Turn variable value into a new variable in bash

Question

I have a function defined below. Each time it runs through the array, I want it to update an existing variable (${1}_FLAG}). Not working as I intended it to. Basically if a new version of said package exist, set the flag to true so I can call another function.

Removed "extra" variables because people are caught up on them. What I need is to update the HTTP_FLAG PHP_FLAG and MOD_QOS variables when they're being looped through.

HTTP_FLAG=false
PHP_FLAG=false
MOD_QOS=false

PKGS=(HTTP PHP MOD_QOS);

check_new_version() {        
    # If a new version is available, download the source file
    if [ ${!check_version} != ${!current_version} ]
    then
        ...
        ...
        ${1}_FLAG=true
    fi
} 

for i in "${PKGS[@]}"
do
    check_new_version $i
done

if ${HTTP_FLAG}; then
    ...
    ...
fi

Where are you setting HTTP_VER, HTTP_CURRENT, HTTP_LINK, et. al.? — twalberg
– twalberg, Commented Nov 19, 2012 at 17:28
Above the script is rather long, I didn't think they would be relevant. An example would be HTTP_LINK="http://apache.ziply.com/httpd/${HTTP_VER}" — sdot257
– sdot257, Commented Nov 19, 2012 at 17:30
please edit your question to include demonstration of "Not working as I intended it to", with required output VS current output. OR use set -vx and set +vx to turn on/off shell debugging trace to see how each cmd/block is being processed with variable substitutions include. Good luck. — shellter
– shellter, Commented Nov 19, 2012 at 17:31

Jonathan Leffler · Accepted Answer · 2012-11-19 17:46:35Z

1

I think you need eval:

$ set -- FOO
$ FOO_FLAG=false
$ eval ${1}_FLAG=true
$ echo $FOO_FLAG
true
$

In bash, you'd like to use the ${!var} notation, but I didn't manage to find the direct invocation that would work. This gets close via the var variable, but not quite there:

$ var=${1}_FLAG
$ FOO_FLAG=false
$ ${!var}=true
bash: false=true: command not found
$ : ${!var}=true
$ echo $FOO_FLAG
false
$

answered Nov 19, 2012 at 17:46

Jonathan Leffler

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2 Comments

sdot257 Over a year ago

That is exactly the issue I'm having.

sdot257 Over a year ago

I got around this by using export "${1}_FLAG"=true

glenn jackman · Accepted Answer · 2012-11-19 22:34:32Z

0

bash has indirect parameter expansion, so you can do:

varname=${1}_LINK
value=${!varname}

example

$ foo_LINK="hello world"
$ set -- foo
$ varname=${1}_LINK
$ echo ${!varname}
hello world

answered Nov 19, 2012 at 22:34

glenn jackman

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