Shell: insert variable into a command [duplicate]

Question

I'm trying to find a way the use a variable in this command to replace -10 with n_days var:

   n_days= -10
   date_prefix=$(date -d '-10 day' +%Y/%m/%d)

I tried this way but it didn't work:

   date_prefix=$(date -d '${n_days} day' +%Y/%m/%d)

Possible duplicate of Why does a space in a variable assignment give an error in Bash?, Expansion of variable inside single quotes in a command in Bash, When to wrap quotes around a shell variable?, Bash command substitution with a variable, etc. — jww
– jww, Commented Aug 30, 2018 at 12:38

arco444 · Accepted Answer · 2018-08-30 12:35:41Z

7

Two things:

Declare your variable properly (there is a space in your example)
Use double quotes in place of single quotes to allow the variable to be interpolated

So:

n_days=-10
date_prefix=$(date -d "$n_days day" +%Y/%m/%d)

answered Aug 30, 2018 at 12:35

arco444

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