How to pass variable name from a file and use that variable in script in unix

Question

I have a config file in unix as

Variable_name:sequences:Status
ABC:1234:/path/txt

and I have a script in unix as

$ABC="select * from table"

I want to pass the variable name from file and use the variable in script with the value in script.

so I need to run script like

for i in /path/file_config;
 $Variable=ABC
 echo $variable # need result "select* from table"

Please help me in this

$ABC= and $Variable= are invalid. That's not how you assign to variables. — KamilCuk
– KamilCuk, Commented May 31, 2020 at 7:22
The answers given are different from how I understand your question. I think you have one variable name in the config file, and you want that string for selecting the SQL statement in the script. Something like var=ABC; grep "^\$${var}=" /path/config | cut -d '"' -f2 . Is that correct? — Walter A
– Walter A, Commented Jun 1, 2020 at 21:42

Shawn · Accepted Answer · 2020-05-31 09:30:41Z

1

In bash (But not other shells), ${!foo} is an indirect reference that expands to the value of the variable named in $foo.

So,

ABC="select * from table"
name=ABC
echo "${!name}"

will display select * from table

answered May 31, 2020 at 9:30

Shawn

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3 Comments

Please put the sql in single quotes, I don't want * replaces by filenames. (off-topic: I understand the question differently)

@WalterA Why would it get replaced with double quotes?

Sorry, my echo ${ABC} was wrong. I should have done echo "${ABC}".

Parthiban Sekar · Accepted Answer · 2020-05-31 10:20:41Z

0

Adding a slight improvement to Shawn's response above:

for Variable_name in `awk -F: '{ print $1 }' /path/file_config`; do
Variable=$Variable_name
echo ${!Variable}
done

answered May 31, 2020 at 10:20

Parthiban Sekar

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