Declaring different variables using for loop in a shell script-CodePudding

I am trying to write a function to input info about different people, for instance, five students. I intend to do it using for loop and without declaring an array.

function student() {
    for (( i = 0; i < 5; i   )); do
        echo "Name"
        read name$i
    done
   }

I believe that at every iteration it would be a different variable name$i the input is read for. Now when I try to display the same data using a for loop

function student() {
    for (( i=0;i < 5; i  )); do
    echo $(name$i)
    done
}

it shows an error name0: command not found

Now when I display the values explicitly like echo "$name0" echo "$name1", it all works fine which tells me that the variables are all declared and input is populated. The problem is that I may not be accessing the variable names right while trying to display the values because when I write echo "$name$i", the output is name0. It means that it takes "name" as a string and the values of i. I have tried to concatenate the string "name" with values of "i" using a couple of ways. There is a way to bind both like namen="${i}name" but I want the integer values after the string so it didn't work for me. I am new to shell scripting, so all your worthy suggestions would be helpful.

CodePudding user response：

Setup:

$ name0=0; name1=1; name2=2; name3=x; name4=y

Running a web search on bash dynamic variables should provide a good number of options, among them:

indirect reference

student() {
for ((i=0; i<5; i  ))
do
    x="name${i}"
    echo "${!x}"
done
}

nameref

student() {
for ((i=0; i<5; i  ))
do
    declare -n x="name${i}"
    echo "${x}"
done
}

Both of these generate:

$ student
0
1
2
x
y