jquery - 如何重新计算选项：第一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. 当我从上到下选择（回答）所有内容时，计算工作正常。如果我更改“日期”，它会重新计算，但是当我之后更改“box2”时，它不会重新计算。有解决办法吗？

2. 因为这是我的第一个脚本，所以我想知道它是否有用？

   $(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.

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

2. 您的代码看起来不错，但是因为您没有很多元素实例，我建议您放弃将每个选择器定义为变量。