Setting a Var= to a variable that is incremented in for loop.

[Bob Fohl]

Example:
i=1
for me in myfile
do
MYVAR$i=`who -u | grep $me'
ME$i=$MYVAR$i < i = i + 1
done

probably over thinking this one...any ideas?

[Mel Burslan]

you'd rather be looking into array type variables if you are using ksh or posix

example command

# for i in 1 2 3
> do
> let a[$i]=$i+$i
> echo ${a[$i]}
> done
2
4
6

so, yours should look like
i=1
for me in myfile
do
MYVAR[$i]=`who -u | grep $me'
ME[$i]=$MYVAR[$i]
let i=$i+1
done

now you have series of variables as follows
ME[1]="some string"
ME[2]="some other string"
...
ME[n]="yet another string"

to do whatever your heart desires

to reference them, use the following notation:

string=${ME[$i]}

where i is your array index

hope this helps

[Rodney Hills]

Use "eval"

t=`who -u | grep $me`
eval "MYVAR$i=$t"
eval "ME$i=$MYVAR$i"

HTH

-- Rod Hills

[Hein van den Heuvel]

Bob, if the prior replies did not help, then be sure to explain in more detail what you are trying to eccomplish. Maybe some sample input/output? What problem are you really trying to solve, once you have that 'who' output in those variables?

- is myfile meant to be a file with names perhaps? In that case check out grep -f and reading its output: who | grep -f myfile | ...

If/when `who -u | grep $me` returns multple lines, should those become individually named/numbered variables?

fwiw,
Hein.

[Muthukumar_5]

You can use while-do-done loop instead for-do-done.

#!/bin/ksh
i=1

while read me;
do

let VAR[$i]=`who -u | grep $me`
let ME[$i]=${VAR[$i]}
let i=i+1

done < myfile

It will do it.

[Eknath]

Hi Bob,

you last line should look like
i=`expr $i +1`

Anyway pls let us know what is your aim behind this. maybe a simple script would solve the problem

Cheers !!!
eknath