摘要: 1. (x) -> 'bar foo'.replace( /(..) (...)/, '$2 $1' ) -> "bfoo ar" 2. (x) -> "foo bar foo bar".match(/(bar) (foo)/, '$2 $1'); -> ["bar foo", "bar", "fo 阅读全文
posted @ 2017-03-10 11:29 戴杭林 阅读(137) 评论(0) 推荐(0)