CODEWITHSUNDEEP
linuxbashvariables
gkr2d2
gkr2d2

Reputation: 773

Variable comparison in bash

I am trying to find out whether disk is SSD or HDD using bash script.

logical_name="/dev/sda"
type="" 
disk=$(basename $logical_name)
x=`cat $filename | grep "${disk}" | awk '{print $2}'`
if [ ! -z "$x" ]
then
        if [ "$x" = "0" ]
        then
                type="SSD"
        fi  
        if [ "$x" = "1" ]
        then
                type="HDD"
        fi  
fi

echo $type

Value of x is correct, 0 or 1. But after comparison, it's not assigning any value to variable type. It prints as empty. Can anyone point out what am I doing wrong here?

More information:

$filename is a file that contains output of sudo lsblk -d -o name,rota

NAME ROTA
sda     1
sdd     1
sdc     0

Upvotes: 2

Views: 113

Answers (2)

Mike Holt
Mike Holt

Reputation: 4612

The lsblk command lets you specify a device, so you shouldn't have to grep through the lsblk output to find the device you're interested in. This also means you don't need the name column. Plus you can disable the header with -n, and the -r option (raw output) gets rid of the leading and trailing whitespace:

> hdtype() { lsblk -drno rota "$1" | sed 's/1/HDD/;s/0/SSD/'; }

> hdtype /dev/sda
HDD

As far as why your code isn't working, I'm not sure. It worked just fine in my bash terminal.

Upvotes: 1

Anubis
Anubis

Reputation: 7445

While I don't see any problem with the posted code, following would be a more simplified and maintainable version for doing the same.

DISK_NAMES=(SSD HDD)  # names are resolved by index

filename="in1.txt"
logical_name="/dev/sda"
disk="$(basename $logical_name)"
x="$(sed -n 's/'$disk' *\([0-9]*\)/\1/p' $filename)"
# check if $x is empty here, if required
echo "$x -> ${DISK_NAMES[$x]}"

Upvotes: 1

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