替换 bash 中两个括号之间的空格和 /

Question

我有这个（文本）文件：

[check consumer TAIL_SCE_O_P]
[CheckCo on IEB02]
[CheckSubscribers consumers for ne/os/jms/queue/FT_PN_IN]
[CheckSubscribers consumers for ne/os/jms/queue/SO_IN]
Please do not change/me/thank/you
another bracketless text

我想将空格 ( ) 和 / 替换为括号之间的 -

示例输出：

[check-consumer-TAIL_SCE_O_P]
[CheckCo-on-IEB02]
[CheckSubscribers-consumers-for-ne-os-jms-queue-FT_PN_IN]
[CheckSubscribers-consumers-for-ne-os-jms-queue-SO_IN]
Please do not change/me/thank/you
another bracketless text

我试过这个：

cat output.txt |grep "\[" | sed 's/\ /-/g'

但是改了很多东西。

Answer 1

输入数据：

$ cat output.txt
[check consumer TAIL_SCE_O_P]
[CheckCo on IEB02]
[CheckSubscribers consumers for ne/os/jms/queue/FT_PN_IN]
[CheckSubscribers consumers for ne/os/jms/queue/SO_IN]
Please do not change/me/thank/you
another bracketless text

一个sed解法：

$ sed '/\[/,/]/{s/[ /]/-/g}' output.txt

其中：

/\[/ + , + /]/ - 范围的开始是左括号；范围的结尾是右括号
{s/[ /]/-/g} - 在范围内用连字符 (-)

/

以上生成：

[check-consumer-TAIL_SCE_O_P]
[CheckCo-on-IEB02]
[CheckSubscribers-consumers-for-ne-os-jms-queue-FT_PN_IN]
[CheckSubscribers-consumers-for-ne-os-jms-queue-SO_IN]
Please do not change/me/thank/you
another bracketless text

Answer 2

您可以使用

sed '/^\[.*]$/s/[[:space:]\/]/-/g' file
# => 
# [check-consumer-TAIL_SCE_O_P]
# [CheckCo-on-IEB02]
# [CheckSubscribers-consumers-for-ne-os-jms-queue-FT_PN_IN]
# [CheckSubscribers-consumers-for-ne-os-jms-queue-SO_IN]
# Please do not change/me/thank/you
# another bracketless text

意思是：

/^\[.*]$/ - 查找以 [ 开头并以 ]
s/[[:space:]\/]/-/g - 在该行中，将空格或 / 替换为 -.

见an online sed demo。

替换 bash 中两个括号之间的空格和 /

Replace spaces and / between two brackets in bash

bash

replace

brackets

sed