Assign variables inside for loops

问题

I am trying a small code which is

for(( i =0;i<2;i++ )); do p$i=\"pra$i\"; done

expected output is: Variable must be assigned

p0="pra0"
p1="pra1"

But bash is taking that as command and am getting output as

p0="pra0": command not found
p1="pra1": command not found

Thanks

回答1:

Use eval to have the value evaluated and stored as you want:

$ for(( i =0;i<2;i++ )); do eval p$i=\"pra$i\"; done
$ echo $p1
pra1

---

Or better with declare (thanks chepner as always!):

$ for(( i =0;i<2;i++ )); do declare "p$i=pra$i"; done
$ echo $p1
pra1

回答2:

for (( i =0;i<2;i++ )); do
  printf -v "p$i" '%s' "pra$i"
done

来源：https://stackoverflow.com/questions/22036966/assign-variables-inside-for-loops