regex - 在grep补码中转义右括号字符

Question

如何在字符类中转义右括号字符？根据手册，要匹配文字]，应将其放在列表的第一位，但我需要匹配由方括号组成的类的补码。

Most meta-characters lose their special meaning inside bracket expressions. To include a literal ] place it first in the list. Similarly, to include a literal ^ place it anywhere but first. Finally, to include a literal - place it last.

Answer 1

如何在字符类中转义右括号字符？
]通常是POSIX character class的结尾，或者它表示字符类的结尾。

如果]是类中的第一个字符(可以跟在未转义的插入号后面)，则它不表示该类的末尾，并且被评估为可以匹配而无需转义的字符集的一部分。

echo "foo [22-Jun-2014 04:11:37 UTC] bar" | grep -o '\[[^]]*\]'
# [22-Jun-2014 04:11:37 UTC]