I need to convert expressions of the form:
return *;
into:
return filter(*);
It seems simple enough to express it with wildcards, however, in visual studio's search & replace dailog, there's no way to associate the first asterisk with the second one. I suppose a regex can do this quite easily, however I know very little about regexes.
How do I express this criteria in regex?
A capture group when searching/replacing with regex in VS can be given by enclosing something with curly braces.
A backreference can be given simply by using \1. There is also a menu to the right of the input fields, containing building blocks.
So you would be simply replacing
return {[^;]+};
by
return filter(\1);
The [^;]+ specifies that you want at least one character that is not a semicolon, so unless you return delegates or anonymous methods this should work fine.
Related
Is there way on visual studio to find and replace text but keeping the number in the string same?
For example, lets say I have a code that saids
fields[0].Value;
fields[1].Value;
And now I would like to replace it with
reader.GetString(0);
reader.GetString(1);
Without manually replacing every single lines of code, I was hoping to do it through find and replace dialog.
Is there any ways of doing this?
Thank you
If you want to replace part of the expression but keep a part (like the number in your case) you can use the search and replace function (ctrl+h) set to use regular expressions (alt+e) and use these expressions:
Search: fields\[(.)\].Value;
Replace: reader.GetString($1);
This will replace all expressions on the form fields[n].Value; with reader.GetString(n); where n is any single character. If you want to restrict it to keep numbers only use fields\[(\d)\].Value;
For more information see: Using Regular Expressions in Visual Studio
I tried it with VS2013 and it worked as expected.
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 having an issue trying to capture a group on a string:
"type=gist\nYou need to gist this though\nbecause its awesome\nright now\n</code></p>\n\n<script src=\"https://gist.github.com/3931634.js\"> </script>\n\n\n<p><code>Not code</code></p>\n"
My regex currently looks like this:
/<code>([\s\S]*)<\/code>/
My goal is to get everything in between the code brackets. Unfortunately, it's matching up to the 2nd closing code bracket Is there a way to match everything inside the code brackets up until the first occurrence of ending code bracket?
All repetition quantifiers in regular expressions are greedy by default (matching as many characters as possible). Make the * ungreedy, like this:
/<code>([\s\S]*?)<\/code>/
But please consider using a DOM parser instead. Regex is just not the right tool to parse HTML.
And I just learned that for going through multiple parts, the
String.scan( /<code>(.*?)<\/code>/ ){
puts $1
}
is a very nice way of going through all occurences of code - but yes, getting a proper parser is better...
I want to transform the following text
This is a , here is another 
into
This is a , here is another 
In other words I want to find all the image paths that are enclosed between brackets (the text is in Markdown syntax) and replace them with other paths. The string containing the new path is returned by a separate real_path function.
I would like to do this using String#gsub in its block version. Currently my code looks like this:
re = /!\[.*?\]\((.*?)\)/
rel_content = content.gsub(re) do |path|
real_path(path)
end
The problem with this regex is that it will match  instead of just foto.jpeg. I also tried other regexen like (?>\!\[.*?\]\()(.*?)(?>\)) but to no avail.
My current workaround is to split the path and reassemble it later.
Is there a Ruby regex that matches only the path inside the brackets and not all the contextual required characters?
Post-answers update: The main problem here is that Ruby's regexen have no way to specify zero-width lookbehinds. The most generic solution is to group what the part of regexp before and the one after the real matching part, i.e. /(pre)(matching-part)(post)/, and reconstruct the full string afterwards.
In this case the solution would be
re = /(!\[.*?\]\()(.*?)(\))/
rel_content = content.gsub(re) do
$1 + real_path($2) + $3
end
A quick solution (adjust as necessary):
s = 'This is a '
s.sub!(/!(\[.*?\])\((.*?)\)/, '\1(/folder1/\2)' )
p s # This is a [foto](/folder1/foto.jpeg)
You can always do it in two steps - first extract the whole image expression out and then second replace the link:
str = "This is a , here is another "
str.gsub(/\!\[[^\]]*\]\(([^)]*)\)/) do |image|
image.gsub(/(?<=\()(.*)(?=\))/) do |link|
"/a/new/path/" + link
end
end
#=> "This is a , here is another "
I changed the first regex a bit, but you can use the same one you had before in its place. image is the image expression like , and link is just the path like foto.jpeg.
[EDIT] Clarification: Ruby does have lookbehinds (and they are used in my answer):
You can create lookbehinds with (?<=regex) for positive and (?<!regex) for negative, where regex is an arbitrary regex expression subject to the following condition. Regexp expressions in lookbehinds they have to be fixed width due to limitations on the regex implementation, which means that they can't include expressions with an unknown number of repetitions or alternations with different-width choices. If you try to do that, you'll get an error. (The restriction doesn't apply to lookaheads though).
In your case, the [foto] part has a variable width (foto can be any string) so it can't go into a lookbehind due to the above. However, lookbehind is exactly what we need since it's a zero-width match, and we take advantage of that in the second regex which only needs to worry about (fixed-length) compulsory open parentheses.
Obviously you can put real_path in from here, but I just wanted a test-able example.
I think that this approach is more flexible and more readable than reconstructing the string through the match group variables
In your block, use $1 to access the first capture group ($2 for the second and so on).
From the documentation:
In the block form, the current match string is passed in as a parameter, and variables such as $1, $2, $`, $&, and $' will be set appropriately. The value returned by the block will be substituted for the match on each call.
As a side note, some people think '\1' inappropriate for situations where an unconfirmed number of characters are matched. For example, if you want to match and modify the middle content, how can you protect the characters on both sides?
It's easy. Put a bracket around something else.
For example, I hope replace a-ruby-porgramming-book-531070.png to a-ruby-porgramming-book.png. Remove context between last "-" and last ".".
I can use /.*(-.*?)\./ match -531070. Now how should I replace it? Notice
everything else does not have a definite format.
The answer is to put brackets around something else, then protect them:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1.')
# => "a-ruby-porgramming-book.png"
If you want add something before matched content, you can use:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1-2019\2.')
# => "a-ruby-porgramming-book-2019-531070.png"
can any body tell me how to use regex for negation of string?
I wanna find all line that start with public class and then any thing except first,second and finally any thing else.
for example in the result i expect to see public class base but not public class myfirst:base
can any body help me please??
Use a negative lookahead:
public\s+class\s+(?!first|second).+
If Peter is correct and you're using Visual Studio's Find feature, this should work:
^:b*public:b+class:b+~(first|second):i.*$
: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 $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)