bash - How to take path from variable Bash

Question

I can't get path from variable in bash. How do it correct? For example:

my@PC:~$ a="~/.bashrc"
my@PC:~$ cat $a
cat: ~/.bashrc: No such file or directory

didn't work, but

cat .bashrc

and

cat ".bashrc"

Works well.

---

Here is right answer from fedorqui

cat $(eval echo $a)

Answer 1

问题的原因是波浪号被 shell 扩展到主目录。当您将其存储在变量中时，波浪号不会展开，并且 cat 在文件夹 ~ （而不是您的主目录）中查找文件 .bashrc

有两种方法可以解决这个问题：建议的 eval 和使用 $HOME：

a="$HOME/.bashrc"