vi中如何删除行末的空格及替换法则

本文主要是介绍vi中如何删除行末的空格及替换法则，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

删除行尾空格

:%s//s/+$//g

删除行尾空格和回车

%s//(/d/+/s/+/l/+/)/s/+/_s/+$//$1/ 

 

简写及意思解释：

       item    matches                 equivalent
        /d      digit                   [0-9]
        /D      non-digit               [^0-9]
        /x      hex digit               [0-9a-fA-F]
        /X      non-hex digit           [^0-9a-fA-F]
        /s      white space             [       ]     (<Tab> and <Space>)
        /S      non-white characters    [^      ]     (not <Tab> and <Space>)
        /l      lowercase alpha         [a-z]
        /L      non-lowercase alpha     [^a-z]
        /u      uppercase alpha         [A-Z]
        /U      non-uppercase alpha     [^A-Z]

 

重复匹配：

        pattern         match count
        /{,4}           0, 1, 2, 3 or 4
        /{3,}           3, 4, 5, etc.
        /{0,1}          0 or 1, same as /=
        /{0,}           0 or more, same as *
        /{1,}           1 or more, same as /+
        /{3}            3

 

 

贪婪匹配和最小匹配

        /a./{-}b

This matches "axb" in "axbxb".  If this pattern would be used:

        /a.*b

It would try to match as many characters as possible with ".*", thus it
matches "axbxb" as a whole.

 

这篇关于vi中如何删除行末的空格及替换法则的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---