How to update variables inside another variable in BASH?

Question

I have a variable, $var_c which contains a lot of complex syntax. I'd rather not need to rewrite this many times, as it is easy to make a mistake. But sometimes the data changes.

#!/bin/bash

var_a="cat"
var_b="dog"

var_c="<div class=\"pet_a\">$var_a</div><div class=\"pet_b\">$var_b</div>"

echo $var_c

var_a="fish"
var_b="whale"

echo $var_c

This should first <div class="pet_a">cat</div><div class="peg_b">dog</div> then <div class="pet_a">fish</div><div class="peg_b">whale</div>, however, after changing the variables $var_a and $var_b, $var_c is not updated.

How can I tell BASH to update this variable, without having to completely provide all of the syntax again?

Todd A. Jacobs · Accepted Answer · 2014-10-30 06:29:46Z

Use a Shell Function Instead

In Bash, interpolation in an assignment happens at the time of assignment. So, you can't really do what you're trying to do with a variable. Instead, you should use a shell function. For example:

var_a="cat"
var_b="dog"
var_c () {
    echo "<div class=\"pet_a\">$var_a</div><div class=\"pet_b\">$var_b</div>"
}
echo $(var_c)
# <div class="pet_a">cat</div><div class="pet_b">dog</div>

var_a="fish"
var_b="whale"
echo $(var_c)
# <div class="pet_a">fish</div><div class="pet_b">whale</div>

Note that in this case you could call var_c directly, rather than echo $(var_c), since the function already includes an echo statement. I deliberately used that in the example to maintain your current semantics, and to show how you can use a shell function like a variable.

Todd A. Jacobs · Accepted Answer · 2014-10-30 06:41:41Z

Use a Printf Format String

In Bash, interpolation in an assignment happens at the time of assignment. So, you can't really do what you're trying to do with a variable. One way to address this is with a format string and the printf builtin. The builtin supports a -v flag, which says:

  -v var    assign the output to shell variable VAR rather than
            display it on the standard output

For example:

fmt="<div class=\"pet_a\">%s</div><div class=\"pet_b\">%s</div>"
var_a="cat"
var_b="dog"
printf -v var_c "$fmt" "$var_a" "$var_b"
echo "$var_c"
# <div class="pet_a">cat</div><div class="pet_b">dog</div>

var_a="fish"
var_b="whale"
printf -v var_c "$fmt" "$var_a" "$var_b"
echo "$var_c"
# <div class="pet_a">fish</div><div class="pet_b">whale</div>

Christopher Bottoms · Accepted Answer · 2014-12-11 20:34:35Z

Variables contain data, not logic. When you set a variable, you're setting it to a particular value, not the rules for computing that value. If you want something that returns different results at different times, you need something more like a function:

func_c() {
    echo "<div class=\"pet_a\">$var_a</div><div class=\"pet_b\">$var_b</div>"
}

var_a="cat"
var_b="dog"
echo "$(func_c)"  # echoes '<div class="pet_a">cat</div><div class="pet_b">dog</div>'

var_a="fish"
var_b="whale"
echo "$(func_c)"  # echoes '<div class="pet_a">fish</div><div class="pet_b">whale</div>'

Note that executing a function uses the syntax $(functionname) (this can actually be used with any command), instead of just $variablename for variables. Also, you should almost always put both function and variable references in double-quotes (as I did above). Also also, if you're really just printing the result, don't use echo "$(functionname)" as I did above, just use functionname directly. Finally, it's generally bad practice to make functions depend on global variables (var_a and var_b) -- they should receive the data they're going to operate on as arguments:

func_c() {
    # Usage: func_c var_a var_b
    echo "<div class=\"pet_a\">$1</div><div class=\"pet_b\">$2</div>"
}

func_c "cat" "dog"  # echoes '<div class="pet_a">cat</div><div class="pet_b">dog</div>'
func_c "fish" "whale"  # echoes '<div class="pet_a">fish</div><div class="pet_b">whale</div>'
var_d="$(func_c "ferret" "weasel")"  # sets var_d to '<div class="pet_a">ferret</div><div class="pet_b">weasel</div>'

(I used more double-quotes than are strictly necessary in the above example -- but it's better to use them when not needed than to leave them off when they are needed.)

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How to update variables inside another variable in BASH?

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Use a Shell Function Instead

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Use a Printf Format String

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