[SOLVED] How to get the last mount point in bash shell script?

Issue

Objective: to get last mount point in bash shell script

Code:

#!/bin/bash
export mt_pt=`lsblk |grep part|tail -1`
awk '{print (($7))}' | $mt_pt

Expected:

/run/media/ava/KINGSTON

Actual:

./time.sh: line 12: └─sdb1: command not found

lsblk output:

[ava@srvr0 ~]$ lsblk
NAME    MAJ:MIN RM   SIZE RO TYPE MOUNTPOINT
sda       8:0    0 298.1G  0 disk 
├─sda1    8:1    0     2G  0 part /boot
├─sda2    8:2    0    32G  0 part /
├─sda3    8:3    0    32G  0 part /data
├─sda4    8:4    0     1K  0 part 
├─sda5    8:5    0    32G  0 part /home
├─sda6    8:6    0    32G  0 part /log
├─sda7    8:7    0    32G  0 part /opt
├─sda8    8:8    0    32G  0 part /tmp
├─sda9    8:9    0    32G  0 part /usr
├─sda10   8:10   0    32G  0 part /var
├─sda11   8:11   0    16G  0 part [SWAP]
└─sda12   8:12   0  24.1G  0 part /index
sdb       8:16   1  28.9G  0 disk 
└─sdb1    8:17   1  28.9G  0 part /run/media/ava/KINGSTON
sr0      11:0    1  1024M  0 rom  

Please guide me in getting last mount point.

Solution

You could save the 7th field in a variable for each line where the 6th field equals part and print the saved value in the END block.

lsblk | awk '$6=="part"{ p=$7 } END{ print p }'

Answered By - Freddy

Answer Checked By - Marie Seifert (WPSolving Admin)