对于后&QUOT提取一切常规的前pression;:"在bash命令行输出 [英] regular expression for extracting everything after ":" of a command line output in bash
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问题描述
我有当前存储的命令行输出到一个变量bash脚本。我想执行常规的前pression抽取一切最后的冒号之后(:
)。
I have a bash script that currently stores a command line output into a variable. I want to perform regular expression to extract everything after the last colon (:
).
这是我的code:
line="$(command line)"
echo "$line" | awk '/[^:]+$/'
这不幸的是不起作用。我应该如何解决?
This unfortunately doesn't work. How should I fix it?
推荐答案
使用GNU bash和它的参数扩展:
With GNU bash and its Parameter Expansion:
line="root:x:0:0:root:/root:/bin/bash"
echo "${line/*:/}"
输出:
/bin/bash
使用GNU bash和正则表达式:
With GNU bash and regex:
line="root:x:0:0:root:/root:/bin/bash"
[[ $line =~ ([^:]*)$ ]] && echo "${BASH_REMATCH[1]}"
输出:
/bin/bash
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