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Full Discussion: Array help (variable substitution).
Top Forums Shell Programming and Scripting Array help (variable substitution). Post 302113971 by cfajohnson on Wednesday 11th of April 2007 04:38:33 PM
Old 04-11-2007
Quote:
Originally Posted by dsimpg1
No, I'm trying to do this:

cmlist01 has a b c d e
cmlist02 has f g h i
cmlist03 has j k l

For the total in cmlist01, I should see 5 items.
For the total in cmlist02, I should see 4 items.
For the total in cmlist03, I should see 3 items.

As a result, I could do:

print "There are ${#cmlist01[@]} items in $listnbr:"
which prints "There are 5 itmes in cmlist01."

However, I want to do all of this in a for loop and have it derive the cmlistnn from the variable $listnbr (which is does just fine as evidenced in the output from the print statement). However, if I try to get the count (number of items in the array) using listnbr (instead of hardcoding cmlistnn in each statement), all I ever get is 1.

If I set the variable listnbr=cmlist01, I should see 5 items, but am only seeing 1 when I do:

print "There are ${#listnbr[@]} items in $listnbr:"
which prints "There are 1 itmes in cmlist01."

The main part of the code that I need this for is:

Code:
while [ $i -lt ${#cmlist03[@]} ]
     do
          ID=${cmlist03[$i]}
          echo "Processing $i with ${cmlist03[$i]}"

Maybe you just can't do what I'm trying to do...

Code:
cmlist01=( a b c d e)
cmlist02=( f g h i )
cmlist03=( j k l )

for listnbr in cmlist01 cmlist02 cmlist03
do
  eval "printf 'There are %d items in %s:\n' \"\${#${listnbr}[@]}\" \"$listnbr\""
  eval "printf '\t%s\n' \"\${$listnbr[@]}\""

done

This User Gave Thanks to cfajohnson For This Post:
Arai
 

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