在函数将两个输入相加在一起时，从总字段中删除NaN

Question

我将两个字段相加在一起以得到总数，但是当我在第一个字段中键入时，总字段显示NaN，而我不需要它。我尝试了5种不同的选项来解决这个问题，但是在函数运行时，似乎没有一个可以删除NaN。

$j(function() {

    var new_fees = 0.00; 
    var referral_fees = 0.00;

    $j('#new_fees').keyup(function() {  
        updateTotal();
    });

    $j('#referral_fees').keyup(function() {  
        updateTotal();
    });

    var net_fees=parseFloat($j('#net_fees').val());    
        if(isNaN(net_fees)){
    net_fees=0;
}

    var updateTotal = function () {
      new_fees = parseFloat($j('#new_fees').val());
      referral_fees = parseFloat($j('#referral_fees').val());
      $j('#net_fees').val(new_fees + referral_fees);

    var output_total = $j('#net_fees');
    var total = new_fees + referral_fees;

    output_total.val(total);
    };
 });

这里是琴杆，但是它在我的页面上所做的工作中似乎根本不起作用。

Answer 1

我更新了你的"updateTotal“方法，请试试。

var updateTotal = function () {
     new_fees = parseFloat($('#new_fees').val());
     referral_fees = parseFloat($('#referral_fees').val());
     $('#net_fees').val(new_fees + referral_fees);
     var output_total = $('#net_fees');
     var total = new_fees + referral_fees;
     if(isNaN(total)){
         total=0;
     }
     output_total.val(total);
};

试试https://jsfiddle.net/ydLtcrya/1/

Answer 2

尝尝这个

​

var new_fees = $("#new_fees").val();
var referral_fees = $("#referral_fees").val();

$('#new_fees').keyup(function() {  
        updateTotal();
});

$('#referral_fees').keyup(function() {  
        updateTotal();
});

var updateTotal = function () {
   new_fees = parseFloat($('#new_fees').val());
   referral_fees = parseFloat($('#referral_fees').val());
   if(isNaN(new_fees))
   net_fees=0;
   if(isNaN(referral_fees))
   referral_fees=0;
   $('#net_fees').val(new_fees + referral_fees);
	  
   var output_total = $('#net_fees');

   var total = new_fees + referral_fees;
   if(isNaN(total ))
   total=0.00;
   output_total.val(total);
};

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="col-xs-12 col-sm-2 nopadding">
  <label for="template_range" class="col-xs-11 col-sm-11 col-md-11 col-lg-11 nopadding">New Fees<br><small>(net)</small></label>
  <input type="text" class=" col-xs-5 form-control" id="new_fees" placeholder="0.00">
</div>															
<div class="col-xs-12 col-sm-2 nopadding">
  <label for="template_range" class="col-xs-11 col-sm-11 col-md-11 col-lg-11 nopadding">Referral Fees<br> <small>(net)</small></label>
  <input type="text" class="col-xs-5 form-control" id="referral_fees" placeholder="0.00" >
</div>														
<div class="col-xs-12 col-sm-2 nopadding">
  <label for="template_range" class="col-xs-11 col-sm-11 col-md-11 col-lg-11 nopadding">Total Fees <br><small>(net)</small></label>
  <input type="text" class="col-xs-5 form-control" value="0.00" id="net_fees" readonly>
</div>
<div class="col-xs-12 col-sm-2 nopadding">
  <label for="template_range" class="col-xs-11 col-sm-11 col-md-11 col-lg-11 nopadding">Total Fees <br><small>(gross)</small></label>
  <input type="text" class="col-xs-5 form-control" value="0.00" id="gross_fees" readonly>
</div>

​

Answer 3

如果parseFloat返回NaN，则默认为0。

var updateTotal = function () {
  new_fees = parseFloat($j('#new_fees').val()) || 0;
  referral_fees = parseFloat($j('#referral_fees').val()) || 0;
  $j('#net_fees').val(new_fees + referral_fees);

  var output_total = $j('#net_fees');
  var total = new_fees + referral_fees;

  output_total.val(total);
};