
The rest is standard regex syntax:
^ for beginning of line
$ for end of line
. for any character
* for zero or more
+ for one or more
| for alternation
Both the other two answers come close, but probably fail for different reasons.
public\s+class\s+(?:(?!first|second).)+
Note how there is a (non-capturing) group around the negative lookahead, to ensure it applies to more than just the first position.
And that group is less restrictive - since . excludes newline, it's using that instead of \S, and the $ is not necessary - this will exclude the specified words and match others.
No slashes wrapping the expression since those aren't required in everything and may confuse people that have only encountered string-based regex use.
If this still fails, post the exact content that is wrongly matched or missed, and what language/ide you are using.
Update:
Turns out you're using Visual Studio, which has it's own special regex implementation, for some unfathomable reason. So, you'll be wanting to try this instead:
public:b+class:b+~(first|second)+$
I have no way of testing that - if it doesn't work, try dropping the $, but otherwise you'll have to find a VS user. Or better still, the VS engineer(s) responsible for this stupid non-standard regex.
Here is something that should work for you
/public\sclass\s(?:[^fs\s]+|(?!first|second)\S)+(?=\s|$)/
The second look a head could be changed to a $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)
Related
This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*
When I was practice in rubular.com, I've be trying to match with a regular expression that checks if a word starts with a non-consonant. My approach it's check cases how that begins with a non-letter, or starts with a number or underscore, or checks the empty string
I've founded a strange behaviour:
My regex /^[aeiou_0-9\W]|^$/i match the k and s consonants!. I don't understand why.
Any ideas?
A link to example -> http://rubular.com/r/0zt0VPmcwr
This is very funny because you have stumbled across a bug specifically for just the letters k and s when using \W with /i (it's like a perfect storm).
Here is the link that explains the bug: https://bugs.ruby-lang.org/issues/4044
Perhaps this was patched in a later version of ruby, but if you don't feel like going through the hassle of going to a new version of ruby, then you can just explicitly make an inverted character class of all the consonants:
/^[^bcdfghjklmnpqrstvwxyz]|^$/i
Here is the rubular link: http://rubular.com/r/URgsWP3suQ
Edit:
So, something else I noticed about your regex is that your regex (and the regex I provided above) matches only the first letter of the words where as the regex that I provided matches the whole word. I don't know if this makes a difference for you, but I felt it was worth pointing out. Please see the difference in the highlighting in the rubular link above and the one below (See how the link above only highlights the first letter of the words where as the link below highlights the whole words):
^[^bcdfghjklmnpqrstvwxyz].*|^$
http://rubular.com/r/IVJ03uOK4h
It is a bug in Ruby regex in some versions. Select version 1.8.7 in the dropdown and you will see your regex works properly.
Edit. Check the docs at http://ruby-doc.org/core-2.1.5/Regexp.html. More specifically, in the metacharacters section:
/\W/ - A non-word character ([^a-zA-Z0-9_]). Please take a look at Bug #4044 if using /\W/ with the /i modifier.
I'm trying to come up with a regex that will elegantly match everything in an URL AFTER the domain name, and before the first ?, the last slash, or the end of the URL, if neither of the 2 exist.
This is what I came up with but it seems to be failing in some cases:
regex = /[http|https]:\/\/.+?\/(.+)[?|\/|]$/
In summary:
http://nytimes.com/2013/07/31/a-new-health-care-approach-dont-hide-the-price/ should return
2013/07/31/a-new-health-care-approach-dont-hide-the-price
http://nytimes.com/2013/07/31/a-new-health-care-approach-dont-hide-the-price?id=2 should return
2013/07/31/a-new-health-care-approach-dont-hide-the-price
http://nytimes.com/2013/07/31/a-new-health-care-approach-dont-hide-the-price should return
2013/07/31/a-new-health-care-approach-dont-hide-the-price
Please don't use Regex for this. Use the URI library:
require 'uri'
str_you_want = URI("http://nytimes.com/2013/07/31/a-new-health-care-approach-dont-hide-the-price").path
Why?
See everything about this famous question for a good discussion of why these kinds of things are a bad idea.
Also, this XKCD really says why:
In short, Regexes are an incredibly powerful tools, but when you're dealing with things that are made from hundred page convoluted standards when there is already a library for doing it faster, easier, and more correctly, why reinvent this wheel?
If lookaheads are allowed
((2[0-9][0-9][0-9].*)(?=\?\w+)|(2[0-9][0-9][0-9].*)(?=/\s+)|(2[0-9][0-9][0-9].*).*\w)
Copy + Paste this in http://regexpal.com/
See here with ruby regex tester: http://rubular.com/r/uoLLvTwkaz
Image using javascript regex, but it works out the same
(?=) is just a a lookahead
I basically set up three matches from 2XXX up to (in this order):
(?=\?\w+) # lookahead for a question mark followed by one or more word characters
(?=/\s+) # lookahead for a slash followed by one or more whitespace characters
.*\w # match up to the last word character
I'm pretty sure that some parentheses were not needed but I just copy pasted.
There are essentially two OR | expressions in the (A|B|C) expression. The order matters since it's like a (ifthen|elseif|else) type deal.
You can probably fix out the prefix, I just assumed that you wanted 2XXX where X is a digit to match.
Also, save the pitchforks everyone, regular expressions are not always the best but it's there for you when you need it.
Also, there is xkcd (https://xkcd.com/208/) for everything:
I am trying to grasp the concept of Regular Expressions but seem to be missing something.
I want to ensure that someone enters a string that ends with .wav in a field. Should be a pretty simple Regular Expression.
I've tried this...
[RegularExpression(#"$.wav")]
but seem to be incorrect. Any help is appreciated. Thanks!
$ is the anchor for the end of the string, so $.wav doesn't make any sense. You can't have any characters after the end of the string. Also, . has a special meaning for regex (it just means 'any character') so you need to escape it.
Try writing
\.wav$
If that doesn't work, try
.*\.wav$
(It depends on if the RegularExpression attribute wants to match the whole string, or just a part of it. .* means 'any character, 0 or more times')
Another thing you should consider is what to do with extra whitespace in the field. Users have a terrible habit of adding extra white space in inputs - its why various .Trim() functions are so important. Here, RegularExpressionAttribute might be evaluated before you can trim the input, so you might want to write this:
.*\.wav[\s]*$
The [\s]* section means 'any whitespace character (tabs, space, linebreak, etc) 0 or more times'.
You should read a tutorial on regex. It's not so hard to understand for simple problems like this. When I was learning I found this site pretty handy: http://www.regular-expressions.info/
The format I'm trying to match is:
# (Apple push notification codes)
"11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7"
The simplest expression I can think of is: /((\w{8}\s){7}\w{8})/i
Can anyone think of a simpler one?
(I'm using Ruby regular expressions)
UPDATE - thanks to user1096188, I've removed \d - this is included in \w
You can detect a word boundary using \b, and use (?: to prevent capturing groups
/(?:\w{8}\b\s?){8}/
You could do this if the end of the match is the end of the whole string.
(\w{8}(:?\s|$)){7}
Taking #zapthedingbat's solution one stage further, it looks like the code only contains hexadecimal characters (0-9 and a-f) and spaces. So you could possibly sacrifice a little simplicity for accuracy.
I'm making an assumption, but I suspect letters g to z are invalid.
If the format is hexadecimal only (you should check Apple's documentation to be sure), a tighter match would be:
/(?:[0-9a-f]{8}\b\s?){8}/
EDIT
In fact, in Ruby, it looks like you should be able to do:
/(?:\h{8}\b\s?){8}/
> "11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7".match(/((\w{8}\s)+)/)
> $&
=> "11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7"