One\n
Two\n
Three\n
Four\n
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remove_lines(2)将删除前两行,保留字符串:
Three\n
Four\n
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Dig*_*oss 43
>> s = "One\nTwo\nThree\nFour\n"
=> "One\nTwo\nThree\nFour\n"
>> s.to_a[2..-1].join
=> "Three\nFour\n"
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s = "One\nTwo\nThree\nFour"
lines = s.lines
> ["One\n", "Two\n", "Three\n", "Four"]
remaining_lines = lines[2..-1]
> ["Three\n", "Four"]
remaining_lines.join
> "Three\nFour"
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String#lines 将字符串转换为行数组(保留每个字符串末尾的换行符)[2..-1] 指定要返回的行范围,在本例中是第三行到最后一行Array#join 将行重新连接在一起,没有任何空格(但由于行仍然包含换行符,我们不需要分隔符)在一行中:
s.lines[2..-1].join
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class String
def remove_lines(i)
split("\n")[i..-1].join("\n")
end
end
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打电话"One\nTwo\nThree\nFour\n".remove_lines(2)会导致"Three\nFour".如果需要尾随"\n",则需要相应地扩展此方法.
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