如何保存滑块jQuery UI的更改值

Question

我有课：

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();
        }
    });
};
}

并具有检查所有滑块总和的功能：

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);
}

函数sprawdzSume（）更改最后第五个滑块的值。 我想将滑块的值传递给此函数，然后使用该值进行一些操作。 在这里，我创建了一些对象：

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);

如何获取每个滑块的更改后的值，然后对它们进行一些操作？

Answer 1

var selection = $（“＃slider1”）.slider（“ value”）;