title: How can I access variables that are named sequentially by a loop while still inside of the loop?
body: <p>I'm trying to understand if it's possible to create a set of variables that are numbered based on another variable (using eval) in a loop, and then call on it before the loop ends.</p>
  
  <p>As an example I've written a script called question (The fist command is to show what is the contents of the variable $tab)</p>
  
  <pre><code>(23:32:12\[deco@S.Man)
  [~/bin]$ listQpsk 40|grep -w [1-4]
  40 SMANHUBAQPSK1          1    1344      1195        88
  40 SMANHUBAQPSK1          2    1668      1470        88
  40 SMANHUBAQPSK1          3    1881      1539        81
  40 SMANHUBAQPSK1          4    1686      1409        83
  
  
  (23:18:42\[deco@S.Man)
  [~/bin]$ cat question
  #!/usr/bin/bash
  tab=`listQpsk 40|grep -w [1-4]`
  seq=1
  num=4
  until [[ $seq -gt $num ]];do
  eval count$seq=`echo "$tab"|grep -w $seq|awk '{print $5}'`
  seq=$(($seq+1))
  done
  echo $count1
  echo $count2
  echo $count3
  echo $count4
  </code></pre>
  
  <p>When I run this I get </p>
  
  <pre><code>(23:32:23\[deco@S.Man)
  [~/bin]$ ./question 
  1195
  1471
  1538
  1409
  </code></pre>
  
  <p>Which is exactly what I would expect, but is there a way to move the echo commands inside of the until loop so that part of the loop is echoing the value of the variable that was just created. Something like:</p>
  
  <pre><code>until [[ $seq -gt $num ]];do
  eval count$seq=`echo "$tab"|grep -w $seq|awk '{print $5}'`
  seq=$(($seq+1))
  echo "$count$seq"
  done
  </code></pre>
  
  <p>PS: Sorry if my formatting is off...first time posting here, and I only know markdown from reddit.</p>