Regex to check alphanumeric string in ruby - ruby

Question

I am trying to validate strings in ruby.
Any string which contains spaces,under scores or any special char should fail validation.
The valid string should contain only chars a-zA-Z0-9
My code looks like.
def validate(string)
regex ="/[^a-zA-Z0-9]$/
if(string =~ regex)
return "true"
else
return "false"
end
I am getting error:
TypeError: type mismatch: String given.
Can anyone please let me know what is the correct way of doing this?

Answer 1

If you are validating a line:
def validate(string)
!string.match(/\A[a-zA-Z0-9]*\z/).nil?
end
No need for return on each.

Answer 2

You can just check if a special character is present in the string.
def validate str
chars = ('a'..'z').to_a + ('A'..'Z').to_a + (0..9).to_a
str.chars.detect {|ch| !chars.include?(ch)}.nil?
end
Result:
irb(main):005:0> validate "hello"
=> true
irb(main):006:0> validate "_90 "
=> false

Answer 3

def alpha_numeric?(char)
if (char =~ /[[:alpha:]]/ || char =~ /[[:digit:]]/)
true
else
false
end
end
OR
def alpha_numeric?(char)
if (char =~ /[[:alnum:]]/)
true
else
false
end
end
We are using regular expressions that match letters & digits:
The above [[:alpha:]] ,[[:digit:]] and [[:alnum:]] are POSIX bracket expressions, and they have the advantage of matching Unicode characters in their category. Hope this helps.
checkout the link below for more options:
Ruby: How to find out if a character is a letter or a digit?

Answer 4

No regex:
def validate(str)
str.count("^a-zA-Z0-9").zero? # ^ means "not"
end

Answer 5

Great answers above but just FYI, your error message is because you started your regex with a double quote ". You'll notice you have an odd number (5) of double quotes in your method.
Additionally, it's likely you want to return true and false as values rather than as quoted strings.

Answer 6

Similar to the very efficient regex-ish approach mentioned already by #steenslag and nearly just as fast:
str.tr("a-zA-Z0-9", "").length.zero?
OR
str.tr("a-zA-Z0-9", "") == 0
One benefit of using tr though is that you could also optionally analyze the results using the same basic formula:
str = "ABCxyz*123$"
rejected_chars = str.tr("a-zA-Z0-9", "")
#=> *$
is_valid = rejected_chars.length.zero?
#=> false

Answer 7

Similar to #rohit89:
VALID_CHARS = [*?a..?z, *?A..?Z, *'0'..'9']
#=> ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m",
# "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z",
# "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
# "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z",
# "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
def all_valid_chars?(str)
a = str.chars
a == a & VALID_CHARS
end
all_valid_chars?('a9Z3') #=> true
all_valid_chars?('a9 Z3') #=> false

Answer 8

Use .match? in Ruby 2.4+.
Ruby 2.4 introduced a convenient boolean-returning .match? method.
In your case, I would do something like this:
# Checks for any characters other than letters and numbers.
# Returns true if there are none. Returns false if there are one or more.
#
def valid?( string )
!string.match?( /[^a-zA-Z0-9]/ ) # NOTE: ^ inside [] set turns it into a negated set.
end