Vim:用 n+1 替换 n [英] Vim: Replace n with n+1

问题描述

如何用 n+1 替换与特定模式匹配的每个数字 n?例如.我想用值+1 替换一行中括号中的所有数字.

How do I replace every number n that matches a certain pattern with n+1? E.g. I want to replace all numbers in a line that are in brackets with the value+1.

1 2 <3> 4 <5> 6 7 <8> <9> <10> 11 12

应该变成

1 2 <4> 4 <6> 6 7 <9> <10> <11> 11 12

推荐答案

%s/<\zs\d\+\ze>/\=(submatch(0)+1)/g

说明:

%s          " replace command
"""""
<           " prefix
\zs         " start of the match
\d\+        " match numbers
\ze         " end of the match
>           " suffix
"""""
\=          " replace the match part with the following expression
(
submatch(0) " the match part
+1          " add one
)
"""""
g           " replace all numbers, not only the first one

如果只想替换特定行，将光标移动到该行，然后执行

If you only want to replace in specific line, move your cursor on that line, and execute

s/<\zs\d\+\ze>/\=(submatch(0)+1)/g

或使用

LINENUMs/<\zs\d\+\ze>/\=(submatch(0)+1)/g

(用实际的行号替换LINENUM，例如13)

(replace LINENUM with the actual line number, eg. 13)

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