my var in bash doesn't work: ARRAY_$a_$b="test"

SlackerJack · 01-12-2008, 02:42 PM

Hi,

I'm trying to kind of emulate an array by getting this to work:

Code:

for MONTH in jan feb mar
do
  for DAY in 1 2 3
  do
      
    # numbers work fine
    let ARRAY_$MONTH_$DAY=$DAY
    
    # strings don't work <<<<<<<<<<<<<<<<<
    #ARRAY_$MONTH_$DAY="test"
    #set ARRAY_$MONTH_$DAY="test"
    #ARRAY_${MONTH}_${DAY}="test"
    #set ARRAY_${MONTH}_${DAY}="test"
    # help please
    
    # print contents
    eval echo \$ARRAY_$MONTH_$DAY
    
  done
done

So, do you know how to get this to work: ARRAY_$MONTH_$DAY="test"

Thanks

colucix · 01-12-2008, 04:58 PM

You may find very useful to use the -x option of bash for debugging: if you'd run the script as in

Code:

bash -x script.sh

you would be aware that it doesn't work in the way you expected (neither with numbers nor with strings). To assign variables (without the let statement) you can't use other variables to build the name in the left-hand side of the assignment, otherwise the entire line is interpreted as a command. If you really want to build the variable name using the value of any other variable, you have to consider the eval statement, as you have done for the "print section". For example:

Code:

my_var=hello
eval ${my_var}_world=75      # hello_world=75
echo $hello_world            # 75

Then be careful when you put together (concatenate) variables and strings: the statement

Code:

let ARRAY_$MONTH_$DAY=$DAY

has the string "ARRAY_", the value of a variable $MONTH_ (undefined) and the value of the variable $DAY. In this case the syntax $MONTH cannot be used: the most correct syntax ${MONTH} avoids errors. In summary, a correct version of the script could be

Code:

for MONTH in jan feb mar
do
  for DAY in 1 2 3
  do  
    eval ARRAY_${MONTH}_${DAY}=${DAY}
    eval echo \$ARRAY_${MONTH}_${DAY}
    
    eval ARRAY_${MONTH}_${DAY}="test"
    eval echo \$ARRAY_${MONTH}_${DAY}
    
  done
done

where the brackets in red are mandatory! Also, when you have the basics clear, you may consider to use arrays in bash... without trying to emulate them!

SlackerJack · 01-12-2008, 06:40 PM

Thank you for a lengthy and (very!) informative answer.

Your answer obviously solved my problems.

So close, yet so far away :-)

Quote:

You may find very useful to use the -x option of bash for debugging

Nice, that will come in handy.

Quote:

consider to use arrays in bash

Bash can't handle two-dimentional/multi-dimentional arrays, can it? That's why I had to do something of my own.

Again, great reply...

colucix · 01-13-2008, 03:45 AM

Quote:

Originally Posted by SlackerJack

Thank you for a lengthy and (very!) informative answer.

You're welcome!

Quote:

Bash can't handle two-dimentional/multi-dimentional arrays, can it?

It can't indeed. You should consider other languages like perl or gawk, even if they manage multi-dimensional arrays with artifacts. For example, here's an excerpt from "GAWK: Effective AWK Programming" guide:

Quote:

Multidimensional arrays are supported in awk through concatenation of indices into one string. awk converts the indices into strings (see Section 5.4 [Conversion of Strings and Numbers], page 77) and concatenates them together, with a separator between them. This creates a single string that describes the values of the separate indices. The combined string is used as a single index into an ordinary, one-dimensional array. The separator used is the value of the built-in variable SUBSEP.

that sounds very similar to what you're trying to accomplish in bash. Obviously before spending time to learn some other language, you have to evaluate if it's worth to do, depending on your prime objectives.

One thing I forgot to mention in my previous post: it is correct to use the let statement to assign a variable. Indeed it evaluates the expression and let build the name of the variables from other names or let arithmetic expansion like

Code:

my_var=8
let my_var=my_var+10
echo $my_var          # 18

another_var=68
let my_var="75 - $another_var"
echo $my_var          # 7 and so on...

Ciao and good luck!