'Regex to check alphanumeric string in ruby

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?

rubyregexruby-on-rails-3


Solution 1:[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.

Solution 2:[2]

def alpha_numeric?(char)  
 
   if((char =~ /[[:alpha:]]) || (char =~ [[:digits:]]))
      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?

Solution 3:[3]

No regex:

def validate(str)
  str.count("^a-zA-Z0-9").zero?  # ^ means "not"
end

Solution 4:[4]

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.

Solution 5:[5]

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

Solution 6:[6]

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

Solution 7:[7]

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

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2
Solution 3 steenslag
Solution 4 Aaron Washburn
Solution 5 Cary Swoveland
Solution 6 Joshua Pinter
Solution 7

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