string = 'xabcdexfghijk'
In the example above, 'x' appears twice. I want to capture everything between the first 'x' and the next 'x'. Thus, the desired result is a new string that equals 'xabcdex'. Any ideas?
You could use a simple regular expression: /x.*?x/. This basically means "match any characters in between two x characters, as few times as possible (non-greedy)".
The matched text can be extracted with String#[regexp]
string = 'xabcdexfghijk'
string[/x.*?x/] # => "xabcdex"
Related
I need to match all the alphabets and numbers in a string str.
This is my code.
str.match(/^(AB)(\d+)([A-Za-z][0-9])?/)
When str = AB57933A [sic], it matches only AB57933, and not the characters appended after the numbers.
If I try with str = AB57933AbC [sic], it matches only AB57933; it only matches up to the last number, and not the characters after that.
In the way you have written it:
/^(AB)(\d+)([A-Za-z][0-9])/
you impose that the last character is between 0 and 9, you can replace it depending on your needs by if you do not expect digits after the last letter
/^(AB)(\d+)([A-Za-z]+)/
or by
/^(AB)(\d+)([A-Za-z0-9]+)/
if AB57933AbC12 are also accepted as valid input.
Last but not least, if you do not use back references you can omit the parenthesis as you do not need capturing groups
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 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 "{some|words|are|here}" or "{another|set|of|words}"
So in general the string consists of an opening curly bracket,words delimited by a pipe and a closing curly bracket.
What is the most efficient way to get the selected word of that string ?
I would like do something like this:
#my_string = "{this|is|a|test|case}"
#my_string.get_column(0) # => "this"
#my_string.get_column(2) # => "is"
#my_string.get_column(4) # => "case"
What should the method get_column contain ?
So this is the solution I like right now:
class String
def get_column(n)
self =~ /\A\{(?:\w*\|){#{n}}(\w*)(?:\|\w*)*\}\Z/ && $1
end
end
We use a regular expression to make sure that the string is of the correct format, while simultaneously grabbing the correct column.
Explanation of regex:
\A is the beginnning of the string and \Z is the end, so this regex matches the enitre string.
Since curly braces have a special meaning we escape them as \{ and \} to match the curly braces at the beginning and end of the string.
next, we want to skip the first n columns - we don't care about them.
A previous column is some number of letters followed by a vertical bar, so we use the standard \w to match a word-like character (includes numbers and underscore, but why not) and * to match any number of them. Vertical bar has a special meaning, so we have to escape it as \|. Since we want to group this, we enclose it all inside non-capturing parens (?:\w*\|) (the ?: makes it non-capturing).
Now we have n of the previous columns, so we tell the regex to match the column pattern n times using the count regex - just put a number in curly braces after a pattern. We use standard string substition, so we just put in {#{n}} to mean "match the previous pattern exactly n times.
the first non skipped column after that is the one we care about, so we put that in capturing parens: (\w*)
then we skip the rest of the columns, if any exist: (?:\|\w*)*.
Capturing the column puts it into $1, so we return that value if the regex matched. If not, we return nil, since this String has no nth column.
In general, if you wanted to have more than just words in your columns (like "{a phrase or two|don't forget about punctuation!|maybe some longer strings that have\na newline or two?}"), then just replace all the \w in the regex with [^|{}] so you can have each column contain anything except a curly-brace or a vertical bar.
Here's my previous solution
class String
def get_column(n)
raise "not a column string" unless self =~ /\A\{\w*(?:\|\w*)*\}\Z/
self[1 .. -2].split('|')[n]
end
end
We use a similar regex to make sure the String contains a set of columns or raise an error. Then we strip the curly braces from the front and back (using self[1 .. -2] to limit to the substring starting at the first character and ending at the next to last), split the columns using the pipe character (using .split('|') to create an array of columns), and then find the n'th column (using standard Array lookup with [n]).
I just figured as long as I was using the regex to verify the string, I might as well use it to capture the column.