I have an string of an email that looks like "<luke#example.com>"
I would like to use regex for deleting "<" and ">", so I wanted something like
"<luke#example.com>".sub /<>/, ""
The problem is quite clear, /<>/ doesn't wrap what I want. I tried with different regex, but I don't know how to choose < AND >, it is there any and operator where I can say: "wrap this and this"?
As written, your regex matches the literal substring "<>" only. You need to use [] to make them a character class so that they're matched individually, and gsub to replace all matches:
"<luke#example.com>".gsub(/[<>]/, "") # => "luke#example.com"
"<luke#example.com>".gsub /[<>]/, ""
http://regex101.com/r/hP3sY2
If you only ever want to strip the < and > from the start and end only, you can use this:
'<luke#example.com>'.sub(/\A<([^<>]+)>\z/, '\1')
You don't need, nor should you use, a regex.
string[1..-2]
is enough.
Related
Using a regex, how can I match strings that end with exactly one . as:
This is a string.
but not those that end with more than one . as:
This is a string...
I have a regex that detects a single .:
/[\.]{1}\z/
but I do not want it to match strings that end in ....
What you want is a 'negative lookbehind' assertion:
(?<!\.)\.\z
This looks for a period at the end of a string that isn't preceded by a period. The other answers won't match the following string: "."
Also, you may need to look out for unicode ellipsis characters…
You can detect this like so: str =~ /\u{2026}/
You can use:
[^\.][\.]\z
You are looking for a string that before the last dot there is a char that is not a dot.
I like Regexr a lot!
Solution similar to Dekel:
[^.]+[.]
Live demo
I have a string and I want to remove all non-word characters and whitespace from it. So I thought Regular expressions would be what I need for that.
My Regex looks like that (I defined it in the string class as a method):
/[\w&&\S]+/.match(self.downcase)
when I run this expression in Rubular with the test string "hello ..a.sdf asdf..," it highlioghts all the stuff I need ("hellloasdfasdf") but when I do the same in irb I only get "hello".
Has anyone any ideas about why that is?
Because you use match, with returns one matching element. If you use scan instead, all should work properly:
string = "hello ..a.sdf asdf..,"
string.downcase.scan(/[\w&&\S]+/)
# => ["hello", "a", "sdf", "asdf"]
\w means [a-zA-Z0-9_]
\S means any non-whitespace character [a-zA-Z_-0-9!##$%^&*\(\)\\{}?><....etc]
so using a \w and \S condition is ambiguous.
Its like saying What is an intersection of India and Asia. Obviously its going to be India. So I will suggest you to use \w+.
and you can use scan to get all matches as mentioned in the second answer :
string = "hello ..a.sdf asdf..,"
string.scan(/\w+/)
I have some strings like this
../..//somestring
../somestring
../..somestring
./somestring
How to write a regular expression in ruby to extract "somestring" from above strings. Before some strings it can be any combination of . and /
Thanks for you help
Do this:
string.sub(/\A[.\/]+/, "")
"../../test/file/cases".sub(/\A[.\/]+/, "")
# => "test/file/cases"
Just match letters:
str = "../..//somestring" # or "../somestring", "../..somestring", "./somestring"
str[/[a-z]+/] # somestring
RE: your comment
If you just want to remove leading dots and slashes, use
str.gsub(/[.\/]/, '')
It looks like you're dealing with file paths. If so, there are more appropriate tools than regexps
File.basename('../..//somestring')
# "somestring"
regexp = /^[.\/]+(.*?)$/i
result = subject.scan(regexp)
http://rubular.com/r/tMYKPe78uc
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]
I have a string, like this:
"yellow-corn-(corn-on-the-cob)"
and I would like to strip the parenthesis from the string to get something like this:
"yellow-corn-corn-on-the-cob"
I believe you would use gsub to accomplish this, but I'm not sure what pattern I would need to match the parenthesis. Something like:
clean_string = old_string.gsub(PATTERN,"")
Without regular expression:
"yellow-corn-(corn-on-the-cob)".delete('()') #=> "yellow-corn-corn-on-the-cob"
Try this:
clean_string = old_string.gsub(/[()]/, "")
On a side note, Rubular is awesome to test your regular expressions quickly.