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Dynamic variable assignment

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# 1  
Old 02-15-2014
Dynamic variable assignment
My Code :
--------------------------------------------
Code:
#!/bin/bash

for i in `echo server1 server2`
do
eval ${i}_name = "apache"
echo ${i}_name
done

--------------------------------------------

Current output :
Code:
>./test.sh
./test.sh: line 5: server1_name: command not found
server1_name
./test.sh: line 5: server2_name: command not found
server2_name

--------------------------------------------

Expected output :

Code:
apache
apache

--------------------------------------------


Basically, I need to have variable name depending on arguments passed in for loop etc.
Please assist.
Last edited by Scrutinizer; 02-15-2014 at 04:49 AM.. Reason: code tags
# 2  
Old 02-15-2014
Try :

Code:
#!/bin/bash

for i in server1 server2
do
	eval ${i}_name="apache"
	eval echo \$${i}_name
done

Code:
$ bash tester
apache
apache

This User Gave Thanks to Akshay Hegde For This Post:
sameermohite
# 3  
Old 02-15-2014
Since you are using bash, you might as well use arrays.

Code:
server_name=( apache apache )
echo "${server_name[0]}"
echo "${server_name[1]}"

Last edited by Scrutinizer; 02-15-2014 at 08:33 AM..
# 4  
Old 02-15-2014
Another method:

Code:
#!/bin/bash
for i in server1 server2
do
   var=${i}_name
   printf -v $var "%s" "apache"
   echo ${!var}
done

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