使用sed从输出中删除不需要的字符 [英] Using sed to remove unwanted characters from output

问题描述

我正在bash shell中运行以下命令，以尝试仅返回 [] 中的十六进制字符和紧随其后的整数值.第一次sed调用将替换所有内容，包括 [在内的所有内容.第二个替换:和] .该命令会产生所需的结果，但是我想在一个sed语句中完成此操作，即无需再次通过sed进行遍历.我已经尝试了这两个regex表达式的很多组合，但是找不到能够产生所需结果的组合.

I am running the following command in a bash shell in an attempt to return only the hexadecimal characters inside the [] and the integer value immediately following. The first sed call replaces everything up to and including the [. The second replaces : and ]. The command produces the desired result however I'd like to accomplish this in a single sed statement, i.e. without having to pipe through sed a second time. I've attempted a good many combinations of these two regex expressions but have not been able to find a combination that produces the desired result.

echo 'dot1dTpFdbPort[00:02:6f:d9:16:ca] 12' | sed 's/^.*\[//g' | sed 's/[][\:]//g'
00026fd916ca 12

推荐答案

也许您可以尝试通过一个唯一的sed命令使用两个正则表达式:

maybe you can try using the two regex with one unique sed command:

echo 'dot1dTpFdbPort[00:02:6f:d9:16:ca] 12' | sed -e 's/^.*\[//g' -e 's/[][\:]//g'

该命令可以完全按照您的要求产生所需的结果:仅在一次sed指令中即可.

That command produces your desired result exactly what you want: in only one sed stetment.

礼炮.

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