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Nicolás Arévalo
Asked: 2020-07-22 11:25:47 +0800 CST 2020-07-22 11:25:47 +0800 CST

Why doesn't the counter work?

  • 772

My code:

#! /bin/bash
contador = 0
while :
do
        echo -n "Escriba el nmero:"
        read numero
        case $numero in
        (""|*[^0-9]*) printf "\nError: [numero] no es un caracter valido\n";;
        (*) printf "\nEl primer digito es: ${numero:0:1}";;
        esac
        echo -n "Si quiere volver a intentarlo presione 1, de lo contrario presiona cualquier otra cosa"
        read volver
        if [[ $volver != 1 ]]
        then
             echo "El nmero de vces usado fue $contador"
             echo "Adios"
             sleep 2
             break
        fi
        contador = contador + 1
done

When I run it, it tells me:

line 26: counter: not found.

I thought it was 0 by default and that's why at first I did it without defining the counter outside of the while. Since it didn't help, I defined it outside but nothing, the same thing continues, the counter had it up and I passed it to the end of the while but nothing, the same.

Thanks in advance.

variables

1 Answers

  1. Best Answer

    The assignment cannot have spaces around the equals, that is, line 2 must be:

    contador=0

    And on the other hand, the line in which you try to increment the counter has several problems. On one side, again the spaces around the equal. On the other hand, on the right you should use $contadorinstead of contadorto get the value it had, but finally, most important of all, the shell language is not a programming language, but a command execution language, and it tends to consider the variable values ​​as strings instead of numbers. The line:

    contador=$contador+1

    it will not produce the expected result, but, for example, if the loop is executed three times, in the end it $contadorwill be valued 0+1+1(as a string).

    For mathematical calculations you must use letwith a syntax like the following:

    let "contador = $contador + 1"

    Notice that letyou pass a string to it (all in quotes), but the $contadorwill be expanded to its value inside the quotes, so it letwill see something like contador = 0 + 1the first iteration, and it will mathematically evaluate the expression, which will return it 1, and will assign to contador. When you use let, by enclosing the expression in quotes, you can use spaces more freely.

    • 5