Bash, integer variable inside of a string variable

Question

Greetins, I need to do the following:

string_a="string_a${int}"
string_b="string_b${int}"
int=1

String variables are in a text file, other than file which contains bash script and where variable int is declared. I need that I write string_a${int} in a text file, and sed that line from this file so that it becomes a string variable with nested int variable ${int}. Do you know any nice solution for that? :)

Maybe I try to write codes in both files:

File "string"

aaaaaaaaa${i}

File "program"

#!/bin/bash

int=1
i=1

string=$(sed -n "${int}p" string)

echo ${string}

As you can see, I try to sed first line from file "string", but as a result I get aaaaaaaaa${i} instead of aaaaaaaaa1.

String variables are in a text file, other than variable int. Can you clarify it. — anubhava
– anubhava, Commented Apr 7, 2014 at 7:00
@user3162968 After foo=bar and bar=5, $foo produces "bar", while ${!foo} produces 5. The exclamation tells bash to use the value of foo as the name of the parameter to expand using $. It's called indirect parameter expansion. — chepner
– chepner, Commented Apr 7, 2014 at 14:41

user3162968 · Accepted Answer · 2014-04-07 07:34:11Z

1

I've got a solution. I need to write eval echo ${string} instead of echo ${string}. I hope that it helps someone.

edited Apr 7, 2014 at 7:34

answered Apr 7, 2014 at 7:21

user3162968

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3 Comments

user3162968 Over a year ago

Can you explain me in a comment how I can write part of a single line in a code block? Four spaces does not work. I would like to write eval echo ${string} inside of a code block.

Henk Langeveld Over a year ago

Use backticks. (`) - it's the last example in the first line (mini-Markdown) if you click help.

chepner Over a year ago

Never use eval when another technique will suffice.

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