How to echo a dynamic variable's content in shell script

Question

How to echo a dynamic variable's content in shell script ?

i=1
declare  x$i=help
echo  $(echo x$i)
echo $x$i

Current output:

x1  
1

Desired output:

help

[bash-dynamic-variable-names](http://stackoverflow.com/questions/16553089/bash-dynamic-variable-names) — Eugeniu Rosca, Jul 10 '15 at 13:04
As you're using `declare`, so you probably are using bash, just use an array. — Tom Fenech, Jul 10 '15 at 13:05
I really wish I could use arrays here, but due to some system level conflicts I have to go with this raw way. — , Jul 10 '15 at 13:14

score 0 · Accepted Answer · answered Jul 11 '15 at 10:18

0

Use eval:

i=1
declare  x$i=help
eval "echo \$x$i"

answered Jul 11 '15 at 10:18

user464502

score 0 · Answer 2 · answered Jul 12 '15 at 13:42

0

This should work:

echo  $(tmpvar=x$i && echo ${!tmpvar})

Example:

i=1
declare  x$i=help
echo  "$(tmpvar=x$i && echo ${!tmpvar})"

Output:

help

answered Jul 12 '15 at 13:42

Jahid

2 Answers2