在bash中解析输出

我的文件看起来像

    //
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---

所以在第一行之后，每一行都以括号中的数字开始。我需要grep括号中的数字并将其输出到一个新文件中（不带[）…如何实现>

grep-Po'（？如果您的
grep
不支持
-p
选项，但
awk
可用，另一种可能性是：

grep -Po '(?<=\[)\d+(?=\])' file > new_file

awk -F '[][]' '{ if ($2 != "") print $2 }' file > new_file

-F
告诉
awk
接受
]
和
[
作为字段分隔符，
$2
然后包含所需的数字并打印出来。
使用简单命令分三步进行：

grep -v "//" inputfile |  cut -d"[" -f2  | cut -d"]" -f1

在sed中，是否可以删除[]之外的所有内容：

grep -v "//" inputfile |  sed 's/.*\[\(.*\)].*/\1/'

它给了我以下警告：
escape sequence'\['被视为普通'['
，它不会打印任何内容。我总是以相反的顺序使用它们，解析器正确地读取它们以避免出现空字符类，如：
awk-F'[]“…”
我没有收到Bash4.3.30下awk版本1.3.3的警告。但是你的版本更可读，我会编辑我的答案。我的
bash
与你的相同，但是我的
awk
是
GNU
版本：
GNU awk 4.1.1
。这可能会有所不同。