需要RegExp帮助Linux Bash grep命令来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，我无法成功过滤掉带方括号的行：

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

请帮帮忙？

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

Answer 2

 $cat log |  grep -v '\[.*\]' | grep ABC