Shell Programming and Scripting

Thanks for the response!!

Ok... For example, I want this script to read a file and find some values I am interested in.... But I want to create a more flexible script (if possible) and set this values that I am interested in as input....

the file is like:

var1 value1 value2 value3
var2 value1 value2 value3
var3 value1 value2 value3
var4 value1 value2 value3
.....

I want to run the script like : script.sh var2 var4 or next time script.sh var1 var2 var3 var6 or etc....

The idea is to read the number of arguments (or parameters) and save them to a variable in order to use them later.... Unfortunately, the var{i} names may not be always the same....

Is it more clear now?

How it is possible to echo the variables in the example of aigles inside the for loop?

The other common idiom for a list of random length is of course

Code:

for arg in "$@"; do
  echo another $arg
done

Thanks era, this is very interesting... but can I save these arguments to different variables?

Hmm, this feels a bit like going in circles. Can you explain what problem you are trying to solve?

When you start a script, its arguments are available in $1 $2 $3 $4 etc. The variable $# will tell you how many they are (you obviously already knew this) and "$@" will contain them all.

aigles' loop with the eval or the code joeyg posted will allow you to copy their values to other variables, using slightly different techniques.

If you want to use the index in the variable's name, the syntax you tried doesn't work directly, because of the way the shell parses strings. The eval command exists for situations where you want the name of a variable to include the value of another variable, for example, so it's exactly the right tool for this. It can be used if you want to echo things, too.

Code:

for ((i=1; i<=$#; ++i))
do
  eval VAR$i=\$$i
  eval echo VAR$i is \$$i
done

My comment was more of a sidetrack, in case you were really only looking for a way to loop over them all, without necessarily caring exactly how many times you loop.

First of all, really, thanks for your patience....

I think your last solution will work fine...

It does what I want... it reads all the arguments... but it doesn't save them to a variable... Maybe if there was a way to do something like:

Code:

i=0
for arg in "$@"; do
  let "  i = $i + 1 "
  VAR${i} = $arg
  echo VAR${i} 
done

I want to use the different names of the arguments as variables inside the script....

But it does save them to a variable. The first is VAR1, and so forth.

The syntax in your latest attempt is almost correct, but the double quotes are wrong, and of course, you still need to use "eval" like above. There's also a minor other nit regarding the let statement. But the for loop I posted above already does exactly this.

Ah, yeah... I think you're right!!!!! It works fine...

Thanks a lot, for both your patience and your valuable time!!!! I really appreciate it!!

cheers!