获取与常规的前pressions分离器之间的东西 [英] Get something between separators with regular expressions

问题描述

我想从一个寄存器是这样的提取的中间值：

I am trying to extract the middle value from a register that is something like this:

NAME=PATH=USER=DATE

我需要得到的路径部分和所有我想出了是这样的：

I need to get the path part and all I came up with is this:

=[^=]*=

的问题是，它包括=在开始和结束时。我怎么能不包括他们？

the problem is that it includes the = at the beginning and at the end. How could i not include them?

此外，它的UNIX命令我应该使用解压呢？我在想SED，但我通常用它来替换字符串没有得到他们。用grep可能？

Also, which unix command should I use to extract it? I was thinking sed, but I usually use it to replace strings not to get them. grep maybe?

我bash编程是新...

I am new in bash programming...

有没有在那里我可以学到一些正则表达式的任何地方吗？因为我觉得我需要真正知道如何与他们合作。

Is there any place where I can learn some regex? because I think I will need to really know how to work with them.

推荐答案

使用捕获组：

if [[ $str =~ =([^=]+)= ]]
then
    echo "Part between = and = is ${BASH_REMATCH[1]}."
fi

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