script help needed

chatguy2020 · #1 (**permalink**) 04-06-04, 15:47

I want to initialize variables in a loop as shown below (commented). It gives errors Ex: COUNT1=1: not found. Similarily the statement below it gives errors: 1=COUNT1: not found. etc...

If I say for example,

COUNT1=`echo "$ALLCOUNTS" | cut -d" " -f1`
COUNT2=`echo "$ALLCOUNTS" | cut -d" " -f2` the variables initialize well.

Can somebody please tell why?

Code:

#
i=1 
while [ $i -lt $COUNT_DESIRED ]
do
    #COUNT$i=`echo "$ALLCOUNTS" | cut -d" " -f$i`
    [ -z $COUNT$i ] && COUNT$i=11
    ${allcounts[$i-1]}=COUNT$i
    i=`expr $i + 1`
done

aigles · #2 (**permalink**) 04-06-04, 17:01

You can do something like this :

Code:

i=1 
while [ $i -le $COUNT_DESIRED ]
do
    #COUNT$i=`echo "$ALLCOUNTS" | cut -d" " -f$i`
    eval Count=\$COUNT$i
    allcounts[$((i-1))]=${Count:-11}
    (( i += 1 ))
done

chatguy2020 · #3 (**permalink**) 04-09-04, 11:10

Aigles, Thanks for the reply.

i=1
while [ $i -le 5 ]
do
COUNT$i=$i
(( i+= 1 ))
done

Here my idea is just to initialize 5 variables dynamically. The above approach won't work. I tried using

i=1
while [ $i -le 5 ]
do
((COUNT$i=$i))
(( i+= 1 ))
done

it won't work either. Any ideas? Thanks.

aigles · #4 (**permalink**) 04-09-04, 13:03

First approach

Code:

i=1
while [ $i -le 5 ]
do
   eval COUNT$i=$i
   (( i+= 1 ))
done

The second approach works fine for me, what is your problem ?
execute with -x option to verify variable creation.

Note: The two solutions works only with bash and ksh due to '(( ))' syntax usage.
(( i+= 1)) can be replaced by i=`expr $i + 1`

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