获取URL

-2

一个斜线后，我有一个简单的URL像这样获取URL

http://siteurl.com/2014/09/net-special-report-9ice-father-of-3-successful-entertainer-with-eyes-on-house-of-reps/

我怎么能是

结尾的斜线之后得到的最后一个值的值
net-special-report-9ice-father-of-3-success-entertainer-with-eyes-on-house-of-reps

任何想法。谢谢

来源

2014-09-30 Uchenna

使用拆分来做到这一点。如下所示：

str = "http://siteurl.com/2014/09/net-special-report-9ice-father-of-3-successful-entertainer-with-eyes-on-house-of-reps/" 
#=> "http://siteurl.com/2014/09/net-special-report-9ice-father-of-3-successful-entertainer-with-eyes-on-house-of-reps/" 
a = str.split('/') 
#=> ["http:", "", "siteurl.com", "2014", "09", "net-special-report-9ice-father-of-3-successful-entertainer-with-eyes-on-house-of-reps"] 
a.last 
#=> "net-special-report-9ice-father-of-3-successful-entertainer-with-eyes-on-house-of-reps"

来源

2014-09-30 13:23:01

谢谢。猜我只是懒惰。 – Uchenna 2014-09-30 13:24:13

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