I am trying to write a Ruby regex that will return a set of named matches. If the first element (defined by slashes) is found anywhere later in the string then I want the match to return that 2nd match onward. Otherwise, return the whole string. The closest I've gotten is (?<p1>top_\w+).*?(?<hier>\k<p1>.*) which doesn't work for the 3rd item. I've tried regex ifthen-else constructs but Rubular says it's invalid. I've tried (?<p1>[\w\/]+?)(?<hier>\k<p1>.*) which correct splits the 1st and 4th lines but doesn't work for the others. Please note: I want all results to return as the same named reference so I can iterate through "hier".
Input:
top_cat/mouse/dog/top_cat/mouse/dog/elephant/horse
top_ab12/hat[1]/top_ab12/hat[1]/path0_top_ab12/top_ab12path1/cool
top_bat/car[0]
top_2/top_1/top_3/top_4/top_2/top_1/top_3/top_4/dog
Output:
hier = top_cat/mouse/dog/elephant/horse
hier = top_ab12/hat[1]/path0_top_ab12/top_ab12path1/cool
hier = top_bat/car[0]
hier = top_2/top_1/top_3/top_4/dog
Problem
The reason it does not match the second line is because the second instance of hat does not end with a slash, but the first instance does.
Solution
Specify that there is a slash between the first and second match
Regex
(top_.*)/(\1.*$)|(^.*$)
Replacement
hier = \2\3
Example
Regex101 Permalink
More info on the Alternation token
To explain how the | token works in regex, see the example: abc|def
What this regex means in plain english is:
Match either the regex below (attempting the next alternative only if this one fails)
Match the characters abc literally
Or match the regex below (the entire match attempt fails if this one fails to match)
Match the characters def literally
Example
Regex: alpha|alphabet
If we had a phrase "I know the alphabet", only the word alpha would be matched.
However, if we changed the regex to alphabet|alpha, we would match alphabet.
So you can see, alternation works in a left-to-right fashion.
paths = %w(
top_cat/mouse/dog/top_cat/mouse/dog/elephant/horse
top_ab12/hat/top_ab12/hat[1]/path0_top_ab12/top_ab12path1/cool
top_bat/car[0]
top_2/top_1/top_3/top_4/top_2/top_1/top_3/top_4/dog
test/test
)
paths.each do |path|
md = path.match(/^([^\/]*).*\/(\1(\/.*|$))/)
heir = md ? md[2] : path
puts heir
end
Output:
top_cat/mouse/dog/elephant/horse
top_ab12/hat[1]/path0_top_ab12/top_ab12path1/cool
top_bat/car[0]
top_2/top_1/top_3/top_4/dog
test
Related
I have a string https://stackverflow.com. I want a new string that contains the domain from the given string using regular expressions.
Example:
x = "https://stackverflow.com"
newstring = "stackoverflow.com"
Example 2:
x = "https://www.stackverflow.com"
newstring = "www.stackoverflow.com"
"https://stackverflow.com"[/(?<=:\/\/).*/]
#⇒ "stackverflow.com"
(?<=..) is a positive lookbehind.
If string = "http://stackoverflow.com",
a really easy way is string.split("http://")[1]. But this isn't regex.
A regex solution would be as follows:
string.scan(/^http:\/\/(.+)$/).flatten.first
To explain:
String#scan returns the first match of the regex.
The regex:
^ matches beginning of line
http: matches those characters
\/\/ matches //
(.+) sets a "match group" containing any number of any characters. This is the value returned by the scan.
$ matches end of line
.flatten.first extracts the results from String#scan, which in this case returns a nested array.
You might want to try this:
#!/usr/bin/env ruby
str = "https://stackoverflow.com"
if mtch = str.match(/(?::\/\/)(/S)/)
f1 = mtch.captures
end
There are two capturing groups in the match method: the first one is a non-capturing group referring to your search pattern and the second one referring to everything else afterwards. After that, the captures method will assign the desired result to f1.
I hope this solves your problem.
I'm trying to create a regex pattern to match particular sets of text in my string.
Let's assume this is the string ^foo{bar}#Something_Else
I would like to match ^foo{} skipping entirely the content of the brackets.
Until now i figured out how to get all everything with this regex here \^(\w)\{([^\}]+)} but i really don't know how to ignore the text inside the curly brackets.
Anyone has an idea? Thanks.
Update
This is the final solution:
puts script.gsub(/(\^\w+)\{([^}]+)(})/, '[BEFORE]\2[AFTER]')
Though I'd prefer this with fewer groups:
puts script.gsub(/\^\w+\{([^}]+)}/, '[BEFORE]\1[AFTER]')
Original answer
I need to replace the ^foo{} part with something else
Here is a way to do it with gsub:
s = "^foo{bar}#Something_Else"
puts s.gsub(/(.*)\^\w+\{([^}]+)}(.*)/, '\1SOMETHING ELSE\2\3')
See demo
The technique is the same: you capture the text you want to keep and just match text you want to delete, and use backreferences to restore the text you captured.
The regex matches:
(.*) - matches and captures into Group 2 as much text as possible from the start
\^\w+\{ - matches ^, 1 or more word characters, {
([^}]+) - matches and captures into Group 2 1 or more symbols other than }
} - matches the }
(.*) - and finally match and capture into Group 3 the rest of the string.
If you mean to match ^foo{} by a single match against a regex, it is impossible. A regex match only matches a substring of the original string. Since ^foo{} is not a substring of ^foo{bar}#Something_Else, you cannot match that with a single match.
I have a string something like:
test:awesome my search term with spaces
And I'd like to extract the string immediately after test: into one variable and everything else into another, so I'd end up with awesome in one variable and my search term with spaces in another.
Logically, what I'd so is move everything matching test:* into another variable, and then remove everything before the first :, leaving me with what I wanted.
At the moment I'm using /test:(.*)([\s]+)/ to match the first part, but I can't seem to get the second part correctly.
The first capture in your regular expression is greedy, and matches spaces because you used .. Instead try:
matches = string.match(/test:(\S*) (.*)/)
# index 0 is the whole pattern that was matched
first = matches[1] # this is the first () group
second = matches[2] # and the second () group
Use the following:
/^test:(.*?) (.*)$/
That is, match "test:", then a series of characters (non-greedily), up to a single space, and another series of characters to the end of the line.
I am guessing you want to remove all the leading spaces before the second match too, hence I have \s+ in the expression. Otherwise, remove the \s+ from the expression, and you'll have what you want:
m = /^test:(\w+)\s+(.*)/.match("test:awesome my search term with spaces")
a = m[1]
b = m[2]
http://codepad.org/JzuNQxBN
I don't think I'll even try to explain this, I don't know the words to, but I'd like to achieve the following:
Given a string like this:
+++>><<<--
I'd like a match to give me: +++, but also match if any of the other characters were in the string consecutively like they are. So if the +++ wasn't there, I'd like to match >>.
I tried using the following regular expression:
([><\-\+]+)
However, given the string above, it would match the entire string, and not the first list of consecutive characters.
If it makes a difference, this is in Ruby (1.9.3).
Not sure about the ruby bit, but you can do this with backreferences in the pattern:
(.)\1+
What this does is to use a capturing group () to capture any character . followed by any number + of the same character \1. The \1 is a backreference to the the first captured group; in a pattern with more capturing groups \2 would be the second captured group and so on.
Java Example
Pattern p = Pattern.compile("(.)\\1+");
Matcher m = p.matcher("aaabbccaa");
m.find();
System.out.println(m.group(0)); // prints "aaa"
Ruby Example
# Return an array of matched patterns.
string = '+++>><<<--'
string.scan( /((.)\2+)/ ).collect { |match| match.first }
I have a string like this:
http://www.example.com/value/1234/different-value
How can I extract the 1234?
Note: There may be a slash at the end:
http://www.example.com/value/1234/different-value
http://www.example.com/value/1234/different-value/
/([^/]+)(?=/[^/]+/?$)
should work. You might need to format it differently according to the language you're using. For example, in Ruby, it's
if subject =~ /\/([^\/]+)(?=\/[^\/]+\/?\Z)/
match = $~[1]
else
match = ""
end
Use Slice for Positional Extraction
If you always want to extract the 4th element (including the scheme) from a URI, and are confident that your data is regular, you can use Array#slice as follows.
'http://www.example.com/value/1234/different-value'.split('/').slice 4
#=> "1234"
'http://www.example.com/value/1234/different-value/'.split('/').slice 4
#=> "1234"
This will work reliably whether there's a trailing slash or not, whether or not you have more than 4 elements after the split, and whether or not that fourth element is always strictly numeric. It works because it's based on the element's position within the path, rather than on the contents of the element. However, you will end up with nil if you attempt to parse a URI with fewer elements such as http://www.example.com/1234/.
Use Scan/Match for Pattern Extraction
Alternatively, if you know that the element you're looking for is always the only one composed entirely of digits, you can use String#match with look-arounds to extract just the numeric portion of the string.
'http://www.example.com/value/1234/different-value'.match %r{(?<=/)\d+(?=/)}
#=> #<MatchData "1234">
$&
#=> "1234"
The look-behind and look-ahead assertions are needed to anchor the expression to a path. Without them, you'll match things like w3.example.com too. This solution is a better approach if the position of the target element may change, and if you can guarantee that your element of interest will be the only one that matches the anchored regex.
If there will be more than one match (e.g. http://www.example.com/1234/5678/) then you might want to use String#scan instead to select the first or last match. This is one of those "know your data" things; if you have irregular data, then regular expressions aren't always the best choice.
Javascript:
var myregexp = /:\/\/.*?\/.*?\/(\d+)/;
var match = myregexp.exec(subject);
if (match != null) {
result = match[1];
}
Works with your examples... But I am sure it will fail in general...
Ruby edit:
if subject =~ /:\/\/.*?\/.*?\/(.+?)\//
match = $~[1]
It does work.
I think this is a little simpler than the accepted answer, because it doesn't use any positive lookahead (?=), but rather simply makes the last slash optional via the ? character:
^.+\/(.+)\/.+\/?$
In Ruby:
STDIN.read.split("\n").each do |nextline|
if nextline =~ /^.+\/(.+)\/.+\/?$/
printf("matched %s in %s\n", $~[1], nextline);
else
puts "no match"
end
end
Live Demo
Let's break down what's happening:
^: start of the line
.+\/: match anything (greedily) up to a slash
Since we're going to later match at least 1, at most 2 more slashes, this slash will be either the second last slash (as in http://www.example.com/value/1234/different-value) or the third last slash as in (http://www.example.com/value/1234/different-value/)
Up to this point we've matched http://www.example.com/value/ (due to greediness)
(.+)\/: Our capturing group for 1234 indicated by the parenthesis. It's anything followed by another slash.
Since the previous match matched up to the second or third last slash, this will match up to the last slash or second last slash, respectively
.+: match anything. This would be after our 1234, so we're assuming there are characters after 1234/ (different-value)
\/?: optionally match another slash (the slash after different-value)
$: match the end of the line
Note that in a url, you probably won't have spaces. I used the . character because it's easily distinguished, but perhaps you might use \S instead to match non-spaces.
Also, you might use \A instead of ^ to match start of string (instead of after line break) and \Z instead of $ to match end of string (instead of at line break)