I'm looking to check if a string is all capitals in Rails.
How would I go about doing that?
I'm writing my own custom pluralize helper method and I would something be passing words like "WORD" and sometimes "Word" - I want to test if my word is all caps so I can return "WORDS" - with a capital "S" in the end if the word is plural (vs. "WORDs").
Thanks!
Do this:
str == str.upcase
E.g:
str = "DOG"
str == str.upcase # true
str = "cat"
str == str.upcase # false
Hence the code for your scenario will be:
# In the code below `upcase` is required after `str.pluralize` to transform
# DOGs to DOGS
str = str.pluralize.upcase if str == str.upcase
Thanks to regular expressions, it is very simple. Just use [[:upper:]] character class which allows all upper case letters, including but not limited to those present in ASCII-8.
Depending on what do you need exactly:
# allows upper case characters only
/\A[[:upper:]]*\Z/ =~ "YOURSTRING"
# additionally disallows empty strings
/\A[[:upper:]]+\Z/ =~ "YOURSTRING"
# also allows white spaces (multi-word strings)
/\A[[:upper:]\s]*\Z/ =~ "YOUR STRING"
# allows everything but lower case letters
/\A[^[:lower:]]*\Z/ =~ "YOUR 123_STRING!"
Ruby doc: http://www.ruby-doc.org/core-2.1.4/Regexp.html
Or this:
str =~ /^[A-Z]+$/
e.g.:
"DOG" =~ /^[A-Z]+$/ # 0
"cat" =~ /^[A-Z]+$/ # nil
Related
How can I check if a string include symbol? It looks confusing to me:
class Beba
def initialize
while true
puts "Qfar emri deshironi ti vnoni bebes?"
##emri = gets.chomp.capitalize
if ##emri.scan(/\d+/).empty? && ##emri.scan(/\/./).empty?
puts "Ti e emertove beben me emrin: #{##emri}"
break
else
puts "Emri nuk mund te jete me numra/simbole, provoni perseri."
end
end
end
end
As you can see, at if##emri.scan(/\d+/).empty? && ##emri.scan(/\/./).empty?, I don't know what to do, like which method can I use for ##emri.scan(/\\.\).empty? to check if my string doesn't include any symbol?
For the specific characters you asked for, you can use this:
##emri.scan(/[!##$%^&*()_+{}\[\]:;'"\/\\?><.,]/).empty?
Will return true if no special character is found.
str !~ /[!##$%^&*()_+{}\[\]:;'"\/\\?><.,]/
returns true if and only if the string str contains none of the characters in the regex's character class (else false is returned).
Its seems you are looking for special characters.
Use something like
"Hel#lo".index( /[^[:alnum:]]/ )
It will return nil if no special charatcters.
[:alnum:] includes all 0-9, a-z, A-Z.
IF YOU WANT TO GO FOR SPECIFIC CHARATCERS
place all characters in a string & create regex like
characters = "!##$%^&*()_+{}[]:;'\"\/?><.,"
regex = /[#{characters.gsub(/./){|char| "\\#{char}"}}]/
& than use this regex to see if any of them exist in string like
if some_string =~ regex
I'm trying to write a regular expression in Ruby where I want to see if the string contains a certain word (e.g. "string"), followed by a url and link name in parenthesis.
Right now I'm doing:
string.include?("string") && string.scan(/\(([^\)]+)\)/).present?
My input in both conditionals is a string. In the first one, I'm checking if it contains the word "link" and then I will have the link and link_name in parenthesis, like this:
"Please go to link( url link_name)"
After validating that, I extract the HTML link.
Is there a way I can combine them using regular expressions?
The most important improvement you can make is to also test that the word and the parentheseses have the correct relationship. If I understand correctly, "link(url link_name)" should be a match but "(url link_name)link" or "link stuff (url link_name)" should not. So match "link", the parentheses, and their contents, and capture the contents, all at once:
"stuff link(url link_name) more stuff".match(/link\((\S+?) (\S+?)\)/)&.captures
=> ["url", "link_name"]
(&. is Ruby 2.3; use Rails' .try :captures in older versions.)
Side note: string.scan(regex).present? is more concisely written as string =~ regex.
Checking If a Word Is Contained
If you want to find matches that contain a specific word somewhere in the string, you can accomplish this through a lookahead :
# This will match any string that contains your string "{your-string-here}"
(?=.*({your-string-here}).*).*
You could consider building a string version of your expression and passing the word you are looking for using a variable :
wordToFind = "link"
if stringToTest =~ /(?=.*(#{wordToFind}).*).*/
# stringToTest contains "link"
else
# stringToTest does not contain "link"
end
Checking for a Word AND Parentheses
If you also wanted to ensure that somewhere in your string you had a set of parentheses with some content in them and your previous lookahead for a word, you could use :
# This will match any strings that contain your word and contain a set of parentheses
(?=.*({your-string-here}).*).*\([^\)]+\).*
which might be used as :
wordToFind = "link"
if stringToTest =~ /(?=.*(#{wordToFind}).*).*\([^\)]+\).*/
# stringToTest contains "link" and some non-empty parentheses
else
# stringToTest does not contain "link" or non-empty parentheses
end
def has_both?(str, word)
str.scan(/\b#{word}\b|(?<=\()[^\(\)]+(?=\))/).size == 2
end
has_both?("Wait for me, Wild Bill.", "Bill")
#=> false
has_both?("Wait (for me), Wild William.", "Bill")
#=> false
has_both?("Wait (for me), Wild Billy.", "Bill")
#=> false
has_both?("Wait (for me), Wild bill.", "Bill")
#=> false
has_both?("Wait (for me, Wild Bill.", "Bill")
#=> false
has_both?("Wait (for me), Wild Bill.", "Bill")
#=> true
has_both?("Wait ((for me), Wild Bill.", "Bill")
#=> true
has_both?("Wait ((for me)), Wild Bill.", "Bill")
#=> true
These are the calculations for
word = "Bill"
str = "Wait (for me), Wild Bill."
r = /
\b#{word}\b # match the value of the variable 'word' with word breaks for and aft
| # or
(?<=\() # match a left paren in a positive lookbehind
[^\(\)]+ # match one or more characters other than parens
(?=\)) # match a right paren in a positive lookahead
/x # free-spacing regex definition mode
#=> /
\bBill\b # match the value of the variable 'word' with word breaks for and aft
| # or
(?<=\() # match a left paren in a positive lookbehind
[^\(\)]+ # match one or more characters other than parens
(?=\)) # match a right paren in a positive lookahead
/x
arr = str.scan(r)
#=> ["for me", "Bill"]
arr.size == 2
#=> true
I would go with something like this regex:
/link\s*\(([^\)\s]+)\s*([^\)]+)?\)/i
This will find any match starting with the word link, followed by any number of spaces, then a url followed by a link name, both in parentheses. In this regex, the link name is optional, but the url is not. The matching is case-insensitive, so it will match link and LINK exactly the same.
You can use the Regexp#match method to compare the regex to a string, and check the result for matches and captures, like so:
m = /link\s*\(([^\)\s]+)\s*([^\)]+)?\)/i.match("link (stackoverflow.com StackOverflow)")
if m # the match array is not nil
puts "Matched: #{m[0]}"
puts " -- url: {m[1]}"
puts " -- link-name: #{m[2] || 'none'}"
else # the match array is nil, so no match was found
puts "No match found"
end
If you'd like to use different strings to identify the match, you can use a non-capturing group, where you change link to something like:
(?:link|site|website|url)
In this case, the (?: syntax says not to capture this part of the match. If you want to capture which term matched, simply change that from (?: to (, and adjust the capture indexes by 1 to account for the new capture value.
Here's a short Ruby test program:
data = [
[ true, "link (http://google.com Google)", "http://google.com", "Google" ],
[ true, "LiNk(ftp://website.org)", "ftp://website.org", nil ],
[ true, "link (https://facebook.com/realstanlee/ Stan Lee) linkety link", "https://facebook.com/realstanlee/", "Stan Lee" ],
[ true, "x link (https://mail.yahoo.com Yahoo! Mail)", "https://mail.yahoo.com", "Yahoo! Mail" ],
[ false, "link lunk (http://www.com)", nil, nil ]
]
data.each do |test_case|
link = /link\s*\(([^\)\s]+)\s*([^\)]+)?\)/i.match(test_case[1])
url = link ? link[1] : nil
link_name = link ? link[2] : nil
success = test_case[0] == !link.nil? && test_case[2] == url && test_case[3] == link_name
puts "#{success ? 'Pass' : 'Fail'}: '#{test_case[1]}' #{link ? 'found' : 'not found'}"
if success && link
puts " -- url: '#{url}' link_name: '#{link_name || '(no link name)'}'"
end
end
This produces the following output:
Pass: 'link (http://google.com Google)' found
-- url: 'http://google.com' link_name: 'Google'
Pass: 'LiNk(ftp://website.org)' found
-- url: 'ftp://website.org' link_name: '(no link name)'
Pass: 'link (https://facebook.com/realstanlee/ Stan Lee) linkety link' found
-- url: 'https://facebook.com/realstanlee/' link_name: 'Stan Lee'
Pass: 'x link (https://mail.yahoo.com Yahoo! Mail)' found
-- url: 'https://mail.yahoo.com' link_name: 'Yahoo! Mail'
Pass: 'link lunk (http://www.com)' not found
If you want to allow anything other than spaces between the word 'link' and the first paren, simply change the \s* to [^\(]* and you should be good to go.
I am having trouble writing this so that it will take a sentence as an argument and perform the translation on each word without affecting the punctuation.
I'd also like to continue using the partition method.
It would be nice if I could have it keep a quote together as well, such as:
"I said this", I said.
would be:
"I aidsay histay", I said.
def convert_sentence_pig_latin(sentence)
p split_sentence = sentence.split(/\W/)
pig_latin_sentence = []
split_sentence.each do |word|
if word.match(/^[^aeiou]+/x)
pig_latin_sentence << word.partition(/^[^aeiou]+/x)[2] + word.partition(/^[^aeiou]+/x)[1] + "ay"
else
pig_latin_sentence << word
end
end
rejoined_pig_sentence = pig_latin_sentence.join(" ").downcase + "."
p rejoined_pig_sentence.capitalize
end
convert_sentence_pig_latin("Mary had a little lamb.")
Your main problem is that [^aeiou] matches every character outside that range, including spaces, commas, quotation marks, etc.
If I were you, I'd use a positive match for consonants, ie. [b-df-hj-np-tv-z] I would also put that regex in a variable, so you're not having to repeat it three times.
Also, in case you're interested, there's a way to make your convert_sentence_pig_latin method a single gsub and it will do the whole sentence in one pass.
Update
...because you asked...
sentence.gsub( /\b([b-df-hj-np-tv-z])(\w+)/i ) { "#{$2}#{$1}ay" }
# iterate over and replace regexp matches using gsub
def convert_sentence_pig_latin2(sentence)
r = /^[^aeiou]+/i
sentence.gsub(/"([^"]*)"/m) {|x| x.gsub(/\w+/) {|y| y =~ r ? "#{y.partition(r)[2]}#{y.partition(r)[1]}ay" : y}}
end
puts convert_sentence_pig_latin2('"I said this", I said.')
# define instance method: String#to_pl
class String
R = Regexp.new '^[^aeiou]+', true # => /^[^aeiou]+/i
def to_pl
self.gsub(/"([^"]*)"/m) {|x| x.gsub(/\w+/) {|y| y =~ R ? "#{y.partition(R)[2]}#{y.partition(R)[1]}ay" : y}}
end
end
puts '"I said this", I said.'.to_pl
sources:
http://www.ruby-doc.org/core-2.1.0/Regexp.html
http://ruby-doc.org/core-2.0/String.html#method-i-gsub
I have a string and I need to check whether the last character of that string is *, and if it is, I need to remove it.
if stringvariable.include? "*"
newstring = stringvariable.gsub(/[*]/, '')
end
The above does not search if the '*' symbol is the LAST character of the string.
How do i check if the last character is '*'?
Thanks for any suggestion
Use the $ anchor to only match the end of line:
"sample*".gsub(/\*$/, '')
If there's the possibility of there being more than one * on the end of the string (and you want to replace them all) use:
"sample**".gsub(/\*+$/, '')
You can also use chomp (see it on API Dock), which removes the trailing record separator character(s) by default, but can also take an argument, and then it will remove the end of the string only if it matches the specified character(s).
"hello".chomp #=> "hello"
"hello\n".chomp #=> "hello"
"hello\r\n".chomp #=> "hello"
"hello\n\r".chomp #=> "hello\n"
"hello\r".chomp #=> "hello"
"hello \n there".chomp #=> "hello \n there"
"hello".chomp("llo") #=> "he"
"hello*".chomp("*") #=> "hello"
String has an end_with? method
stringvariable.chop! if stringvariable.end_with? '*'
You can do the following which will remove the offending character, if present. Otherwise it will do nothing:
your_string.sub(/\*$/, '')
If you want to remove more than one occurrence of the character, you can do:
your_string.sub(/\*+$/, '')
Of course, if you want to modify the string in-place, use sub! instead of sub
Cheers,
Aaron
You can either use a regex or just splice the string:
if string_variable[-1] == '*'
new_string = string_variable.gsub(/[\*]/, '') # note the escaped *
end
That only works in Ruby 1.9.x...
Otherwise you'll need to use a regex:
if string_variable =~ /\*$/
new_string = string_variable.gsub(/[\*]/, '') # note the escaped *
end
But you don't even need the if:
new_string = string_variable.gsub(/\*$/, '')
unless (place =~ /^\./) == 0
I know the unless is like if not but what about the condtional?
=~ means matches regex
/^\./ is a regular expression:
/.../ are the delimiters for the regex
^ matches the start of the string or of a line (\A matches the start of the string only)
\. matches a literal .
It checks if the string place starts with a period ..
Consider this:
p ('.foo' =~ /^\./) == 0 # => true
p ('foo' =~ /^\./) == 0 # => false
In this case, it wouldn't be necessary to use == 0. place =~ /^\./ would suffice as a condition:
p '.foo' =~ /^\./ # => 0 # 0 evaluates to true in Ruby conditions
p 'foo' =~ /^\./ # => nil
EDIT: /^\./ is a regular expression. The start and end slashes denotes that it is a regular expression, leaving the important bit to ^\.. The first character, ^ marks "start of string/line" and \. is the literal character ., as the dot character is normally considered a special character in regular expressions.
To read more about regular expressions, see Wikipedia or the excellent regular-expressions.info website.