2013-05-11 61 views

回答

3

您可以使用正则表达式^[a-z](?<=^|[\r\n])[a-z]在每一行,选择第一个非资本性质

+0

也许'(<=^| [\ r \ n] {1,2})[az]'? (我不确定) – HamZa 2013-05-11 13:19:18

+2

@HamZaDzCyber​​DeV在** lookbehind **中,像'*','+','{n}这样的量词是不被Ruby支持的 – Anirudha 2013-05-11 13:25:10

+0

啊谢谢指出:) – HamZa 2013-05-11 13:34:07

3

让我们把它棘手通过引入前导空格:

str = "New line\r\n\r\n also a new line\r\n\r\n\r\n\r\n\r\nanother new line" 
str.gsub(/^\s*[a-z]/i){|x| x.upcase} 
#=> "New line\r\n\r\n Also a new line\r\n\r\n\r\n\r\n\r\nAnother new line" 
2
"New line\r\n\r\n also a new line\r\n\r\n\r\n\r\n\r\nanother new line" 
.gsub(/^\W*./, &:upcase) 
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