passing shell variable into parameter in shell script [duplicate]

Question

I'm running this command to find the mounted network folder :

df | awk '/[email protected]/ { print $1 }'

This code is working fine but rather using 'andi' i want to pass it using 'USER' variable. It showing the correct value if I echo the USER var.

echo $USER

andi

But it's not working when I used the var . Seems the single quot pair is not working with $VAR :

df | awk '/[email protected]/ { print $1 }'

How should I use the var in this case ?

Something like awk -v user=$USER '$1 ~ user {print $1}' will make the bash variable into an awk variable. — Mark Setchell
– Mark Setchell, Commented Nov 6, 2023 at 8:43
... though you'd probably want to define a field separator to split username from mount, i.e. awk -F@ ... — Mark Setchell
– Mark Setchell, Commented Nov 6, 2023 at 9:04

user1934428 · Accepted Answer · 2023-11-06 08:43:43Z

0

As you found out, parameter expansion is not performed between single quotes. Hence $USER must be outside of single quotes, while $1 must be inside:

df | awk /[email protected]/' { print $1 }'

To cope for the possibility, that the name stored on USER contains a space, you could alternatively write it as

df | awk "/[email protected]/"' { print $1 }'

since parameter expansion is done within double quotes. Just make sure that awk sees the program as single parameter. Therefore we can't have a space in front of the opening single quote.

edited Nov 6, 2023 at 8:43

answered Nov 6, 2023 at 8:38

user1934428

22.8k9 gold badges57 silver badges108 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

passing shell variable into parameter in shell script [duplicate]

1 Answer 1

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Linked

Related