How can I access variables that are named sequentially by a loop while still inside of the loop?
Question
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.
As an example I've written a script called question (The fist command is to show what is the contents of the variable $tab)
(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
When I run this I get
(23:32:23\[deco@S.Man)
[~/bin]$ ./question
1195
1471
1538
1409
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:
until [[ $seq -gt $num ]];do
eval count$seq=`echo "$tab"|grep -w $seq|awk '{print $5}'`
seq=$(($seq+1))
echo "$count$seq"
done
PS: Sorry if my formatting is off...first time posting here, and I only know markdown from reddit.
Solution
Use indirection:
until [[ $seq -gt $num ]];do
var="count$seq"
eval $var=$(echo "$tab"|awk -v seq=$seq '$3==seq {print $5}') # awk loves to do grep's job
let seq+=1 # another way
echo "${!var}" # indirection
done
No need for another eval.
OTHER TIPS
Not exactly answering your question, but... did you know bash has array variables?
seq=1
num=4
until [[ $seq -gt $num ]];do
count[$seq]=`echo "$tab"|grep -w $seq|awk '{print $5}'`
seq=$(($seq+1))
done
echo ${count[*]}
Or without arrays:
seq=1
num=4
until [[ $seq -gt $num ]];do
eval count$seq=`echo "$tab"|grep -w $seq|awk '{print $5}'`
eval echo \$count$seq
seq=$(($seq+1))
done
Yes, like this (in bash at least):
$ count1=xyz
$ seq=1
$ value=`eval echo \$\{count$seq\}`
$ echo $value
Try this:
varName=a
for i in 1 2 3
do
eval $varName$i=stuff
eval var=\$$varName$i
echo $var
done
some versions of shell also let you use ${!var} to indirectly reference variables