For loop with an array in bash script

I want to use for loop with an array. I using the following command for this:

#! /bin/bash
rm -f /orch/list
arrayVM=($(cat /orch/servers | grep $1 | awk '{print $1}'))
for i in $arrayVM
do echo ${arrayVM[$i]}>>list
done

But when I check the list file I see only the first element of arrayVM array. What's wrong with my command?

2 Answers

The principle error is that for i in $arrayVM sets i to the first element in arrayVM, since there is no index. I am surprised that this does not give an error on the echo command, unless the first array element is numeric.

What you need is the iterative form of for:-

for (( i=0; i<${#arrayVM[*]}; ++i ))
do echo ${arrayVM[$i]}>>list
done

However, this is unnecessarily long-winded: much simpler is:-

for e in "${arrayVM[@]}"
do echo $e>>list
done

This assigns e to each element in turn, without enumerating them.

In the light of Fedorqui's answer, if arrayVM is not needed elsewhere, then there is a much simpler way to create the list file:-

cat /orch/servers | grep $1 | awk '{print $1}' >/orch/list

Or, since the cat is unnecessary:-

grep $1 </orch/servers | awk '{print $1}' >/orch/list

If you don't need the data to be stored in an array but you also want to use every record for other things, you can loop normally through the data with a process substitution:

while IFS= read -r value _;
do echo "$value" >> list
done < <(grep "$1" /orch/servers)

With read -r value _ we are storing the first field in $value and the rest in the throw away variable $_.