Shell脚本在for循环中接受多组数组

Question

Hi can some one help me here on a shell script?

I have 2 hosts and 4 servers, I want to pass using for loop

host1 -->  host1_server1 host1_server2 
host2 -->  host2_server1 host2_server2 

I want to use a for loop to print both sets, outputting both sets:

set 1 output

hostname : host1 servers: host1_server1
hostname : host1 servers: host1_server2

set 2 output

hostname : host2 servers: host2_server1
hostname : host2 servers: host2_server2

How can I achieve this using a Bash shell script?

Answer 1

One Bash solution is to use Bash arrays, Indirect References , and printf .

#!/bin/bash

declare -a hosts=(host1 host2)     # array of variable names; quoting not needed

declare -a host1=("host1_server1" "host1_server2")
declare -a host2=("as-host2_server1" "as-host2_server2")

for Host in "${hosts[@]}"; do
   Indirect="$Host[@]"
   printf "hostname : $Host servers: %s\n" "${!Indirect}"
done

The above generates the following output:

hostname : host1 servers: host1_server1
hostname : host1 servers: host1_server2
hostname : host2 servers: as-host2_server1
hostname : host2 servers: as-host2_server2

This solution also takes advantage of the printf command's behavior to reuse the format to consume all of the arguments. From man bash , see the printf section:

The format is reused as necessary to consume all of the arguments

As a point of interest, note that the Bash idiom "${Variable[@]}" preserves any spaces within array elements. Also, if there are no elements in the array, this idiom expands to nothing. This in many circumstances is very useful.

These approaches, combined, allow for a single for loop to accomplish what you are asking for.