bash - loop over variable array names

talanis · 02-18-2009, 04:59 PM

Hello,

i have a little problem concerning variable variable names in bash.
I am doing a loop to fill multiple arrays.

if i want to output the arrays, i can, however, do it only one by one; not in another loop.

Quote:

#!/bin/bash

for (( n=0; n<5; n++ )); # create 5 arrays
do
let array$n[0]="432859"

done

echo ${array0[*]} # arrays 0 to 4 can be easily displayed.
echo ${array1[*]} # but i want it as a loop (necessary in mightier programs)
echo ${array2[*]} # and so on...

#this loop does not work.
#line 16: ${array$y[*]}: bad substitution
for (( y=0; y<$n; y++ ));
do
echo ${array$y[*]} |sed s/"533862"/" "/g
done

Since i need to loop over the output, so i really only get the lines that where filled, any help is really appreciated!

Hko · 02-18-2009, 07:35 PM

Code:

#!/bin/bash

array0=(Hello Hallo Hola Aloha)
array1=(one two three four)
array2=(1 2 3 4)

echo "Printing them the normal way:"
echo "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"
echo ${array0[*]}
echo ${array1[*]}
echo ${array2[*]}
echo

echo "Printing them from a loop:"
echo "~~~~~~~~~~~~~~~~~~~~~~~~~~"
for (( y=0; y<3; y++ )); do
    echo $(eval echo \${array$y[*]})
done

talanis · 02-19-2009, 12:09 PM

Thanks man!
It works like a charm. Perfect ^^