How to generate a random string in Ruby

I'm currently generating an 8-character pseudo-random uppercase string for "A" .. "Z":

value = ""; 8.times{value  << (65 + rand(25)).chr}

but it doesn't look clean, and it can't be passed as an argument since it isn't a single statement. To get a mixed-case string "a" .. "z" plus "A" .. "Z", I changed it to:

value = ""; 8.times{value << ((rand(2)==1?65:97) + rand(25)).chr}

but it looks like trash.

Does anyone have a better method?

(0...8).map { (65 + rand(26)).chr }.join

I spend too much time golfing.

(0...50).map { ('a'..'z').to_a[rand(26)] }.join

And a last one that's even more confusing, but more flexible and wastes fewer cycles:

o = [('a'..'z'), ('A'..'Z')].map(&:to_a).flatten
string = (0...50).map { o[rand(o.length)] }.join

If you want to generate some random text then use the following:

50.times.map { (0...(rand(10))).map { ('a'..'z').to_a[rand(26)] }.join }.join(" ")

this code generates 50 random word string with words length less than 10 characters and then join with space

Why not use SecureRandom?

require 'securerandom'
random_string = SecureRandom.hex

# outputs: 5b5cd0da3121fc53b4bc84d0c8af2e81 (i.e. 32 chars of 0..9, a..f)

SecureRandom also has methods for:

• base64
• random_bytes
• random_number

see: http://ruby-doc.org/stdlib-1.9.2/libdoc/securerandom/rdoc/SecureRandom.html

I use this for generating random URL friendly strings with a guaranteed maximum length:

string_length = 8
rand(36**string_length).to_s(36)

It generates random strings of lowercase a-z and 0-9. It's not very customizable but it's short and clean.

This solution generates a string of easily readable characters for activation codes; I didn't want people confusing 8 with B, 1 with I, 0 with O, L with 1, etc.

# Generates a random string from a set of easily readable characters
def generate_activation_code(size = 6)
  charset = %w{ 2 3 4 6 7 9 A C D E F G H J K M N P Q R T V W X Y Z}
  (0...size).map{ charset.to_a[rand(charset.size)] }.join
end

Others have mentioned something similar, but this uses the URL safe function.

require 'securerandom'
p SecureRandom.urlsafe_base64(5) #=> "UtM7aa8"
p SecureRandom.urlsafe_base64 #=> "UZLdOkzop70Ddx-IJR0ABg"
p SecureRandom.urlsafe_base64(nil, true) #=> "i0XQ-7gglIsHGV2_BNPrdQ=="

The result may contain A-Z, a-z, 0-9, “-” and “_”. “=” is also used if padding is true.

Since Ruby 2.5, it's really easy with SecureRandom.alphanumeric:

len = 8
SecureRandom.alphanumeric(len)
=> "larHSsgL"

It generates random strings containing A-Z, a-z and 0-9 and therefore should be applicable in most use-cases. And they are generated randomly secure, which might be a benefit, too.

---

This is a benchmark to compare it with the solution having the most upvotes:

require 'benchmark'
require 'securerandom'

len = 10
n = 100_000

Benchmark.bm(12) do |x|
  x.report('SecureRandom') { n.times { SecureRandom.alphanumeric(len) } }
  x.report('rand') do
    o = [('a'..'z'), ('A'..'Z'), (0..9)].map(&:to_a).flatten
    n.times { (0...len).map { o[rand(o.length)] }.join }
  end
end

---

                   user     system      total        real
SecureRandom   0.429442   0.002746   0.432188 (  0.432705)
rand           0.306650   0.000716   0.307366 (  0.307745)

So the rand solution only takes about 3/4 of the time of SecureRandom. That might matter if you generate a lot of strings, but if you just create some random string from time to time I'd always go with the more secure implementation since it is also easier to call and more explicit.

[*('A'..'Z')].sample(8).join

Generate a random 8 letter string (e.g. NVAYXHGR)

([*('A'..'Z'),*('0'..'9')]-%w(0 1 I O)).sample(8).join

Generate a random 8 character string (e.g. 3PH4SWF2), excludes 0/1/I/O. Ruby 1.9

I can't remember where I found this, but it seems like the best and the least process intensive to me:

def random_string(length=10)
  chars = 'abcdefghjkmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ0123456789'
  password = ''
  length.times { password << chars[rand(chars.size)] }
  password
end

require 'securerandom'
SecureRandom.urlsafe_base64(9)

If you want a string of specified length, use:

require 'securerandom'
randomstring = SecureRandom.hex(n)

It will generate a random string of length 2n containing 0-9 and a-f

Array.new(n){[*"0".."9"].sample}.join, where n=8 in your case.

Generalized: Array.new(n){[*"A".."Z", *"0".."9"].sample}.join, etc.

From: "https://stackoverflow.com/questions/26076011/rails-generate-pseudo-random-string-a-z-0-9/26076063#26076063";.

Here is one line simple code for random string with length 8:

 random_string = ('0'..'z').to_a.shuffle.first(8).join

You can also use it for random password having length 8:

random_password = ('0'..'z').to_a.shuffle.first(8).join

require 'sha1'
srand
seed = "--#{rand(10000)}--#{Time.now}--"
Digest::SHA1.hexdigest(seed)[0,8]

Ruby 1.9+:

ALPHABET = ('a'..'z').to_a
#=> ["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"]

10.times.map { ALPHABET.sample }.join
#=> "stkbssowre"

# or

10.times.inject('') { |s| s + ALPHABET.sample }
#=> "fdgvacnxhc"

Here is one simple code for random password with length 8:

rand_password=('0'..'z').to_a.shuffle.first(8).join

Be aware: rand is predictable for an attacker and therefore probably insecure. You should definitely use SecureRandom if this is for generating passwords. I use something like this:

length = 10
characters = ('A'..'Z').to_a + ('a'..'z').to_a + ('0'..'9').to_a

password = SecureRandom.random_bytes(length).each_char.map do |char|
  characters[(char.ord % characters.length)]
end.join

Another method I like to use:

 rand(2**256).to_s(36)[0..7]

Add ljust if you are really paranoid about the correct string length:

 rand(2**256).to_s(36).ljust(8,'a')[0..7]

SecureRandom.base64(15).tr('+/=lIO0', 'pqrsxyz')

Something from Devise

I think this is a nice balance of conciseness, clarity and ease of modification.

characters = ('a'..'z').to_a + ('A'..'Z').to_a
# Prior to 1.9, use .choice, not .sample
(0..8).map{characters.sample}.join

Easily modified

For example, including digits:

characters = ('a'..'z').to_a + ('A'..'Z').to_a + (0..9).to_a

Uppercase hexadecimal:

characters = ('A'..'F').to_a + (0..9).to_a

For a truly impressive array of characters:

characters = (32..126).to_a.pack('U*').chars.to_a

Just adding my cents here...

def random_string(length = 8)
  rand(32**length).to_s(32)
end

You can use String#random from the Facets of Ruby Gem facets.

It basically does this:

class String
  def self.random(len=32, character_set = ["A".."Z", "a".."z", "0".."9"])
    characters = character_set.map { |i| i.to_a }.flatten
    characters_len = characters.length
    (0...len).map{ characters[rand(characters_len)] }.join
  end
end

This solution needs external dependency, but seems prettier than another.

1. Install gem faker
2. Faker::Lorem.characters(10) # => "ang9cbhoa8"

I was doing something like this recently to generate an 8 byte random string from 62 characters. The characters were 0-9,a-z,A-Z. I had an array of them as was looping 8 times and picking a random value out of the array. This was inside a Rails app.

str = ''
8.times {|i| str << ARRAY_OF_POSSIBLE_VALUES[rand(SIZE_OF_ARRAY_OF_POSSIBLE_VALUES)] }

The weird thing is that I got good number of duplicates. Now randomly this should pretty much never happen. 62^8 is huge, but out of 1200 or so codes in the db i had a good number of duplicates. I noticed them happening on hour boundaries of each other. In other words I might see a duple at 12:12:23 and 2:12:22 or something like that...not sure if time is the issue or not.

This code was in the before create of an ActiveRecord object. Before the record was created this code would run and generate the 'unique' code. Entries in the DB were always produced reliably, but the code (str in the above line) was being duplicated much too often.

I created a script to run through 100000 iterations of this above line with small delay so it would take 3-4 hours hoping to see some kind of repeat pattern on an hourly basis, but saw nothing. I have no idea why this was happening in my Rails app.

Given:

chars = [*('a'..'z'),*('0'..'9')].flatten

Single expression, can be passed as an argument, allows duplicate characters:

Array.new(len) { chars.sample }.join

My favorite is (:A..:Z).to_a.shuffle[0,8].join. Note that shuffle requires Ruby > 1.9.

I like Radar's answer best, so far, I think. I'd tweak a bit like this:

CHARS = ('a'..'z').to_a + ('A'..'Z').to_a
def rand_string(length=8)
  s=''
  length.times{ s << CHARS[rand(CHARS.length)] }
  s
end

''.tap {|v| 4.times { v << ('a'..'z').to_a.sample} }

2 solutions for a random string consisting of 3 ranges:

(('a'..'z').to_a + ('A'..'Z').to_a + (0..9).to_a).sample(8).join

([*(48..57),*(65..90),*(97..122)]).sample(8).collect(&:chr)*""

One Character from each Range.

And if you need at least one character from each range, such as creating a random password that has one uppercase, one lowercase letter and one digit, you can do something like this:

( ('a'..'z').to_a.sample(8) + ('A'..'Z').to_a.sample(8) + (0..9).to_a.sample(8) ).shuffle.join 
#=> "Kc5zOGtM0H796QgPp8u2Sxo1"

My 2 cents:

  def token(length=16)
  	chars = [*('A'..'Z'), *('a'..'z'), *(0..9)]
  	(0..length).map {chars.sample}.join
  end

I just write a small gem random_token to generate random tokens for most use case, enjoy ~

https://github.com/sibevin/random_token

We've been using this on our code:

class String
  
  def self.random(length=10)
    ('a'..'z').sort_by {rand}[0,length].join
  end
  
end

The maximum length supported is 25 (we're only using it with the default anyway, so hasn't been a problem).

Someone mentioned that 'a'..'z' is suboptimal if you want to completely avoid generating offensive words. One of the ideas we had was removing vowels, but you still end up with WTFBBQ etc.

With this method you can pass in an abitrary length. It's set as a default as 6.

def generate_random_string(length=6)
  string = ""
  chars = ("A".."Z").to_a
  length.times do
    string << chars[rand(chars.length-1)]
  end
  string
end

try this out

def rand_name(len=9)
  ary = [('0'..'9').to_a, ('a'..'z').to_a, ('A'..'Z').to_a]
  name = ''

  len.times do
    name << ary.choice.choice
  end
  name
end

I love the answers of the thread, have been very helpful, indeed!, but if I may say, none of them satisfies my ayes, maybe is the rand() method. it's just doesn't seems right to me, since we've got the Array#choice method for that matter.

Here is another method:

• It uses the secure random number generator instead of rand()
• Can be used in URLs and file names
• Contains uppercase, lowercase characters and numbers
• Has an option not to include ambiguous characters I0l01

Needs require "securerandom"

def secure_random_string(length = 32, non_ambiguous = false)
  characters = ('a'..'z').to_a + ('A'..'Z').to_a + ('0'..'9').to_a
  
  %w{I O l 0 1}.each{ |ambiguous_character| 
    characters.delete ambiguous_character 
  } if non_ambiguous
  
  (0...length).map{
    characters[ActiveSupport::SecureRandom.random_number(characters.size)]
  }.join
end

If you are on a UNIX and you still must use Ruby 1.8 (no SecureRandom) without Rails, you can also use this:

random_string = `openssl rand -base64 24`

Note this spawns new shell, this is very slow and it can only be recommended for scripts.

Another trick that works with Ruby 1.8+ and is fast is:

>> require "openssl"
>> OpenSSL::Random.random_bytes(20).unpack('H*').join
=> "2f3ff53dd712ba2303a573d9f9a8c1dbc1942d28"

It get's you random hex string. Similar way you should be able to generate base64 string ('M*').

This is based on a few other answers, but it adds a bit more complexity:

def random_password
  specials = ((32..47).to_a + (58..64).to_a + (91..96).to_a + (123..126).to_a).pack('U*').chars.to_a
  numbers  = (0..9).to_a
  alpha    = ('a'..'z').to_a + ('A'..'Z').to_a
  %w{i I l L 1 O o 0}.each{ |ambiguous_character| 
    alpha.delete ambiguous_character 
  }
  characters = (alpha + specials + numbers)
  password = Random.new.rand(8..18).times.map{characters.sample}
  password << specials.sample unless password.join =~ Regexp.new(Regexp.escape(specials.join))
  password << numbers.sample  unless password.join =~ Regexp.new(Regexp.escape(numbers.join))
  password.shuffle.join
end

Essentially it ensures a password that is 8 - 20 characters in length, and which contains at least one number and one special character.

10.times do 
  alphabet = ('a'..'z').to_a
  string += alpha[rand(alpha.length)]
end

For devise secure_validatable you can use this

(0...8).map { ([65, 97].sample + rand(26)).chr }.push(rand(99)).join

Here is a improve of @Travis R answer:

 def random_string(length=5)
    chars = 'abdefghjkmnpqrstuvwxyzABDEFGHJKLMNPQRSTUVWXYZ'
    numbers = '0123456789'
    random_s = ''
    (length/2).times { random_s << numbers[rand(numbers.size)] }
    (length - random_s.length).times { random_s << chars[rand(chars.size)] }
    random_s.split('').shuffle.join
  end

At @Travis R answer chars and numbers were together, so sometimes random_string could return only numbers or only characters. With this improve at least half of random_string will be characters and the rest are numbers. Just in case if you need a random string with numbers and characters

To make your first into one statement:

(0...8).collect { |n| value  << (65 + rand(25)).chr }.join()

`pwgen 8 1`.chomp

Create an empty string or a pre-fix if require:

myStr = "OID-"

Use this code to populate the string with random numbers:

begin; n = ((rand * 43) + 47).ceil; myStr << n.chr if !(58..64).include?(n); end while(myStr.length < 12)

Notes:

(rand * 43) + 47).ceil

It will generate random numbers from 48-91 (0,1,2..Y,Z)

!(58..64).include?(n)

It is used to skip special characters (as I am not interested to include them)

while(myStr.length < 12)

It will generate total 12 characters long string including prefix.

Sample Output:

"OID-XZ2J32XM"

Here's a solution that is flexible and allows dups:

class String
  # generate a random string of length n using current string as the source of characters
  def random(n)
    return "" if n <= 0
    (chars * (n / length + 1)).shuffle[0..n-1].join  
  end
end

Example:

"ATCG".random(8) => "CGTGAAGA"

You can also allow a certain character to appear more frequently:

"AAAAATCG".random(10) => "CTGAAAAAGC"

Explanation: The method above takes the chars of a given string and generates a big enough array. It then shuffles it, takes the first n items, then joins them.

Array.new(8).inject(""){|r|r<<('0'..'z').to_a.shuffle[0]}  # 57
(1..8).inject(""){|r|r<<('0'..'z').to_a.shuffle[0]}        # 51
e="";8.times{e<<('0'..'z').to_a.shuffle[0]};e              # 45
(1..8).map{('0'..'z').to_a.shuffle[0]}.join                # 43
(1..8).map{rand(49..122).chr}.join                         # 34

a='';8.times{a<<[*'a'..'z'].sample};p a

or

8.times.collect{[*'a'..'z'].sample}.join

Use 'SafeRandom' Gem GithubLink

It will provide the easiest way to generate random values for Rails2, Rails 3, Rails 4, Rails 5 compatible.

The following worked well for me

def generate_random_password(min_length, max_length)
    length = SecureRandom.random_number(max_length - min_length) + min_length
    character_sets = [ 
      ('a'..'z').to_a,
      ('A'..'Z').to_a,
      ('0'..'9').to_a,
      "~!@^&*()_-+=[]|:;<,>.?".split('')
    ]   
    retval = []
    #   
    # Add one character from each set
    #   
    character_sets.each do |character_set|
      character = character_set[SecureRandom.random_number(character_set.count)]
      retval.push character
    end 
    #   
    # Fill the rest of the password with a random character from a random set
    #   
    i = character_sets.count - 1 
    while i < length
      character_set = character_sets[SecureRandom.random_number(character_sets.count)]
      character = character_set[SecureRandom.random_number(character_set.count)]
      retval.push character
      i += 1
    end
    retval.shuffle.join
  end

This is almost as ugly but perhaps as step in right direction?

 (1..8).map{|i| ('a'..'z').to_a[rand(26)]}.join

In ruby 1.9 one can use Array's choice method which returns random element from array

I don't know ruby, so I can't give you the exact syntax, but I would set a constant string with the list of acceptable characters, then use the substring operator to pick a random character out of it.

The advantage here is that if the string is supposed to be user-enterable, then you can exclude easily confused characters like l and 1 and i, 0 and O, 5 and S, etc.