Read command in shell script - variable input substitution

Question

I am new to Linux, pardon any wrong terminology used here.

I am using Ubuntu 18.04. I am reading text in a shell script using read command. I have a variable in bash window that I want to substitute in that read command. My shell script looks like this:

echo "Enter Text"
read text
echo "Text is $text"

Then I execute it like this in terminal

user@machine:~$ var="hello world"
user@machine:~$ ./script.sh
Enter text
$var
Text is $var
user@machine:~$

Notice the line in output - Text is $var.. how can I make it read the value of variable var so it stores hello world in variable named text?

NOTE: I know there is an option to pass parameters to scripts, but I find this more intuitive for the user executing script hence wanted to know how to do it this way!

More intuitive, I don't know. Less efficient, certainly. You are preventing name completion. Bash will do completion on the names of environment variables. Try this: execute `export SOME_LONG_VARIABLE_NAME="variable content"`. Then type `echo $SOME_L` and [tab][enter] — xenoid, Jul 30 '19 at 09:45
Possible duplicate of [indirect variable expansion in POSIX as done in bash?](https://unix.stackexchange.com/questions/111618/indirect-variable-expansion-in-posix-as-done-in-bash) — l0b0, Jul 30 '19 at 09:52

score 4 · Answer 1 · answered Jul 30 '19 at 08:56

4

This is called indirection, and the syntax is ${!reference} to substitute the value of the variable whose name is in the reference variable. For example:

name='Jane Doe'
reference='name'
$ echo "${!reference}"
Jane Doe

answered Jul 30 '19 at 08:56

l0b0

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Why the downvote? – l0b0 Jul 30 '19 at 08:59
I did not downvote.. someone else did on both the answers! – Shishir Gupta Jul 30 '19 at 09:01
Also, I changed script.sh last line to `echo "Text is $text ${!text}"` and it threw an error on execution saying "./script.sh: line 3: $var: bad substitution." – Shishir Gupta Jul 30 '19 at 09:06
Then you're most likely not running Bash. Or you're running [pre-4.3](https://mywiki.wooledge.org/BashFAQ/061). – l0b0 Jul 30 '19 at 09:08
1

@Shishir you don't have a shebang in your script, so it's run with sh, not bash – muru Jul 30 '19 at 09:13
@l0b0 oh okay.. I am new to Linux. If it is running with sh, then how do I do it? – Shishir Gupta Jul 30 '19 at 09:32
@muru AFAIK it'll run with your *current* shell, not necessarily `sh`. Do you have a counter-reference? – l0b0 Jul 30 '19 at 09:41
It depends on the shell, IIRC. Bash and ksh run it with itself, other shells may use sh (dash and zsh do). – muru Jul 30 '19 at 09:49
I think it is dash. Does it matter? – Shishir Gupta Jul 30 '19 at 09:50
See [Which shell interpreter runs a script with no shebang?](https://unix.stackexchange.com/q/373223/73093), but the short answer is as muru said. – Michael Homer Aug 01 '19 at 06:59

pLumo · Answer 2 · 2019-07-30T09:19:31.537

0

You need to export your variable first:

export var=test

Then, use envsubst to replace variables inside text:

echo "Enter Text"
read text
printf 'Text is %s\n' "$text" | envsubst

envsubst - substitutes environment variables in shell format strings

edited Jul 30 '19 at 09:19

answered Jul 30 '19 at 08:41

pLumo

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Doesn't work. Empty output! Probably since $var is not an environment variable? – Shishir Gupta Jul 30 '19 at 09:09
yes sure ... see my edit – pLumo Jul 30 '19 at 09:20

Read command in shell script - variable input substitution

2 Answers2