Ruby 将字符串中的所有内容放在特定单词之前？

如何删除特定单词前字符串中的所有内容（或包括第一个空格和后面）

我有这样一个字符串：

12345 Delivered to: Joe Schmoe

我只想

发送给：Joe Schmoe

所以，基本上从第一个空间到后面的任何东西我都不想要

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我正在运行Ruby 1.9.3。

使用正则表达式只选择所需字符串的一部分

"12345 Delivered to: Joe Schmoe"[/Delive.*/]
# => "Delivered to: Joe Schmoe"

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使用正则表达式仅选择所需字符串的一部分

"12345 Delivered to: Joe Schmoe"[/Delive.*/]
# => "Delivered to: Joe Schmoe"

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有很多不同的方法是可能的。以下是几点：

s = '12345 Delivered to: Joe Schmoe'
s.split(' ')[1..-1].join(' ') # split on spaces, take the last parts, join on space
# or
s[s.index(' ')+1..-1] # Find the index of the first space and just take the rest
# or
/.*?\s(.*)/.match(s)[1] # Use a reg ex to pick out the bits after the first space

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有很多不同的方法是可能的。以下是几点：

s = '12345 Delivered to: Joe Schmoe'
s.split(' ')[1..-1].join(' ') # split on spaces, take the last parts, join on space
# or
s[s.index(' ')+1..-1] # Find the index of the first space and just take the rest
# or
/.*?\s(.*)/.match(s)[1] # Use a reg ex to pick out the bits after the first space

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如果交付并不总是第二个词，您可以这样使用：

s_line = "12345 Delivered to: Joe Schmoe"
puts s_line[/\s.*/].strip #=> "Delivered to: Joe Schmoe"

---

如果交付并不总是第二个词，您可以这样使用：

s_line = "12345 Delivered to: Joe Schmoe"
puts s_line[/\s.*/].strip #=> "Delivered to: Joe Schmoe"