How can I access variables that are named sequentially by a loop while still inside of the loop?

问题

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.

回答1:

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.

回答2:

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

回答3:

Yes, like this (in bash at least):

$ count1=xyz
$ seq=1
$ value=`eval echo \$\{count$seq\}`
$ echo $value

回答4:

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

来源：https://stackoverflow.com/questions/1030532/how-can-i-access-variables-that-are-named-sequentially-by-a-loop-while-still-ins

标签

bash

variables

loops

eval