using a variable inside parameter substitutions

[doggybah]

I am using bash shell. My problem is with parameter substitution. I want this script to display four variables (or their value substitutions) using positional arguments $1-$4 as parameter in: "parameter:-value"
but I get a "bad substitution" when I use a "$" inside the curly braces... Is there any way to do this without having the user type "$" before each variable when they run the command?

Here is my code:

#display first variable; if empty or not set, display "NO VAR1"
echo ${1:-"NO VAR1"}

#display second variable; if empty, assign "NOW V2" to it and #display

: ${2:="NOW V2"}
echo ${2}

#if third variable is set, display "VAR3 SET"
echo ${3:+"VAR3 SET"}

#display fourth variable; if empty, display "Oh NO!" and exit
echo ${4: ?"Oh NO!"}

echo all done

[Ravime]

I am using bash shell. My problem is with parameter substitution. I want this script to display four variables (or their value substitutions) using positional arguments $1-$4 as parameter in: "parameter:-value"
but I get a "bad substitution" when I use a "$" inside the curly braces... Is there any way to do this without having the user type "$" before each variable when they run the command?

Here is my code:

#display first variable; if empty or not set, display "NO VAR1"
echo ${1:-"NO VAR1"}

#display second variable; if empty, assign "NOW V2" to it and #display

: ${2:="NOW V2"}
echo ${2}

#if third variable is set, display "VAR3 SET"
echo ${3:+"VAR3 SET"}

#display fourth variable; if empty, display "Oh NO!" and exit
echo ${4: ?"Oh NO!"}

echo all done

Hi You can do as fallows

var1=32
echo $var1
var2=31
echo $var2

[dxqcanada]

There is a problem with one of the statements:

${parameter:=word}
If parameter is unset or null, the expansion of word is assigned to parameter. The value of parameter is then substituted. Positional parameters and special parameters may not be assigned to in this way.