Vim - 在不以'character'开头的行上用'bar'替换'foo'

Question

In Vim regular expression, I know it is possible to replace foo by bar on all lines starting with % using

:g/^%/s/foo/bar/g

but I want to replace foo by bar on all lines NOT starting with % . Is there a way to easily do so?

Answer 1

试试:vglobal ：

:v/^%/s/foo/bar/g

Answer 2

You can just negate the % character using character class : -

:g/^[^%]/s/foo/bar/g

[^%] match any character except % , at the start of the string.

Answer 3

The inverse of :g is :g! , so your example could be expressed:

:g!/^%/s/foo/bar/g

Note that :g! is just another way of writing :v (cf. Jim Davis' answer)

Answer 4

尝试:g/^[^%]/s/foo/bar/g匹配所有不以％开头的行