UNIX for Dummies Questions & Answers

Suppose I assign these variables:
n=5
row=n
let col=n/2
val=17
let table$row[$col]=val

I want to print the last variable. How do I do that? I have tried:
echo ${table$row[$col]}
This is the error message:
ksh: ${table$row[$col]}: bad substitution

I fixed that problem

When you fix problems or find solutions to problems that you post, PLEASE post the solutions and fixes. This adds to the knowledge base and helps everyone greatly! THANK YOU.

Disregard the below solution. It does not work. I am still working on this problem. Any ideas??

The array variable name includes "table" and the value of another variable, $row (thus, table$row). This variable is an array with $col number of elements.

The erroneous command:

echo ${table$row[$col]}

The problem:
The shell attempts substitution, which is not the desired outcome. I want to echo the value of the array element.

The solution:

echo $(table$row[$col])

This command prints to the terminal the value of the $col element in the table$row variable.

hi folks.

i hav a hint for you:

to display the last fieldnumber type:

echo ${#myarray} # shows number of fields

cheers, alex...

see:
https://www.unix.com/unix-for-dummies-questions-and-answers/1225-does-variable-call-work-korn.html?s=

Thanks for the idea, Alex. Do you know how to echo a variable within a variable?

hi.

variable within a variable? i am not shure what you mean (my english si not so good, i am from vienna/austria, living in germany now), but look at this:

VAR1="alex"

VAR2=$($alex)

echo $VAR2


this is for kornshell!

cheers,
alex...