Bash enumerated types

Shell Programming and Scripting
1

I have a variable, vbLevel and want to assign it to one of several values which are "none", "low", "medium", "high".

Then I want to do something like

if [ vbLevel -gt low ]; then
  ...
fi
2

your text comparing doesn't makes sense to me. Please elaborate more
If you like to perform some action based in text value following code will help you

for var in none low medium high;
    do
        case $var in   
             none)      
                    <your statements here>
                    ;;   
             low)       
                     <your statements here>
                     ;;   
             medium)  
                    <your statements here>
                    ;;
             high)       
                    <your statements here> 
                    ;;  
        esac
    done
3

You can do this in ksh93 and bash, without the dollar signs I mean...

none=0 low=1 medium=2 high=3
vbLevel=2
if [[ vbLevel -gt low    ]]; then
  echo hello
fi
4

The assign can also be done with the enumerated type let vbLevel=medium

5

Indeed. let is an arithmetic evaluation. A synonymous notation would be (ksh/bash):

(( vbLevel=medium ))

And you can also do this:

none=0 low=1 medium=2 high=3
vbLevel=2
if (( vbLevel > low )); then
  echo hello
fi
6

So one does not use the $ sign for the variables?

7

Inside the double parentheses you are allowed to leave the $-signs out if you wish..

8

Anything that uses bash expressions does not require the $ sign eg:

let expression
(( expression ))
[[ expression ]]

Anywhere else will eg:

find . -mtime +$vbLevel
if [ $vbLevel -eq $none ]
case $vbLevel in 
    ($medium) echo yes ;;
esac
9

Any idea why

printf "-- Name --"

gives invalid option and how can I make it print without error

---------- Post updated at 07:02 PM ---------- Previous update was at 06:58 PM ----------

This seems to work. I rather set vbLevel to none, medium, ... and do as below. or is it better using

(( vbLevel=medium ))

vbLevel=$medium
if (( vbLevel > low )); then
  echo hello
fi

This works too

vbLevel=$medium
if [ $vbLevel -gt $low ]; then
  echo hello2
fi

---------- Post updated at 07:10 PM ---------- Previous update was at 07:02 PM ----------

I want to stick with the one below. Would this be ok?

vbLevel=$medium
if (( vbLevel > low )); then
  echo hello1
fi

---------- Post updated at 07:23 PM ---------- Previous update was at 07:10 PM ----------

I get the user to supply none, low, medium and high from a user option.

Example

/home/chrisd/srv/sunny/hstmy/bin/bash/raytrac.bash --vblv=high

My code then uses

vbLevel=$medium
if [ $hasArgument_vbLevel == "true" ]; then
  vbLevel=$value_vbLevel
fi

echo "vbLevel = $vbLevel"

if (( vbLevel > medium )); then
  echo "Hello done"
fi

Thus I replace vbLevel by the user string, which is either none, low, medium or high which then gets associated with the number and then I can do

if (( vbLevel > medium )); then

---------- Post updated at 07:31 PM ---------- Previous update was at 07:23 PM ----------

I thought to use indirect substitution for this

vbLevel=${!medium}                    # Indirect substitution
if [ $hasArgument_vbLevel == "true" ]; then
  vbLevel=${!value_vbLevel}        # Indirect substitution
fi

if (( vbLevel > medium )); then
  echo "Hello done"
fi
10

Try:

printf "%s" "-- Name --"