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Old 08-21-2006, 03:57 AM   #1
nickleus
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linux bash - how to use a dynamic parameter in shell parameter expansion expression


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

Last edited by nickleus; 08-21-2006 at 03:59 AM.
 
Old 08-21-2006, 04:37 AM   #2
nickleus
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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
 
Old 08-21-2006, 04:54 AM   #3
unSpawn
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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.
 
  


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