I just want to capture the part of the string in nbnbaasd<sd which appears before any a.
I want it to return nbnb as the match.
/.+(?!a)/.match("nbnbaasd<sd") # returns the whole string
Just use a negated character set:
/[^a]+/.match("nbnbaasd<sd")
It's far more efficient than the look-ahead method.
See it here in action: http://regexr.com?32288
It returns the whole string because indeed, "nbnbaasd<sd" is not followed by an "a".
Try this.
/.+?(?=a)/.match("nbnbaasd<sd")
(You do not actually need to use a lookahead to achieve this, but perhaps you've simplified your problem and in your real problem you do need a zero-width assertion for some reason. So this is a solution as close as possible to the one you've attempted.)
Related
I want to add a validation to input field so that it won't accept zero as valid input. I want to use a regex for that, so it allows everything except zero.
I tested a couple of regex e.g. ^((?!0).)*$ but did not quite get the thing that I was looking for.
Maybe with .(?!0)
? Normally this should work, as . matches any character excepted new line.
Don't forget to add the flag g when you use your regexp, otherwise, it will stop after first matching.
Your final regexp should then be : /.(?!0)/g
I'm trying to take out the the content inside the parenthesis. For example, if the string is "(blah blah) This is stack(over)flow", I want to just take out "(blah blah)" but leave "(over)" alone. I'm trying
/\A\(.*\)/
but returns "(blah blah) This is stack(over)", and I'm sure why it's returning that.
Easiest fix:
/\A\(.*?\)/
Normally, * will try to match as much as it possibly can, so it'll match all the way to the last ) in the line. This is called "greedy" matching. Putting ? after +/*/? makes them non-greedy, and they'll match the shortest possible string.
But note that this won't work for nested parentheses. That's rather more complicated. Given your example, I assume this is for a pretty simple ad-hoc format where nesting isn't a concern.
I'm trying to match some text if it does not have another block of text in its vicinity. For example, I would like to match "bar" if "foo" does not precede it. I can match "bar" if "foo" does not immediately precede it using negative look behind in this regex:
/(?<!foo)bar/
but I also like to not match "foo 12345 bar". I tried:
/(?<!foo.{1,10})bar/
but using a wildcard + a range appears to be an invalid regex in Ruby. Am I thinking about the problem wrong?
You are thinking about it the right way. But unfortunately lookbehinds usually have be of fixed-length. The only major exception to that is .NET's regex engine, which allows repetition quantifiers inside lookbehinds. But since you only need a negative lookbehind and not a lookahead, too. There is a hack for you. Reverse the string, then try to match:
/rab(?!.{0,10}oof)/
Then reverse the result of the match or subtract the matching position from the string's length, if that's what you are after.
Now from the regex you have given, I suppose that this was only a simplified version of what you actually need. Of course, if bar is a complex pattern itself, some more thought needs to go into how to reverse it correctly.
Note that if your pattern required both variable-length lookbehinds and lookaheads, you would have a harder time solving this. Also, in your case, it would be possible to deconstruct your lookbehind into multiple variable length ones (because you use neither + nor *):
/(?<!foo)(?<!foo.)(?<!foo.{2})(?<!foo.{3})(?<!foo.{4})(?<!foo.{5})(?<!foo.{6})(?<!foo.{7})(?<!foo.{8})(?<!foo.{9})(?<!foo.{10})bar/
But that's not all that nice, is it?
As m.buettner already mentions, lookbehind in Ruby regex has to be of fixed length, and is described so in the document. So, you cannot put a quantifier within a lookbehind.
You don't need to check all in one step. Try doing multiple steps of regex matches to get what you want. Assuming that existence of foo in front of a single instance of bar breaks the condition regardless of whether there is another bar, then
string.match(/bar/) and !string.match(/foo.*bar/)
will give you what you want for the example.
If you rather want the match to succeed with bar foo bar, then you can do this
string.scan(/foo|bar/).first == "bar"
following a RoR security tutorial (here), i wrote something along the lines of
##private_re = //
def secure?
action_name =~ ##private_re
end
the idea is that in the base case, this shouldn't match anything, and return nil. problem is that it doesn't. i've worked around for the time being by using a nonsensical string, but i'd like to know the answer.
The empty regular expression successfully matches every string.
Examples of regular expressions that will always fail to match:
/(?=a)b/
/\Zx\A/
/[^\s\S]/
It is meant to not change the behavior of the controller in any way, as // will match every string.
The idea is that ##private is meant to be set in the controller to match things you DO want to be private. Thus, that code is meant to do nothing, but when combined with
##private = /.../ in the controller, gives you a nice privacy mechanism.
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 $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)