如何濃縮這些jQ功能？

Question

//只是在一些計算器上工作

//我確定有一個更簡單的方法，但我只是不記得該怎麼做。

//這差不多，但這個網站希望我把更多的話在這個崗位

// asdfasdfasdfsafs

$(document).ready(function(){ 
    console.log('jQ is working'); 


$(".operator1").click(function(){ 

    var a = document.getElementById('fieldA').value; 
    var b = document.getElementById('fieldB').value; 
    parseInt(a); 
    parseInt(b); 
    add = +a + +b; 
    $('#answer').append(add); 




}); 

$(".operator2").click(function(){ 

    var a = document.getElementById('fieldA').value; 
    var b = document.getElementById('fieldB').value; 
    sub = (a - b); 
    $('#answer').append(sub); 


}); 

$(".operator3").click(function(){ 

    var a = document.getElementById('fieldA').value; 
    var b = document.getElementById('fieldB').value; 
    mult = a * b; 
    $('#answer').append(mult); 

}); 

$("#operator4").click(function(){ 

    var a = document.getElementById('fieldA').value; 
    var b = document.getElementById('fieldB').value; 
    console.log(a/b); 
    divi = a/b; 
    $('divi').appened(divi); 

}); 

function doMath(a, b){ 

} 




}); 

Answer 1

一些data屬性

$(document).ready(function() { 
 
    console.log('jQ is working'); 
 

 
    $('.operator').click(function() { 
 
    var a = document.getElementById('fieldA').value; 
 
    var b = document.getElementById('fieldB').value; 
 
    a = parseFloat(a); 
 
    b = parseFloat(b); 
 
    var res = eval(a +$(this).attr('data-opr')+b); 
 
    $('#answer').append(':'+res); 
 
    }); 
 

 

 
    function doMath(a, b) { 
 

 
    } 
 

 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<button class="operator" data-opr="+">+</button> 
 
<button class="operator" data-opr="-">-</button> 
 
<button class="operator" data-opr="*">*</button> 
 
<button class="operator" data-opr="/">/</button> 
 
<input type="number" value="2" id="fieldA"> 
 
<input type="number" value="9" id="fieldB"> 
 
<p id="answer"><p>

做