从Ruby字符串中删除除字母和数字之外的所有字符

Question

我在表单中有一个字符串输入字段。我在params散列中获得了这个值。我应该如何从字符串中删除除字母和数字之外的所有字符。

Answer 1

只是提醒良好的人“醇tr：

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

这是找到字符范围的补充和转换的字符'。

我很好奇，一个\W字符类是否使用快于范围和gsub与tr：

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[\]^_`abcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
>> ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
>>    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) 

你可以看到一对夫妇的模式都没有捕捉“_”，这是部分\w，结果不符合OP的要求。

Answer 2

=> '^/how/now#(Brown) Cow'.gsub /\W/, '' # or /[\W_]/ 
=> "hownowBrownCow" 

...基于评论更新中...

Answer 3

没有一个正则表达式：

garbage = 'ab_c<>?AB [email protected]#123' 
puts garbage.delete("^a-zA-Z0-9") #=> abcABC123 

其中 '^' 后，它否定了一切。