Save variable with command substitution… command not found

Question

I need to save into a bash variable a string after a grep and sed treatment. Here my code :

echo ${plan} | grep -e '^\S' -e 'Home directory:' | sed -e 's/Home directory: //'

/home/james 

That's what I need to save into a variable... so I tried :

HOME_DIRECTORY=$(${plan} | $(grep -e '^\S' -e 'Home directory:') | $(sed -e 's/Home directory: //'))

and

HOME_DIRECTORY=`${plan} | grep -e '^\S' -e 'Home directory:' | sed -e 's/Home directory: //'`

But both give me :

line 121: Home: command not found

Answer 1

Change you command to ,

HOME_DIRECTORY=$(echo ${plan} | grep -e '^\S' -e 'Home directory:' | sed -e 's/Home directory: //')

That is, you need to include the whole command inside $() .