Question

My function in JavaScript/JQuery doesn’t return anything; I want it to return the result for me but it won’t. As you can see, the return is the $("#resultado").html("O Valor do produto ficara "+resultado); but this doesn’t appear as it should when the function is run:

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<script type="text/javascript">
     $("#target").submit(function(){

    var preco_dolar = $("#preco_dolar").val(); 
    var dolar = $("#dolar").val();
    var frete = $("#frete").val();


    if ($('#drop').is(':checked'))
    {
    var drop = 5.8;
    }else var drop = 0; 


    if(typeof preco_dolar == "undefined"){
        alert("Erro ! Falta preco do produto");
    }

    if(typeof frete == "undefined"){
        var frete = 40;
    }

    if(typeof dolar == "undefined"){
        var dolar = 2.16;
    }




    var resultado = (preco_dolar*dolar)+frete;

    $("#resultado").html("O Valor do produto ficara "+resultado);




});
</script>

Can someone point out where I’m going wrong please?

Answer 1

First you need to wrap your script into

// here $("#target") is nothing

$(document).ready(function () {
    // here $("target") exists
});

Second for your improvement, don’t use

if(typeof dolar == "undefined"){
    dolar = 2.16;
}

Use instead

if(!dolar){
    dolar = 2.16;
}

That is the same as you would do:

if(!dolar && dolar == 0 && dolar == "" && dolar == false && dolar == undefined){
    dolar = 2.16;
}

To make 0 a valid value, use

if (!dolar && dolar !== 0) {
    dolar = 2.16;
}

Also don’t use var dolar = 0; twice. If you declared it already, simply use dolar =, to not overwrite the already declared var.