SED/tr 等.:如何注释掉包含“字符串"的行?在文件中? [英] SED/tr etc.. : How to comment out a line that contains "string" in a file?

问题描述

我的文件包含如下几行:

my file contains lines such as these:

-A INPUT -m state --state NEW -m tcp -p tcp --dport 2000 -j ACCEPT
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2001 -j ACCEPT
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2002 -j ACCEPT

我想注释掉(在前面加上#)

i want to comment out ( add # infront of ) the line that is

-A INPUT -m state --state NEW -m tcp -p tcp --dport 2001 -j ACCEPT

我如何通过 SED 或通过命令行的其他方法执行此操作?

how can i do this via SED or some other method via command line ?

我应该寻找......例如.

should i seek for .. e.g..

2001

然后注释掉整行(在前面加上#)

and then comment out that whole line ( add # infront of )

或者我应该搜索整行，然后用包含 # 的新行替换它?

or should i search for the entire line and then replace it with a new one that contains # ?

最实用的方法是什么?(最快)

what would be the most practical method ? ( fastest )

推荐答案

通过 sed，

sed '/ 2001 /s/^/#/' file

添加内联编辑 -i 选项以保存所做的更改.

Add inline edit -i option to save the changes made.

sed -i '/ 2001 /s/^/#/' file

示例:

$ sed '/ 2001 /s/^/#/' file
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2000 -j ACCEPT
#-A INPUT -m state --state NEW -m tcp -p tcp --dport 2001 -j ACCEPT
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2002 -j ACCEPT

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