如何保存滑塊的jQuery UI的變化值

Question

我有類：

function Slider(name,orientation,range,disabled,min,max,value) { 
this.name = name; 
this.orientation = orientation; 
this.range = range; 
this.disabled = disabled; 
this.min = min; 
this.max = max; 
this.value = value; 
this.setSlider = function() { 
    jQ("#"+this.name).slider({ 
     orientation: ''+this.orientation, 
     range: this.range, 
     disabled: this.disabled, 
     min: this.min, 
     max: this.max, 
     value: value, 
     change: function(event, ui) { 
      this.value = jQ("#"+name).slider("option", "value"); 
      console.log(value); 
      sprawdzSume(); 
     } 
    }); 
}; 
} 

和功能檢查所有滑塊的總和：

function sprawdzSume() { 
var cena = 0; 
if (slider3.value === 1) { 
    cena = slider1.value + slider2.value + slider3.value + 1 + slider4.value; 
} 
else{ 
    cena = slider1.value + slider2.value + slider3.value + slider4.value; 
}; 
if (cena > 7) { 
    jQ(function() { 
     jQ(slider5.name).slider({ 
      value: 3 
     }); 
    }) 
} 
else if (cena > 5 && cena < 8) { 
    jQ(function() { 
     jQ(slider5.name).slider({ 
      value: 2 
     }); 
    }) 
} 
else if (cena > 3 && cena < 6) { 
    jQ(function() { 
     jQ(slider5.name).slider({ 
      value: 1 
     }); 
    }) 
} 
else { 
    jQ(function() { 
     jQ(slider5.name).slider({ 
      value: 0 
     }); 
    }) 
};  
console.log(cena); 
} 

功能sprawdzSume（）最後的變化值第五滑塊。我想將滑塊的值傳遞給此函數，然後使用該值執行一些操作。在這裏，我創建了一些對象：

var slider1 = new Slider('slider1','vertical',false,false,0,1,1); 
var slider2 = new Slider('slider2','vertical',false,false,0,3,3); 
var slider3 = new Slider('slider3','vertical',false,false,0,1,1); 
var slider4 = new Slider('slider4','vertical',false,false,0,3,3); 
var slider5 = new Slider('slider5','vertical',false,true,0,3,3); 

如何獲取每個滑塊的更改值，然後對它們執行一些操作？

Answer 1

var selection = $（「＃slider1」）.slider（「value」）;