How to get first N parts of a path?

问题

Imagine a path like this: /a/b/c/d/e/...

Where ... could be any number of levels further deep, i.e., I don't know, ahead of time, whether there will be 2, 3, or 13 more level deep.

How can I, using a FreeBSD shell, e.g., /bin/sh, extract the first "N" parts of this path? e.g., first 4 levels, so that I would get /a/b/c/d?

回答1:

You can use cut:

s='/a/b/c/d/e/f/g/h/i/j/k'
echo "$s" | cut -d/ -f1-5
/a/b/c/d

Or if you are using BASH then you can use shell array:

IFS=/ arr=($s)

Then print desired elements from array:

IFS=/ echo "${arr[*]:0:5}"
/a/b/c/d

来源：https://stackoverflow.com/questions/41229950/how-to-get-first-n-parts-of-a-path