How do I set a bash variable that contains another variable?

Question

I am trying to set the variable service to be the value of

var1=first/second
echo $var1 | cut -d '/' -f 1
var2=$var1 | cut -d '/' -f 1"
echo $var2

The result of echo $var1 | cut -d '/' -f 1 is "first" which is correct. However, I haven't been able to set the result of this to another variable. In the case above, var2 is empty.

What would the correct syntax for line 3 be so that the part of the string before the \ is returned as the value of var2?

See also Storing output of command in shell variable, Calculate variable and output it to another variable, Bash: Assign output of pipe to a variable, setting output of a command to a variable, and more. — G-Man Says 'Reinstate Monica', Mar 12 '19 at 23:32

score 2 · Accepted Answer · answered Mar 12 '19 at 22:49

You could use command substitution

var2=$(echo "$var1" | cut -d '/' -f 1)

However in this case it would be better to use the shell's parameter substitution directly:

$ var2=${var1%/*}
$ echo "$var2"
first

(removes the shortest trailing substring matching /*) and

$ var3=${var1#*/}
$ echo "$var3"
second

(removes the shortest leading substring matching */) should you need it as well.

How do I set a bash variable that contains another variable?

1 Answers1