Variables inside a bash script subshesll

Question

I am trying to run commands as root and capture the output into a variable. However the variable "loopdev" is returning as empty.

sudo bash << "EOF"
whoami
loopdev=`losetup -f --show "${image}"`
echo "####" $loopdev "####"
EOF

The normal output for this command would be like below

$ image=sdimage.img
$ sudo losetup -f --show "${image}"
/dev/loop0

"<< 'EOF' -> losetup: : failed to set up loop device: Success" (no loop device was created, not sure why it says success) — Bob R, Oct 13 '18 at 21:17
Can u try to set "image" variable value inside the bash script ? As like below sudo bash << "EOF" whoami image=sdimage.img loopdev=`losetup -f --show "${image}"` echo "####" $loopdev "####" EOF — vijayalakshmi d, Oct 13 '18 at 22:15

score 0 · Answer 1 · answered Oct 13 '18 at 22:19

I tried your problem as like below

test=12781278
sudo bash << "EOF"
whoami
loopdev=`echo Hi $test`
echo "####" $loopdev "####"
EOF

And the output is

#### Hi  ####

For your problem, the image variable value is not being passed to the bash shell that you are starting.

score 0 · Accepted Answer · answered Oct 14 '18 at 02:34

0

The solution was to pass in the image variable like this:

image=rpi_2.img
sudo image="${image}" bash << 'EOF'
whoami
loopdev=`losetup -f --show "${image}"`
echo "####" $loopdev "####"
EOF

Output:

root
#### /dev/loop0 ####

answered Oct 14 '18 at 02:34

Bob R

2 Answers2