Question

基本上尝试将变量添加到一起，然后将它们传递给最终变量

//-- answers --//
var groupA1 = 0;
var groupA2 = 0;
var groupA3 = 0;

var answerQuestion1 = 0;

testButton.on('click', function () {
    answerQuestion1 = ((groupA1).val) + ((groupA2).val) + ((groupA3).val);
    console.log("I am being clicked");
    console.log(answerQuestion1);
    console.log(groupA1);
    console.log(groupA2);
    console.log(groupA3);

});

任何人都知道为什么'answerQuestion1'正在安慰'NaN'？

干杯

Answer 1

在添加之前使用parseInt，以便它不会连接为字符串

//-- answers --//
var groupA1 = 0;
var groupA2 = 0;
var groupA3 = 0;

var answerQuestion1 = 0;

$('button').on('click', function () {
    answerQuestion1 = parseInt($('[name="groupA1"]').val()) + parseInt($('[name="groupA2"]').val()) + parseInt($('[name="groupA3"]').val());
    console.log("I am being clicked");
    console.log(answerQuestion1);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="number" value="3" name="groupA1">
<input type="number" value="2" name="groupA2">
<input type="number" value="1" name="groupA3">
<button>Compute</button>

Answer 2

变量未与.val一起使用，并使用parseInt进行计算，否则使用“+”运算符进行字符串连接。

var groupA1 = 0;
var groupA2 = 0;
var groupA3 = 0;

var answerQuestion1 = 0;

$('#testButton').on('click', function() {
  answerQuestion1 = parseInt(groupA1) + parseInt(groupA2) + parseInt(groupA3);
  console.log("I am being clicked");
  console.log(answerQuestion1);
  console.log(groupA1);
  console.log(groupA2);
  console.log(groupA3);

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<input type='button' id='testButton'>

将变量值传递给变量

2 个答案: