bash parameter indirection possible?

kornelix · 11-29-2005, 10:27 AM

I want to get the value of a parameter whose name (unknown) is in another parameter (known).

$ n=2
$ set aaa bbb
$ p="$"$n
$ echo $n $2 $p
2 bbb $2

$p contains the name $2 whose value "bbb" I want to get.
Assume we know only the name of $p.

$ echo $(echo $p)
$2
###### (expect bbb, not $2)

$ pp=$(echo $p)
$ echo $pp
$2
####### (ditto)

####### (try an extra indirection)
$ pp=$(echo $(echo $p))
$ echo $pp
$2

A) can someone explain why it does not work?
B) how can I do what I want to do?

bigearsbilly · 11-29-2005, 10:43 AM

eval

Code:

billym.>a=hello
billym.>b=a
billym.>eval c=\$$b
billym.>echo $c
hello

Hko · 11-29-2005, 11:07 AM

(assuming bash)

Code:

#!/bin/bash

a=hello
b=a
echo ${!b}

Or, for kornelix' example:

Code:

#!/bin/bash

n=2
set aaa bbb
p=${!n}
echo $n $2 $p

bigearsbilly · 11-30-2005, 03:28 AM

doesn't work on my ksh Hko

and you have
#!/bin/bash at the top

SunOS primadtpdev 5.8 Generic_108528-20 sun4u sparc SUNW,Ultra-250

kornelix · 11-30-2005, 07:52 AM

thanks billy and Hko. works fine.

I understand the eval, I think
(interpret (interpret xxxxx))

I need to dig on the use of {!string}.
Can't find it in man bash.

bigearsbilly · 11-30-2005, 08:35 AM

eval will work on sh, bash, ksh,

it's best to not get too bash-specific if you work in
the outside world.