How do I generate a random 10 digit number in ruby?

Additionally, how can I format it as a string padded with zeros?

To generate the number call rand with the result of the expression "10 to the power of 10"

rand(10 ** 10)

To pad the number with zeros you can use the string format operator

'%010d' % rand(10 ** 10)

or the rjust method of string

rand(10 ** 10).to_s.rjust(10,'0')  

I would like to contribute probably a simplest solution I know, which is a quite a good trick.

rand.to_s[2..11] 
 => "5950281724"

This is a fast way to generate a 10-sized string of digits:

10.times.map{rand(10)}.join # => "3401487670"

The most straightforward answer would probably be

rand(1e9...1e10).to_i

The to_i part is needed because 1e9 and 1e10 are actually floats:

irb(main)> 1e9.class
=> Float

DON'T USE rand.to_s[2..11].to_i

Why? Because here's what you can get:

rand.to_s[2..9] #=> "04890612"

and then:

"04890612".to_i #=> 4890612

Note that:

4890612.to_s.length #=> 7

Which is not what you've expected!

To check that error in your own code, instead of .to_i you may wrap it like this:

Integer(rand.to_s[2..9])

and very soon it will turn out that:

ArgumentError: invalid value for Integer(): "02939053"

So it's always better to stick to .center, but keep in mind that:

rand(9) 

sometimes may give you 0.

To prevent that:

rand(1..9)

which will always return something withing 1..9 range.

I'm glad that I had good tests and I hope you will avoid breaking your system.

Random number generation

Use Kernel#rand method:

rand(1_000_000_000..9_999_999_999) # => random 10-digits number

Random string generation

Use times + map + join combination:

10.times.map { rand(0..9) }.join # => random 10-digit string (may start with 0!)

Number to string conversion with padding

Use String#% method:

"%010d" % 123348 # => "0000123348"

Password generation

Use KeePass password generator library, it supports different patterns for generating random password:

KeePass::Password.generate("d{10}") # => random 10-digit string (may start with 0!)

A documentation for KeePass patterns can be found here.

Just because it wasn't mentioned, the Kernel#sprintf method (or it's alias Kernel#format in the Powerpack Library) is generally preferred over the String#% method, as mentioned in the Ruby Community Style Guide.

Of course this is highly debatable, but to provide insight:

The syntax of @quackingduck's answer would be

# considered bad
'%010d' % rand(10**10)

# considered good
sprintf('%010d', rand(10**10))

The nature of this preference is primarily due to the cryptic nature of %. It's not very semantic by itself and without any additional context it can be confused with the % modulo operator.

Examples from the Style Guide:

# bad
'%d %d' % [20, 10]
# => '20 10'

# good
sprintf('%d %d', 20, 10)
# => '20 10'

# good
sprintf('%{first} %{second}', first: 20, second: 10)
# => '20 10'

format('%d %d', 20, 10)
# => '20 10'

# good
format('%{first} %{second}', first: 20, second: 10)
# => '20 10'

To make justice for String#%, I personally really like using operator-like syntaxes instead of commands, the same way you would do your_array << 'foo' over your_array.push('123').

This just illustrates a tendency in the community, what's "best" is up to you.

More info in this blogpost.

I ended up with using Ruby kernel srand

srand.to_s.last(10)

Docs here: Kernel#srand

Here is an expression that will use one fewer method call than quackingduck's example.

'%011d' % rand(1e10)

One caveat, 1e10 is a Float, and Kernel#rand ends up calling to_i on it, so for some higher values you might have some inconsistencies. To be more precise with a literal, you could also do:

'%011d' % rand(10_000_000_000) # Note that underscores are ignored in integer literals

I just want to modify first answer. rand (10**10) may generate 9 digit random no if 0 is in first place. For ensuring 10 exact digit just modify

code = rand(10**10)
while code.to_s.length != 10
code = rand(11**11)

end

Try using the SecureRandom ruby library.

It generates random numbers but the length is not specific.

Go through this link for more information: http://ruby-doc.org/stdlib-2.1.2/libdoc/securerandom/rdoc/SecureRandom.html

Simplest way to generate n digit random number -

Random.new.rand((10**(n - 1))..(10**n))

generate 10 digit number number -

Random.new.rand((10**(10 - 1))..(10**10))

('%010d' % rand(0..9999999999)).to_s

or

"#{'%010d' % rand(0..9999999999)}"

This technique works for any "alphabet"

(1..10).map{"0123456789".chars.to_a.sample}.join
=> "6383411680"

Just use straightforward below.

rand(10 ** 9...10 ** 10)

Just test it on IRB with below.

(1..1000).each { puts rand(10 ** 9...10 ** 10) }

To generate a random, 10-digit string:

# This generates a 10-digit string, where the
# minimum possible value is "0000000000", and the
# maximum possible value is "9999999999"
SecureRandom.random_number(10**10).to_s.rjust(10, '0')

---

Here's more detail of what's happening, shown by breaking the single line into multiple lines with explaining variables:

  # Calculate the upper bound for the random number generator
  # upper_bound = 10,000,000,000
  upper_bound = 10**10

  # n will be an integer with a minimum possible value of 0,
  # and a maximum possible value of 9,999,999,999
  n = SecureRandom.random_number(upper_bound)

  # Convert the integer n to a string
  # unpadded_str will be "0" if n == 0
  # unpadded_str will be "9999999999" if n == 9_999_999_999
  unpadded_str = n.to_s

  # Pad the string with leading zeroes if it is less than
  # 10 digits long.
  # "0" would be padded to "0000000000"
  # "123" would be padded to "0000000123"
  # "9999999999" would not be padded, and remains unchanged as "9999999999"
  padded_str = unpadded_str.rjust(10, '0')

rand(9999999999).to_s.center(10, rand(9).to_s).to_i

is faster than

rand.to_s[2..11].to_i

You can use:

puts Benchmark.measure{(1..1000000).map{rand(9999999999).to_s.center(10, rand(9).to_s).to_i}}

and

puts Benchmark.measure{(1..1000000).map{rand.to_s[2..11].to_i}}

in Rails console to confirm that.

An alternative answer, using the regexp-examples ruby gem:

require 'regexp-examples'

/\d{10}/.random_example # => "0826423747"

There's no need to "pad with zeros" with this approach, since you are immediately generating a String.

This will work even on ruby 1.8.7:

> rand(9999999999).to_s.center(10, rand(9).to_s).to_i

A better approach is use Array.new() instead of .times.map. Rubocop recommends it.

Example:

string_size = 9
Array.new(string_size) do
   rand(10).to_s
end

Rubucop, TimesMap:

https://www.rubydoc.info/gems/rubocop/RuboCop/Cop/Performance/TimesMap

In my case number must be unique in my models, so I added checking block.

  module StringUtil
    refine String.singleton_class do
      def generate_random_digits(size:)
        proc = lambda{ rand.to_s[2...(2 + size)] }
        if block_given?
          loop do
            generated = proc.call
            break generated if yield(generated) # check generated num meets condition
          end
        else
          proc.call
        end
      end
    end
  end
  using StringUtil
  String.generate_random_digits(3) => "763"
  String.generate_random_digits(3) do |num|
    User.find_by(code: num).nil?
  end => "689"(This is unique in Users code)

I did something like this

x = 10  #Number of digit
(rand(10 ** x) + 10**x).to_s[0..x-1]

Random 10 numbers:

require 'string_pattern'
puts "10:N".gen