linux bash - how to use a dynamic parameter in shell parameter expansion expression

nickleus · 08-21-2006, 03:57 AM

say that i want to use parameter expansion to parse out characters 7 and 8 in a string, but i want to generate that string dynamically inside the parameter expansion phrase. how do i do that?

format:

Code:

${parameter:offset:length}

what parameter will look like:
20060822105030

i want what is bold. this works fine on the command line:

Code:

export imorgen=$(date --date=tomorrow +%Y%m%d%H%M);echo ${imorgen:6:2}

or in a script file:

Code:

IMORGEN=$(date --date=tomorrow +%Y%m%d%H%M)
echo "today is day number ${imorgen:6:2}"

but i can't seem to figure out how to do this dynamically (without setting a variable) inside the parameter expansion phrase. i've tried this, but it doesn't work:

Code:

${$(date --date=tomorrow +%Y%m%d%H%M):6:2}

i get a bad substitution error. anybody know how to do this?

thanks in advance =)
Nick

nickleus · 08-21-2006, 04:37 AM

ok, i just realized i could do this to accomplish the same thing:

Code:

$(date --date=tomorrow +d%)

but i'm still curious as to how to dynamically generate a parameter for a shell parameter expansion expression so anybody who knows how to do that please let me know =) thanks.

Nick

unSpawn · 08-21-2006, 04:54 AM

but i can't seem to figure out how to do this dynamically (without setting a variable) inside the parameter expansion phrase.
AFAIK you can't.