How to pass multiple variables in shell script

Question

I have a shell script that I'm trying to write to a file using multiple variables, but one of them is being ignored.

#!/bin/bash

dir=/folder
name=bob
date=`date +%Y`

command > $dir/$name_$date.ext

The $name is being ignored. How can I fix this?

Your code uses the var name_. When you cannot change the last line into fields with {}, you can work around with name_=bob_. — Walter A
– Walter A, Commented Oct 29, 2015 at 15:11
@Karoly: It is evil indeed, I included it for showing what is happening and encouraging the normal method showed by anubhava. — Walter A
– Walter A, Commented Oct 29, 2015 at 15:53

Karoly Horvath · Accepted Answer · 2015-10-29 15:34:25Z

4

Have you noticed that the _ was "ignored" as well? That's a big hint.
If you use set -u, you'll see the following:

-bash: name_: unbound variable

The way bash parses it, the underscore is part of the variable name.

There are several ways to fix the problem.
The cleanest is the ${var} construct which separate the variable name from its surroundings.
You can also use quotation in various ways to force the right parsing, e.g.: "$dir/$name""_$date.ext"

And in case your variables might contain spaces (now, or in the future) use quotation for words.

command >"$dir/${name}_$date.ext"
command >"${dir}/${name}_${date}.ext"

Both these are fine, just pick one style and stick to it.

answered Oct 29, 2015 at 15:34

Karoly Horvath

96.7k11 gold badges123 silver badges181 bronze badges

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

1 Comment

user1052448 Over a year ago

Ah, forgot that _ can be part of a variable. Thanks!

Collectives™ on Stack Overflow

How to pass multiple variables in shell script

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related