blocks|key|877636|text|puts+'abcdefg'.start_with?('abc')++#=>+true|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|877637|在这个问题之前，我不知道这个：start_with接受多个参数。|unstyled|offset|length|style|CODE|877638|'abcdefg'.start_with?(+'xyz',+'opq',+'ab')|877639|entityMap^0|0|F|A|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>puts 'abcdefg'.start_with?('abc') #=&gt; true
</code></pre>

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

<pre><code>'abcdefg'.start_with?( 'xyz', 'opq', 'ab')
</code></pre>

blocks|key|1092374|text|steenslag提到的方法很简洁，考虑到问题的范围，它应该被认为是正确的答案。然而，值得知道的是，这可以通过正则表达式来实现，如果您在Ruby中还不熟悉正则表达式，这是一项需要学习的重要技能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1092375|玩一玩吧：http://rubular.com/|offset|length|1092376|但是在这种情况下，如果左边的字符串以'abc‘开头，下面的ruby语句将返回true。右侧regex文本中的\A表示“字符串的开头”。试一试橡胶--事情是如何工作的将变得清晰起来。|1092377|'abcdefg'+=~++/\Aabc/+|code-block|syntax|javascript|1092378|entityMap|0|LINK|mutability|MUTABLE|url|http://rubular.com/^0|0|5|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@$D|W|E|X|1|Y]]|A|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

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: <a href="http://rubular.com/" rel="nofollow">http://rubular.com/</a>

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.

<pre><code>'abcdefg' =~ /\Aabc/ 
</code></pre>

blocks|key|1092387|text|我喜欢|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1092388|if+('string'[/%5Estr/])+...|code-block|syntax|javascript|1092389|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I like

<pre><code>if ('string'[/^str/]) ...
</code></pre>

blocks|key|1092446|text|由于这里提供了几种方法，所以我想找出哪种方法最快。使用Ruby+1.9.3p362：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1092447|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|code-block|syntax|javascript|1092448|所以看起来start_with?是其中最快的。|offset|length|style|CODE|1092449|使用Ruby+2.2.2p95和更新的机器更新结果：|1092450|require+'benchmark'
Benchmark.bm+do+%7Cx%7C
++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)|1092451|entityMap^0|0|0|5|B|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

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

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

So it looks like <code>start_with?</code> ist the fastest of the bunch.

Updated results with Ruby 2.2.2p95 and a newer machine:

<pre><code>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)
</code></pre>

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

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

在Ruby中，如果一个字符串以另一个字符串开头(没有rails)，最好的方法是什么？

问在Ruby中，如果一个字符串以另一个字符串开头，该如何查找？
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问在Ruby中，如果一个字符串以另一个字符串开头，该如何查找？
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