[英]Extracting group from regex in shell script using grep
I want to extract the output of a command run through shell script in a variable but I am not able to do it. 我想提取通过shell脚本运行的命令的输出变量,但是我无法做到。 I am using
grep
command for the same. 我使用
grep
命令相同。 Please help me in getting the desired output in a variable. 请帮助我获取所需的变量输出。
x=$(pwd)
pw=$(grep '\(.*\)/bin' $x)
echo "extracted is:"
echo $pw
The output of the pwd
command is /opt/abc/bin/
and I want only /root/abc
part of it. pwd
命令的输出是/opt/abc/bin/
,我只希望它的/root/abc
部分。 Thanks in advance. 提前致谢。
使用dirname
获取路径,而不是路径的最后一段。
You can use: 您可以使用:
x=$(pwd)
pw=`dirname $x`
echo $pw
Or simply: 或者简单地:
pw=`dirname $(pwd)`
echo $pw
All of what you're doing can be done in a single echo
: 您正在做的所有事情都可以在单个
echo
:
echo "${PWD%/*}"
$PWD
variable represents current directory and %/*
removes last /
and part after last /
. $PWD
变量代表当前目录和%/*
删除最后/
和最后的后部分/
。
For your case it will output: /root/abc
对于您的情况,它将输出:
/root/abc
The second (and any subsequent) argument to grep
is the name of a file to search, not a string to perform matching against. grep
的第二个(以及任何后续)参数是要搜索的文件的名称,而不是要执行匹配的字符串。
Furthermore, grep
prints the matching line or (with -o
) the matching string, not whatever the parentheses captured. 此外,
grep
打印匹配的行或(使用-o
)匹配的字符串,而不捕获括号。 For that, you want a different tool. 为此,您需要其他工具。
Minimally fixing your code would be 最小修复您的代码将是
x=$(pwd)
pw=$(printf '%s\n' "$x" | sed 's%\(.*\)/bin.*%\1%')
(If you only care about Bash, not other shells, you could do sed ... <<<"$x"
without the explicit pipe; the syntax is also somewhat more satisfying.) (如果只关心Bash,而不关心其他shell,则可以在没有显式管道的情况下进行
sed ... <<<"$x"
;语法也更令人满意。)
But of course, the shell has basic string manipulation functions built in. 但是,当然,shell具有内置的基本字符串操作功能。
pw=${x%/bin*}
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