通变量值到变量

Question

本质尝试添加变量在一起，然后将它们传递到最终的变数

//-- 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); 

}); 

谁知道为什么“answerQuestion1”被安慰的“男”？

干杯

Answer 1

变量还没有与.val使用和使用parseInt进行计算，否则 “+” 之前字符串连接的运算符。

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'>

Answer 2

使用parseInt函数增加，所以它不会串连为字符串

//-- 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>