Bash variable variables passed through functions

Question

I've looked over a couple other variable variable posts here but still seem stuck with what I'm trying to attempt.

I have an existing script that has a series of similar block like

set_name="something"

# Required values
var1=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "key1_${set_name}")
if [[ -z "$var1" ]]; then
    echo "Cannot find required var1"
    exit 1
fi
var2=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "key2_${set_name}")
if [[ -z "$var2" ]]; then
    echo "Cannot find required var2"
    exit 1
fi
# Optional, okay to be empty
var3=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "key3_${set_name}")
var4=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "key4_${set_name}")

I was trying to factor some of the boilerplate checks out to keep this set of lookups assignment easier to read (in my opinion anyway). The last iteration I had attempted (clearly not working) looks like

ZTest () {
    var=$1
    if [[ -z "${!var}" ]]; then
        echo $2
        exit 1
    fi
}

VarRequire () {
    var=$1
    key=$2
    errmsg=$3
    VarLookup ${!var} $key
    ZTest ${!var} $errmsg
}

VarLookup () {
    var=$1
    key=$2
    ${!var}=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "$key")
}

# Required 
VarRequire "var1" "key1_${set_name}" "Cannot find required var1"
VarRequire "var2" "key2_${set_name}" "Cannot find required var2"

# optional 
VarLookup "var3" "key3_${set_name}"
VarLookup "var4" "key4_${set_name}"

The end result is I would be able to reference $var1, $var2, $var3, $var4 down the line in the script just the same as the original.

Is what I'm attempting possible in bash?

you can use local qualifier to limit the scope of function variables. — karakfa
– karakfa, Commented Sep 23, 2015 at 21:30

gniourf_gniourf · Accepted Answer · 2015-09-23 21:58:45Z

2

ZTest is already available as a parameter expansion operator:

: ${var1:?Cannot find required var1}

You are close with VarLookup, though; you need to use the declare command to create the variable. ${!var} is only for accessing the value once the variable exists. (Note that declare requires the -g option to avoid creating a local variable, and that option was only introduced in version 4.2.)

VarLookup () {
  local var=$1
  local key=$2
  declare -g "${var}=$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "$key")"
}

Prior to version 4.2, you can use printf in place of declare:

printf -v "$var" '%s' "$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "$key")"

I would resist the urge to refactor too much in shell, as indirection can be fragile. I'd suggest something more direct like

var_lookup () {
    REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "${1}_$set_name"
}

var1=$(var_lookup key1); : ${var1:?Cannot find required var1}
var2=$(var_lookup key2); : ${var2:?Cannot find required var2}
var3=$(var_lookup key3)
var4=$(var_lookup key4)

edited Sep 23, 2015 at 21:58

gniourf_gniourf

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answered Sep 23, 2015 at 21:33

chepner

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2 Comments

glenn jackman Over a year ago

In a function, declare == local, need -g

atxdba Over a year ago

I wasn't able to get your printf version working for what ever reason but last option seemed good enough for me

glenn jackman · Accepted Answer · 2015-09-23 21:32:46Z

2

Too much indirection in the Var* functions

VarRequire () {
    local var=$1
    local key=$2
    local errmsg=$3
    VarLookup "$var" "$key"
    ZTest "$var" "$errmsg"
}

VarLookup () {
    local var=$1
    local key=$2
    declare -g "$var"="$(REQUIRED_ENV_VAR=/path/to/somewhere mybinary -p "$key")"
}

The declare command allows you to use a variable's value as the variable name. I use the -g option so the variable is global.

The ZTest function does require the indirection.

answered Sep 23, 2015 at 21:32

glenn jackman

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Comments

quickshiftin · Accepted Answer · 2015-09-23 21:33:05Z

From the BASH manpage

   local [option] [name[=value] ...]
          For  each  argument,  a local variable named name is created,
          and assigned value.  The option can be any of the options
          accepted by declare.  When local is used within a function, it
          causes the variable name to have a visible scope restricted to 
          that function and its children.  With no operands, local writes a
          list of local variables  to the  standard  output.   It is an
          error to use local when not within a function.  The return status
          is 0 unless local is used outside a function, an invalid name is
          supplied, or name is a readonly variable.

So the answer is yes. If you declare a variable a local within a function, and it calls another function, that function has access to the variable as well.

Walter A · Accepted Answer · 2015-09-23 22:00:08Z

0

Can you set REQUIRED_ENV_VAR=/path/to/somewhere once at the start? And have your mybinary return 0 when a var is found?

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