在Ruby中如何匹配子字符串之前，如何删除字符串中的所有字符？
说我有一个字符串： Hey what's up @dude, @how's it going?
我想在@how's之前删除所有字符。
或者使用正则表达式：
str = "Hey what's up @dude, @how's it going?" str.gsub!(/.*?(?=@how)/im, "") #=> "@how's it going?"
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使用String＃slice
s = "Hey what's up @dude, @how's it going?" s.slice(s.index("@how")..-1) # => "@how's it going?"
有几种方法可以做到这一点。 以下是我将使用的：
如果要保留原始字符串 ：
str = "Hey what's up @dude, @how's it going?" str2 = str[/@how's.+/mi] p str, str2 #=> "Hey what's up @dude, @how's it going?" #=> "@how's it going?"
如果要改变原始字符串 ：
str = "Hey what's up @dude, @how's it going?" str[/\A.+?(?=@how's)/mi] = '' p str #=> "@how's it going?"
…要么…
str = "Hey what's up @dude, @how's it going?" str.sub! /\A.+?(?=@how's)/mi, '' p str #=> "@how's it going?"
您需要\A来锚定字符串的开头，并使用m标志以确保您在多行之间进行匹配。
也许最简单的是改变原作：
str = "Hey what's up @dude, @how's it going?" str.replace str[/@how's.+/mi] p str #=> "@how's it going?"
String#slice和String#index工作正常，但是会因ArgumentError而爆炸：如果针不在大海捞针中，则范围值不好 。
在这种情况下，使用String#partition或String#rpartition可能会更好：
s.partition "@how's" # => ["Hey what's up @dude, ", "@how's", " it going?"] s.partition "not there" # => ["Hey what's up @dude, @how's it going?", "", ""] s.rpartition "not there" # => ["", "", "Hey what's up @dude, @how's it going?"]
只获得您感兴趣的部分的简单方法。
>> s="Hey what's up @dude, @how's it going?" => "Hey what's up @dude, @how's it going?" >> s[/@how.*$/i] => "@how's it going?"
如果你真的需要改变字符串对象，你总是可以做s=s[...] 。
>> "Hey what's up @dude, @how's it going?".partition("@how's")[-2..-1].join => "@how's it going?"
不区分大小写
>> "Hey what's up @dude, @HoW's it going?".partition(/@how's/i)[-2..-1].join => "@HoW's it going?"
或者使用scan()
>> "Hey what's up @dude, @HoW's it going?".scan(/@how's.*/i)[0] => "@HoW's it going?"
你也可以直接在字符串上调用[] （与slice相同）
s = "Hey what's up @dude, @how's it going?" start_index = s.downcase.index("@how") start_index ? s[start_index..-1] : ""