CODEWITHSUNDEEP
functionshellvariablesunixksh
Javier Salas
Javier Salas

Reputation: 1188

How to print a value of parameter inside of function in Unix Korn Shell

I'm trying to print a value inside a function. But it is not working; it is printing the variable name. Here is my code:

#!/bin/ksh
MyVariable=""

function ValidateVariableValue
{
   eval $1="Working!"
   echo "$1"  #Here is printing the value "MyVariable" instead of "Working!"
}

ValidateVariableValue MyVariable

echo "value is: ${MyVariable}" #Here is printing the correct value that is "Working!"

Do you know how print the value inside the function?

Upvotes: 1

Views: 696

Answers (1)

codeforester
codeforester

Reputation: 43039

When you call your function, the positional variable $1 is set to MyVariable.

The statement

eval $1="Working!"

is creating a new variable whose name is contained in $1. In your case, MyVariable.

So, echo "$1" is correctly printing the value of $1 which is MyVariable. You need to use eval to print the value of the new variable:

eval "echo \$$1"

See this post on Unix & Linux Stack Exchange:

Upvotes: 6

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