我正在使用 Ruby on Rails 3.2.9，我想验证名称的第一个字符（String）是否不是数字（Integer）。我正在尝试使用以下代码：
class User < ActiveRecord::Base
validates_each :name do |record, attr, value|
record.errors.add(attr, 'cannot begin with a number') if ... # the first char is a number
end
end
我怎样才能做到这一点？
【问题讨论】：
标签： ruby-on-rails ruby regex ruby-on-rails-3 validation
record.errors.add(attr, 'cannot begin with a number') if value =~ /^[0-9].*/
将匹配第一个字符为数字的任何字符串
【讨论】：
/\A\d/
[0-9] 和\d 在技术上是相同的，但\d 更简单，更容易让我们一目了然地解析和理解，因此更不容易出错。添加.* 只是表明您不了解模式的工作原理； .* 没有为这个任务完成任何任务，但是为引擎添加了额外的指令。
查看答案，有多种不同的测试方法。有些让我想知道他们是否会神奇地更快，所以，像往常一样，我做了一个基准测试：
require 'benchmark'
puts "Ruby = #{ RUBY_VERSION }"
str = 'foobar'
puts 'correct result should be false...'
puts !!( str =~ /^\d/ )
puts !!( str =~ /\A\d/ )
puts !!( str =~ /^[0-9].*/ )
puts !!( str.split('').first.to_i.is_a?(Fixnum) )
puts !!( (48..57).include?(str[0]) )
puts !!( ('0'..'9') === str[0] )
puts !!( str[/^\d/] )
puts !!( str[/\A\d/] )
puts !!( str[/\A[0-9]/] )
puts !!( str =~ /\A[0-9]/ )
puts
n = 1_000_000
puts "n = 1_000_000"
puts "str = 'foobar'"
Benchmark::bm(17) do |b|
b.report('^\d regex') { n.times { str =~ /^\d/ } }
b.report('\A\d regex') { n.times { str =~ /\A\d/ } }
b.report('^[0-9].* regex') { n.times { str =~ /^[0-9].*/ } }
b.report('start_with?') { n.times { str.start_with?(*('0'..'9')) } }
b.report("split('')") { n.times { str.split('').first.to_i.is_a?(Fixnum) } }
b.report("(48..57).include?") { n.times { (48..57).include?(str[0]) } }
b.report('range') { n.times { ('0'..'9') === str[0] } }
b.report('str[/^\d/]') { n.times { str[/^\d/] } }
b.report('str[/\A\d/]') { n.times { str[/\A\d/] } }
b.report('str[\A[0-9]') { n.times { str[/\A[0-9]/] } }
b.report('\A[0-9] regex') { n.times { str =~ /\A[0-9]/ } }
end
puts
str = 'foobar' * 1000
puts "str = 'foobar' * 1000"
Benchmark::bm(17) do |b|
b.report('^\d regex') { n.times { str =~ /^\d/ } }
b.report('\A\d regex') { n.times { str =~ /\A\d/ } }
b.report('^[0-9].* regex') { n.times { str =~ /^[0-9].*/ } }
b.report('start_with?') { n.times { str.start_with?(*('0'..'9')) } }
b.report("(48..57).include?") { n.times { (48..57).include?(str[0]) } }
b.report('range') { n.times { ('0'..'9') === str[0] } }
b.report('str[/^\d/]') { n.times { str[/^\d/] } }
b.report('str[/\A\d/]') { n.times { str[/\A\d/] } }
b.report('str[\A[0-9]') { n.times { str[/\A[0-9]/] } }
b.report('\A[0-9] regex') { n.times { str =~ /\A[0-9]/ } }
end
测试结果：
Ruby = 1.9.3
correct result should be false...
false
false
false
true
false
false
false
false
false
false
基准测试结果：
n = 1_000_000
str = 'foobar'
user system total real
^\d regex 0.590000 0.000000 0.590000 ( 0.593534)
\A\d regex 0.560000 0.000000 0.560000 ( 0.556304)
^[0-9].* regex 0.580000 0.000000 0.580000 ( 0.577662)
start_with? 4.020000 0.000000 4.020000 ( 4.025604)
split('') 6.850000 0.000000 6.850000 ( 6.872157)
(48..57).include? 17.260000 0.780000 18.040000 ( 18.038887)
range 1.260000 0.000000 1.260000 ( 1.258191)
str[/^\d/] 0.680000 0.000000 0.680000 ( 0.680291)
str[/\A\d/] 0.660000 0.000000 0.660000 ( 0.663305)
str[\A[0-9] 0.670000 0.000000 0.670000 ( 0.670242)
\A[0-9] regex 0.570000 0.000000 0.570000 ( 0.574152)
为了测试\A 是否比^ 快，看看长字符串会有什么影响，我增加了字符串大小。
"split('')" 被拉出，因为它在 60 多秒后没有完成：
str = 'foobar' * 1000
user system total real
^\d regex 15.010000 0.000000 15.010000 ( 15.020488)
\A\d regex 0.540000 0.010000 0.550000 ( 0.539736)
^[0-9].* regex 15.000000 0.000000 15.000000 ( 15.011137)
start_with? 4.010000 0.000000 4.010000 ( 4.010340)
(48..57).include? 17.320000 0.770000 18.090000 ( 18.124795)
range 1.250000 0.000000 1.250000 ( 1.255724)
str[/^\d/] 15.120000 0.010000 15.130000 ( 15.142242)
str[/\A\d/] 0.650000 0.000000 0.650000 ( 0.656198)
str[\A[0-9] 0.650000 0.000000 0.650000 ( 0.652306)
\A[0-9] regex 0.550000 0.000000 0.550000 ( 0.544415)
我用 1.8.7 重新测试：
Ruby = 1.8.7
correct result should be false...
false
false
false
true
false
false
false
false
false
false
n = 1_000_000
str = 'foobar'
user system total real
^\d regex 0.570000 0.000000 0.570000 ( 0.565397)
\A\d regex 0.550000 0.000000 0.550000 ( 0.552270)
^[0-9].* regex 0.570000 0.000000 0.570000 ( 0.574705)
start_with? 38.180000 0.070000 38.250000 ( 39.864171)
split('') 9.750000 0.040000 9.790000 ( 11.025962)
(48..57).include? 0.580000 0.000000 0.580000 ( 0.917499)
range 2.420000 0.020000 2.440000 ( 3.170774)
str[/^\d/] 0.700000 0.000000 0.700000 ( 0.760180)
str[/\A\d/] 0.680000 0.000000 0.680000 ( 0.762636)
str[\A[0-9] 0.660000 0.010000 0.670000 ( 0.795043)
\A[0-9] regex 0.600000 0.000000 0.600000 ( 0.684566)
str = 'foobar' * 1000
user system total real
^\d regex 7.900000 0.040000 7.940000 ( 10.735175)
\A\d regex 0.600000 0.010000 0.610000 ( 0.784001)
^[0-9].* regex 7.850000 0.020000 7.870000 ( 8.251673)
(48..57).include? 0.580000 0.000000 0.580000 ( 0.683730)
range 2.380000 0.020000 2.400000 ( 2.738234)
str[/^\d/] 7.930000 0.010000 7.940000 ( 8.227906)
str[/\A\d/] 0.670000 0.000000 0.670000 ( 0.682169)
str[\A[0-9] 0.680000 0.000000 0.680000 ( 0.697340)
\A[0-9] regex 0.580000 0.000000 0.580000 ( 0.645136)
你们自己讨论。
【讨论】：
range 感到惊讶。
刚刚发现你可以使用 Range：
"1hello".start_with?(*('0'..'9')) #=> true
【讨论】：
"1hello"*100) start_with? 更快
irb(main):004:0> asdf = *('0'..'9') => ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]，然后嘎嘎叫：irb(main):005:0> asdf.class => Array
可以通过数组访问获取字符串的第一个字符并比较ascii值：
1.8.7 :008 > (48..57).include?("5ssdfsdf"[0])
=> true
1.8.7 :009 > (48..57).include?("ssdfsdf"[0])
=> false
1.8.7 :010 > (48..57).include?("0sdfsdf"[0])
=> true
1.8.7 :011 > (48..57).include?("9sdfsdf"[0])
=> true
【讨论】：