[ruby] How do I find if a string starts with another string in Ruby?

What the best way to find if a string starts with another in Ruby (without rails)?

I like

if ('string'[/^str/]) ...

The method mentioned by steenslag is terse, and given the scope of the question it should be considered the correct answer. However it is also worth knowing that this can be achieved with a regular expression, which if you aren't already familiar with in Ruby, is an important skill to learn.

Have a play with Rubular: http://rubular.com/

But in this case, the following ruby statement will return true if the string on the left starts with 'abc'. The \A in the regex literal on the right means 'the beginning of the string'. Have a play with rubular - it will become clear how things work.

'abcdefg' =~  /\Aabc/ 

puts 'abcdefg'.start_with?('abc')  #=> true

[edit] This is something I didn't know before this question: start_with takes multiple arguments.

'abcdefg'.start_with?( 'xyz', 'opq', 'ab')

Since there are several methods presented here, I wanted to figure out which one was fastest. Using Ruby 1.9.3p362:

irb(main):001:0> require 'benchmark'
=> true
irb(main):002:0> Benchmark.realtime { 1.upto(10000000) { "foobar"[/\Afoo/] }}
=> 12.477248
irb(main):003:0> Benchmark.realtime { 1.upto(10000000) { "foobar" =~ /\Afoo/ }}
=> 9.593959
irb(main):004:0> Benchmark.realtime { 1.upto(10000000) { "foobar"["foo"] }}
=> 9.086909
irb(main):005:0> Benchmark.realtime { 1.upto(10000000) { "foobar".start_with?("foo") }}
=> 6.973697

So it looks like start_with? ist the fastest of the bunch.

Updated results with Ruby 2.2.2p95 and a newer machine:

require 'benchmark'
Benchmark.bm do |x|
  x.report('regex[]')    { 10000000.times { "foobar"[/\Afoo/] }}
  x.report('regex')      { 10000000.times { "foobar" =~ /\Afoo/ }}
  x.report('[]')         { 10000000.times { "foobar"["foo"] }}
  x.report('start_with') { 10000000.times { "foobar".start_with?("foo") }}
end

            user       system     total       real
regex[]     4.020000   0.000000   4.020000 (  4.024469)
regex       3.160000   0.000000   3.160000 (  3.159543)
[]          2.930000   0.000000   2.930000 (  2.931889)
start_with  2.010000   0.000000   2.010000 (  2.008162)