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Full Discussion: Variable substitution with arrays
Top Forums Shell Programming and Scripting Variable substitution with arrays Post 303003528 by RudiC on Friday 15th of September 2017 09:14:09 AM
Old 09-15-2017
Quote:
Originally Posted by bakunin
I hate to say it, but: this was to be expected. What you tried was a so-called "associative array". This is an array where the index is not numbers but (arbitrary) strings. There some programming languages which offer this kind of arrays (awk, for instance), but not bash.
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As much as I hate to say it: ever since version 4 (which the OP seems to run), bash HAS associative arrays:

man bash:
Quote:
Arrays
Bash provides one-dimensional indexed and associative array variables.
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Associative arrays are created using declare -A name.
You need to declare / typeset them correctly, though ...

Code:
typeset -A arr=([animals]="dog cat horse penguin cow" [fruits]="orange apple grapes peach mango" [drinks]="juice milk coffee tea coke" [cities]="toronto paris london glasgow sydney" [countries]="canada france england scotland australia")
for i in ${!arr[@]}
  do    echo $i
        TMP=(${arr[$i]})
        for j in ${!TMP[@]}
          do    echo "  " ${TMP[$j]}
          done
  done
animals
     dog
     cat
     horse
     penguin
     cow
drinks
     juice
     milk
     coffee
     tea
     coke
fruits
     orange
     apple
     grapes
     peach
     mango
countries
     canada
     france
     england
     scotland
     australia
cities
     toronto
     paris
     london
     glasgow
     sydney

Last edited by RudiC; 09-15-2017 at 10:32 AM..
These 2 Users Gave Thanks to RudiC For This Post:
bakunin Kingzy
 

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