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Complex variable assignment

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# 1  
Old 05-05-2011
Complex variable assignment
Hi,
I have a requirement as follows, need to call the format of ${$var} form.

For example, i am taking a variable.
Code:
  count=1, 
  ((LIMIT_$count=$count + 1))

Now i have to echo this variable LIMIT_$count. (This is in a loop..)
echo ${LIMIT_$count} - displays as a syntax error.

Can anyone please help me with this.

Thanks,
Abhishek S.

Moderator's Comments:
Mod Comment Please use [code] and [/code] tags when posting code, data or logs etc. to preserve formatting and enhance readability, thanks.
Last edited by zaxxon; 05-05-2011 at 09:24 AM.. Reason: code tags
# 2  
Old 05-05-2011
Code:
count=1
while [ $count -le 10 ]
do
    echo $count
    let count+=1   #or   count=$(( $count + 1 ))
done

# 3  
Old 05-05-2011
I think this is what you want:

Use 'eval'.

Ex:
Code:
#!/usr/bin/ksh93

COUNT=1
eval LIMIT_${COUNT}=$(($COUNT + 1))

echo $LIMIT_1
eval echo \${LIMIT_$COUNT}

Output:

Code:
>
2
2

This User Gave Thanks to purdym For This Post:
abhisheksunkari
# 4  
Old 05-13-2011
MySQL
Hey purdym,
Sorry for the late response, but thats the exact thing i was looking for. Thanks a lot again man.. !! It works for me..
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