Question

尝试使用gsub来转换字符串的前进顺序同意字符以及字符'qu'的组合。如何通过正则表达式((two specific character in a specific order) OR (any character in a set of characters)) any number of times

来说明

我尝试过以下无效：

"stupid".gsub( /([(qu)]*[^aeiou]*)(.*)/, '\2\1') + 'ay' #=> "upidstay" correct!
"uce".gsub( /([(qu)]*[^aeiou]*)(.*)/, '\2\1') + 'ay' #=> "eucay" NOT correct :(

"stupid".gsub( /((qu)*[^aeiou]*)(.*)/, '\2\1') + 'ay' #=> "stay" 

"uce".gsub( /([(qu)|[^aeiou]]*)(.*)/, '\2\1') + 'ay' #=> "eucay" # why does just a "u" match?

我认为它归结为我不知道如何在[]中选择多个特定的连续字符，我认为它是[(qu)]，但为什么只有u得到匹配？

Answer 1

这是你在找什么？

cons = "(?:qu|[^aeiou])"
"uce".gsub(/^(#{cons}+)(.*)/,'\2\1') + 'ay' # => uceay
"stupid".gsub(/^(#{cons}+)(.*)/,'\2\1') + 'ay' # => upidstay
"quduck".gsub(/^(#{cons}+)(.*)/,'\2\1') + 'ay' # => uckquday

Answer 2

字符类[]指定要匹配的字符，以便[(qu)]匹配(，q，u和)

Piglatin RegEx：/((?:(?:qu))*[^aeiou]*)(\w*)/

替换为：'\2\1'并附加'ay'

代码：

"quick".gsub( /((?:(?:qu))*[^aeiou]*)(\w*)/, '\2\1') + 'ay'

参见示例here。

特定的连续字符是红宝石正则表达式的零或更多倍

2 个答案: