ruby - 从 Ruby 字符串中删除除字母和数字以外的所有字符

Question

我在表单中有一个字符串输入字段。我在 params hash 中得到了那个值。我应该如何从该字符串中删除除字母和数字之外的所有字符。

Answer 1

只是为了提醒人们好的'ol tr:

asdf.tr('^A-Za-z0-9', '')

找到字符范围的补码并将字符转换为 ''。

我很好奇使用 \W 字符类是否比范围和 gsub 与 tr 更快:

require 'benchmark'

asdf = [('A'..'z').to_a, ('0'..'9').to_a].join

puts asdf
puts asdf.tr(   '^A-Za-z0-9',    '' )
puts asdf.gsub( /[\W_]+/,        '' )
puts asdf.gsub( /\W+/,           '' )
puts asdf.gsub( /\W/,            '' )
puts asdf.gsub( /[^A-Za-z0-9]+/, '' )
puts asdf.scan(/[a-z\d]/i).join

n = 100_000
Benchmark.bm(7) do |x|
  x.report("tr:")    { n.times do; asdf.tr('^A-Za-z0-9', '');      end }
  x.report("gsub1:") { n.times do; asdf.gsub(/[\W_]+/, '');        end }
  x.report("gsub2:") { n.times do; asdf.gsub(/\W+/, '');           end }
  x.report("gsub3:") { n.times do; asdf.gsub(/\W/, '');            end }
  x.report("gsub4:") { n.times do; asdf.gsub(/[^A-Za-z0-9]+/, ''); end }
  x.report("scan:")  { n.times do; asdf.scan(/[a-z\d]/i).join;     end }
end

>> ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789
>>              user     system      total        real
>> tr:      0.560000   0.000000   0.560000 (  0.557883)
>> gsub1:   0.510000   0.000000   0.510000 (  0.513244)
>> gsub2:   0.820000   0.000000   0.820000 (  0.823816)
>> gsub3:   0.960000   0.000000   0.960000 (  0.955848)
>> gsub4:   0.900000   0.000000   0.900000 (  0.902166)
>> scan:    5.630000   0.010000   5.640000 (  5.630990)

您可以看到一些模式没有捕捉到 '_'，它是 \w 的一部分，因此不符合 OP 的要求。