Question

使用JSON我得到的價值，從thati嘗試使用下面的腳本圓值：

  function onLoad(){ 
      var output = $('#product'); 
      var id = getUrlVars()["cPath"]; 
      $.ajax({ 
       url: 'getprice.php?cPath='+id, 
       data : {type : 'product'}, 
       dataType: 'jsonp', 
       jsonp: 'jsoncallback', 
       timeout: 5000, 
       success: function(data){ 
      $.each(data, function(c,prd){ 

       var price =Math.round(prd.products_price,2); 




          var product = '<span class="normalprice">$' + price + '</span>' ; 
          output.append(product); 
          } 

         }); 
        output.listview("refresh"); 
         //data loaded 
        }, 
        }); 
     } 

//輸出需要

prd.products_price=450.2500 
    price=$450.25 

也嘗試這種格式但不工作：

  1. var num = Math.val(prd.products_price); 
       var price = num.toFixed(2); 

Answer 1

試試這個：

var price = parseFloat(prd.products_price).toFixed(2); 

Answer 2

var num = Math.val(prd.products_price); 
var num = parseFloat(num); 
var price=num.toFixed(2)); 

Answer 3

var price = prd.products_price; 
var product = price.formatMoney(0, "") 

Number.prototype.formatMoney = function(places, symbol, thousand, decimal) { 
     places = !isNaN(places = Math.abs(places)) ? places : 2; 
     symbol = symbol !== undefined ? symbol : "$"; 
     thousand = thousand || ","; 
     decimal = decimal || "."; 
     var number = this; 
     var negative = number < 0 ? "-" : ""; 
     var i = parseInt(number = Math.abs(+number || 0).toFixed(places), 10) + ""; 
     var j = (j = i.length) > 3 ? j % 3 : 0; 

     return symbol 
       + negative 
       + (j ? i.substr(0, j) + thousand : "") 
       + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + thousand) 
       + (places ? decimal + Math.abs(number - i).toFixed(places).slice(2) : ""); 
    }