CODEWITHSUNDEEP
arraysbashshellvariables
Aakash Kushwaha
Aakash Kushwaha

Reputation: 45

initialising a variable with array variables in shell script

I am trying to concatenate strings using array variable but getting error.

declare -a arr
arr=(one two three)
var= "${arr[0]}    ${arr[1]}"
echo $var

expected output

one    two

(4 spaces between one and two)

I am getting following error:-

[wasadmin@gblabvl31 IBM]$ ./test.sh
./test.sh: line 10: one     two: command not found

Does this mean we can't assign a variable with array element (used as a variable)? What is the other way to do this

Upvotes: 1

Views: 417

Answers (2)

user000001
user000001

Reputation: 33317

You have an extra space in the assignment. Replace 

var= "${arr[0]}    ${arr[1]}"
#   ^

with 

var="${arr[0]}    ${arr[1]}"

You should also quote the argument of echo to preserve the whitespace in it 

echo "$var"

The reason for the error message you see, is that when there is a space after the equal sign, bash interprets the command as assigning an empty environment variable named var, and then tries to execute the command "${arr[0]} ${arr[1]}" which is evaluated to one two, and thus the command not found error

Upvotes: 1

redneb
redneb

Reputation: 23850

You must remove the space after the =:

var="${arr[0]}    ${arr[1]}"

Bash supports a syntax that allows you to temporarily set a variable when you call a command. The syntax works like that VARNAME=somevalue command. This will execute the command, having set the (environment) variable VARNAME to somevalue. If you say VARNAME= command then bash interprets that as VARNAME="" command i.e. sets the variable to the empty string. In your case, that causes bash to try to execute the "${arr[0]} ${arr[1]}" part as if it was a command.

Upvotes: 2

Related Questions