bash: function + source + declare = boom
- by Chen Levy
Here is a problem:
In my bash scripts I want to source several file with some checks, so I have:
if [ -r foo ] ; then
source foo
else
logger -t $0 -p crit "unable to source foo"
exit 1
fi
if [ -r bar ] ; then
source bar
else
logger -t $0 -p crit "unable to source bar"
exit 1
fi
# ... etc ...
Naively I tried to create a function that do:
function save_source() {
if [ -r $1 ] ; then
source $1
else
logger -t $0 -p crit "unable to source $1"
exit 1
fi
}
safe_source foo
safe_source bar
# ... etc ...
But there is a snag there.
If one of the files foo, bar, etc. have a global such as --
declare GLOBAL_VAR=42
-- it will effectively become:
function save_source() {
# ...
declare GLOBAL_VAR=42
# ...
}
thus a global variable becomes local.
The question:
An alias in bash seems too weak for this, so must I unroll the above function, and repeat myself, or is there a more elegant approach?
... and yes, I agree that Python, Perl, Ruby would make my file easier, but when working with legacy system, one doesn't always have the privilege of choosing the best tool.
