这个字符串的Ruby正则表达式？

Question

I'm trying to get the first word in this string: Basic (11/17/2011 - 12/17/2011)

So ultimately wanting to get Basic out of that.

Other example string: Premium (11/22/2011 - 12/22/2011)

The format is always "Single-word followed by parenthesized date range" and I just want the single word.

Answer 1

Use this:

str = "Premium (11/22/2011 - 12/22/2011)"
str.split.first # => "Premium"

The split uses ' ' as default parameter if you don't specify any.
After that, get the first element with first

Answer 2

您不需要正则表达式，只需使用

str.split(' ')[0]

Answer 3

我知道您找到了所需的答案，但以防万一将来有人迷失方向，以便从未知长度的大型String中提取所需的值：

word_you_need = s.slice(/(\b[a-zA-Z]*\b \(\d+\/\d+\/\d+ - \d+\/\d+\/\d+\))/).split[0]

Answer 4

此正则表达式将匹配第一个单词而没有尾随空格

"^\w+ ??"

Answer 5

如果您确实想要一个正则表达式，则可以在使用此正则表达式后获得第一组：

(\w*) .*

Answer 6

"Single-word followed by parenthesized date range"

'word' and 'parenthesized date range' should be better defined
as, by your requirement statement, they should be anchors and/or delimeters.

These raw regex's are just a general guess.

\w+(?=\s*\([^)]*\))

or

\w+(?=\s*\(\s*\d+(?:/\d+)*\s*-\s*\d+(?:/\d+)*\s*\))

Answer 7

Actually, all you need is:

s.split[0]

...or...

s.split.first