Question

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。

Answer 1

或使用正则表达式：

str = "Hey what's up @dude, @how's it going?"
str.gsub!(/.*?(?=@how)/im, "") #=> "@how's it going?"

您可以阅读here

的外观

Answer 2

使用String#slice

s = "Hey what's up @dude, @how's it going?"
s.slice(s.index("@how")..-1)
# => "@how's it going?"

Answer 3

有几种方法可以做到这一点。以下是我将使用的：

如果您想保留原始字符串：

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?"

如果您想改变原始字符串：

str = "Hey what's up @dude, @how's it going?"
str[/\A.+?(?=@how's)/mi] = ''
p str
#=> "@how's it going?"

...或...

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锚定在字符串的开头，并m标记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?"

Answer 4

String#slice和String#index工作正常，但如果指针不在大海捞针中，则会出现 ArgumentError：范围错误值。

在这种情况下，使用String#partition或String#rpartition可能效果更好：

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?"]

Answer 5

只获取您感兴趣的部分的简便方法。

>> 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?"

如果您确实需要更改字符串对象，则可以始终执行s=s[...]。

Answer 6

>> "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?"

或使用scan()

>> "Hey what's up @dude, @HoW's it going?".scan(/@how's.*/i)[0]
=> "@HoW's it going?"

Answer 7

您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] : ""

在Ruby中，如何匹配字符串中的所有字符，直到匹配子字符串？

7 个答案: