sed - 注释与特定字符串匹配且尚未注释掉的行 [英] sed - Commenting a line matching a specific string AND that is not already commented out

问题描述

我有以下测试文件

AAA
BBB
CCC

使用以下 sed 我可以注释掉 BBB 行.

Using the following sed I can comment out the BBB line.

# sed -e '/BBB/s/^/#/g' -i file

我只想注释掉开头没有 # 的行.

I'd like to only comment out the line if it does not already has a # at the begining.

# sed -e '/^#/! /BBB/s/^/#/g' file

sed: -e expression #1, char 7: unknown command: `/'

有什么想法可以实现这一目标吗?

Any ideas how I can achieve this?

推荐答案

假设您没有包含多个 # 的任何行，这将起作用:

Assuming you don't have any lines with multiple #s this would work:

sed -e '/BBB/ s/^#*/#/' -i file

注意:您不需要/g，因为您每行最多进行一次替换.

Note: you don't need /g since you are doing at most one substitution per line.

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