How can I generate an n-character pseudo random string containing only A-Z, 0-9 like SecureRandom.base64 without "+", "/", and "="? For example:
(0..n).map {(('1'..'9').to_a + ('A'..'Z').to_a)[rand(36)]}.join
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5 Answers
Array.new(n){[*"A".."Z", *"0".."9"].sample}.join
3 Comments
gr8scott06
curiosity, any reason to do
.sample over [rand(36)] ?
Cary Swoveland
I can think of a couple:
sample is more elegant, and it needn't be changed if the array being sampled is changed in size.
sawa
Curiosity, any reason to do
[rand(36)] over sample?An elegant way to do it in Rails 5 (I don't test it in another Rails versions):
SecureRandom.urlsafe_base64(n)
where n is the number of digits that you want.
ps: SecureRandom uses a array to mount your alphanumeric string, so keep in mind that n should be the amount of digits that you want + 1.
ex: if you want a 8 digit alphanumeric:
SecureRandom.urlsafe_base64(9)
Comments
You can do simply like below:
[*'A'..'Z', *0..9].sample(10).join
Change the number 10 to any number to change the length of string
1 Comment
dawg
You should do
[*'A'..'Z', *"0".."9"].sample(10).join otherwise you will have a mixture between strings and ints...Even brute force is pretty easy:
n = 20
c = [*?A..?Z + *?0..?9]
size = c.size
n.times.map { c[rand(size)] }.join
#=> "IE210UOTDSJDKM67XCG1"
or, without replacement:
c.sample(n).join
#=> "GN5ZC0HFDCO2G5M47VYW"
should that be desired. (I originally had c = [*(?A..?Z)] + [*(?0..?9)], but saw from @sawa's answer that that could be simplified quite a bit.)
Comments
To generate a random string from 10 to 20 characters including just from A to Z and numbers, both always:
require 'string_pattern'
puts "10-20:/XN/".gen
