Bash does not recognize variable assignment

Question

I'm having a problem with a script I'm writing.

clear
echo "Welcome to the dd interface"
echo "NOTE: When this program executes you will have to provide root access"
echo " "
lsblk

echo " "
read -p "Enter the location of the disk to be imaged: " input_location
echo "Input Location Assigned"

echo " "
read -p "Enter the output location: " output_location
echo "Output Location Assigned"

echo "
1 - 512K
2 - 1024K
3 - 2048K
4 - 4096K
"

blocksize = 0

read -p "Select the write block size to be used: " selection
echo "Block write size selected"


#problem is here
if [[ $selection == 1 ]]; then
        blocksize = 512
elif [[ $selection == 2 ]]; then
        blocksize = 1024
elif [[ $selection == 3 ]]; then
        blocksize = 2048
elif [[ $selection == 4 ]]; then
        blocksize = 4096
fi
echo "Options selected:
Input location: $input_location
Output location: $output_location
Block size: $blocksize
"

Bash is seeing the blocksize in the IF statements as a command rather than a variable. How can I fix this?

utdemir · Accepted Answer · 2015-03-15 17:25:12Z

2

You shouldn't put spaces around = on assignments:

blocksize=0

instead of

blocksize = 0

should work.

answered Mar 15, 2015 at 17:25

utdemir

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