bash 逃避grep中的感叹号？
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Escaping the exclamation point in grep?
提问by tim tran
I have this full line (the rpm command into awk below) that I want to grep out from certain files, including the quotes. I can't seem to be able to get the right output when I try grep, and grep -F. I tried deleting part of the tail end of line from the grep statement and it seems like the "!" causing the problems. I tried wrapping the string in single quotes and there is no luck as well. Thank you.
我有这整行（下面 awk 中的 rpm 命令），我想从某些文件（包括引号）中提取出来。当我尝试 grep 和 grep -F 时，我似乎无法获得正确的输出。我尝试从 grep 语句中删除行尾的一部分，它看起来像“！” 造成问题。我尝试将字符串用单引号括起来，但也没有运气。谢谢你。
rpm -qVa | awk '!="c" {print rpm -qVa | awk '$2!="c" {print ##代码##}'
}'
回答by Mark Reed
You have a few options.
你有几个选择。
Use single quotes - when you need a literal single quote, just drop out of single quotes, add one with a backslash, and then go back in:
grep 'rpm -qVa | awk '\''$2!="c" {print $0}'\' filenameUse a POSIX string:
grep $'rpm -qVA | awk \'$2!="c" {print $0}\'' filenameUse a less-specific pattern:
grep 'rpm -qvA | awk .$2.="c" {print $0}.' filenameor, if you have checks for
$2=="c"as well as$2!="c", you could do something like this:grep 'rpm -qvA | awk .$2[^=]="c" {print $0}.' filename
使用单引号——当你需要一个文字单引号时，只需去掉单引号，用反斜杠添加一个，然后返回：
grep 'rpm -qVa | awk '\''$2!="c" {print $0}'\' filename使用 POSIX 字符串：
grep $'rpm -qVA | awk \'$2!="c" {print $0}\'' filename使用不太具体的模式：
grep 'rpm -qvA | awk .$2.="c" {print $0}.' filename或者，如果您有
$2=="c"以及 的检查$2!="c"，您可以执行以下操作：grep 'rpm -qvA | awk .$2[^=]="c" {print $0}.' filename
I would go with the POSIX string, or maybe the plain single-quote option - which has the debatable advantage of working in other shells, like dash.
我会使用 POSIX 字符串，或者可能是普通的单引号选项 - 它具有在其他 shell 中工作的有争议的优势，例如dash.
回答by CuriousMind
A backslash should do the trick
反斜杠应该可以解决问题
##代码##