需要RegExp帮助Linux Bash grep命令过滤掉包含方括号的行

Question

使用以下示例,我需要过滤掉仅包含'ABC'的行,同时跳过与'ABC'匹配的包含方括号的行:

2012-04-04 04:13:48,760~sample1~ABC[TLE 5332.233 2/13/2032 3320392]:CAST
2012-04-04 04:13:48,761~sample2~ABC
2012-04-04 04:13:48,761~sample3~XYZ[BAC.CAD.ABC.CLONE 232511]:TEST

这是我所拥有的,但到目前为止,我无法使用方括号成功过滤掉这些行:

bash-3.00$ cat Metrics.log | grep -e '[^\[\]]' | grep -i 'ABC'

请帮忙？

Answer 1

根据评论编辑：

尝试grep -i 'ABC' Metrics.log | grep -v "[[]" | grep -v "ABC\w"

输入：

2012-04-04 04:13:48,760~sample1~ABC[TLE 5332.233 2/13/2032 3320392]:CAST
2012-04-04 04:13:48,761~sample2~ABC
2012-04-04 04:13:48,761~sample3~XYZ[BAC.CAD.ABC.CLONE 232511]:TEST
2012-04-04 04:13:48,761~sample4~XYZ
2012-04-04 04:13:48,761~sample5~ABCD
2012-04-04 04:13:48,761~sample6~ABC:TEST

输出：

2012-04-04 04:13:48,761~sample2~ABC
2012-04-04 04:13:48,761~sample6~ABC:TEST