variable substitution in bash from different source files [duplicate]

Question

I have small snippet of code in bash

# below variable coming from a source file
test_name_tgt_tbl='abc_123'

# this variable coming from another source --> source 2
red_test='test_name'

# Based on source2 variable I need to find value in source 1
out_put=`echo ${red_test}_tgt_tbl`

echo "target=`echo $out_put`"

The result I am getting is below

target=test_name_tgt_tbl

The result I want is below

target=abc_123

How can I achieve that

foo=`echo $bar` is a really bad way of writing foo=$bar, btw. — Shawn
– Shawn, Commented Feb 24, 2021 at 22:09

Costi Ciudatu · Accepted Answer · 2021-02-24 22:14:27Z

1

You can achieve that indirection in bash by writing ${!some_var_name}. Here's how I would go about it:

var_name=${red_test}_tgt_tbl
echo ${!var_name}

Prints abc_123.

answered Feb 24, 2021 at 22:14

Costi Ciudatu

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