Increment variable name results in command not found

Question

I am trying to increment a variable name based on the input and call the value after the loop is done.

for i in `seq 5`; do
  var$i="number is $i"
done
 echo $var2 $var5

Results in

./test.sh: line 4: var1=number is 1: command not found
./test.sh: line 4: var2=number is 2: command not found
./test.sh: line 4: var3=number is 3: command not found
./test.sh: line 4: var4=number is 4: command not found
./test.sh: line 4: var5=number is 5: command not found

There are 2 things I don't understand:

1. "var1=number is 1" is understood as a command.
2. var2 and var5 are actually generated but they are not displayed outside of the loop.

Answer 1

To achieve the outcome required, the use of arrays are needed and so:

#!/bin/bash
for i in $(seq 5)
do
  var[$i]="number is $i"
done
for i in "${var[@]}"
do
    echo "$i"
done

Set the index and values for the array var accordingly and then loop through the array and print the values.

Answer 2

You cannot use variable names with a number in it like that: you really need to say:

var1=1
var2=2
var3=3
var4=4
var5=5

Another approach is the usage of arrays.

As far as increasing is concerned (this is not part of the question, but I give is anyways), you might use something like this:

Prompt> var1=1
Prompt> var2=$((var1+1))
Prompt> echo $var2
2

(Mind the double brackets for making calculations)

Answer 3

1. "var1=number is 1" command not found because that's not a command. For example, you can write:

for i in `seq 5`; do
  echo var$i="number is $i"
done

And the output would look like

var1=number is 1
var2=number is 2
var3=number is 3
var4=number is 4
var5=number is 5

2. The variables have not been generated, you can't just dynamically generate variables. Try to use arrays