如何重新计算选项：在首次计算后更改选项时选择？

Question

这是我的第一篇文章，我有两个问题：

1. 当我选择（答案）一切从顶部到底部的计算 工作正常。如果我改变“日期”，它会重新计算，但是当我改变“box2”之后，它不会被重新计算。有解决方案吗？

2. 因为这是我的第一个脚本，我想知道它是否有什么好处？

   $(document).ready(function() { 
   
   $("#box").chained("#box2").change(function() { 
   
   var selector = function (dateStr) { 
   var d1 = $('#from').datepicker('getDate'); 
   var d2 = $('#to').datepicker('getDate'); 
   var diff = 0; 
   
       if (d1 && d2) { 
        diff = Math.floor((d2.getTime() - d1.getTime())/86400000); // ms per day 
       } 
   
       $('#days').val(diff); 
   
   var ttl = $('#totaal'); 
   var discon = $('#discount-container'); 
   var disc = $('#discount'); 
   var tax = $('#tax'); 
   var price = $('#box option:selected').attr('data-price'); 
   var total = diff * price; 
   var tax7 = total*0.07; 
   var disc75 = total*0.075; 
   var disc15 = total*0.15; 
   
   
   if (diff >7 && diff <15) { 
       discon.show('slow'); 
       disc.val("- " + disc75.toFixed(2)); 
       tax.val("+ "+ tax7.toFixed(2)); 
       ttl.val("Naf. "+ (total+tax7).toFixed(2)); 
   } 
   else if (diff >14) { 
       discon.show('slow'); 
       disc.val("- " + disc15.toFixed(2)); 
       tax.val("+ "+ tax7.toFixed(2)); 
       ttl.val("Naf. "+ (total+tax7).toFixed(2)); 
   } 
   else { 
       discon.hide('slow'); 
       tax.val("Naf. "+ tax7.toFixed(2)); 
       ttl.val("Naf. "+ (total+tax7).toFixed(2)); 
   } 
   } 
   
       $('#from').datepicker({ 
       dateFormat: 'D dd-mm-yy', 
       minDate: '+1D', 
       maxDate: '+2Y', 
       changeMonth: true, 
       numberOfMonths: 1, 
       onClose: function(selectedDate) { 
       $("#to").datepicker("option", "minDate", selectedDate); 
       } 
   }); 
   $('#to').datepicker({ 
       dateFormat: 'D dd-mm-yy', 
       minDate: '+1D', 
       maxDate: '+2Y', 
       changeMonth: true, 
       numberOfMonths: 1, 
       onClose: function(selectedDate) { 
       $('#from').datepicker("option", "maxDate", selectedDate); 
       } 
   }); 
   
   $('#from,#to').change(selector) 
   
   
   }); 
   
   }); 
   

Answer 1

1.

昌$("#box").chained("#box2").change(function() {
到$("#box, #box2").change(function() {

2. 您的代码看起来不错，但因为你没有的元素很多情况下，我建议你放弃将每个选择器定义为一个变量。