Question

根据测验的得分，任务是产生几根浅色和深色羽毛。浅色羽毛代表正确的点（light_feather），黑色羽毛是不正确的点（dark_feather）（每个都被跟踪）。所有的羽毛都应该在一条线上排成一行，这意味着第一根羽毛，然后是黑色的羽毛。我得到了测验动力学，我在这里发布的功能仅适用于他们按下结束测验的时候。

var light_feather:LightFeather = new LightFeather();
var dark_feather:DarkFeather = new DarkFeather();

var good_answers:uint = 0;
var bad_answers:uint = 0;

function avsluttFunc (evt:MouseEvent) 
{   
    var sum_LightFeatherX:Number = 0;
    for (var i = 0; i < good_answers; i++) {
        addChild(light_feather);
        light_feather.x += 12 + (i*16);
        light_feather.y = 0;
        trace("Lys X-verdi: " + light_feather.x);
        sum_LightFeatherX += Number(light_feather.x);
        return sum_LightFeatherX;
    }
    trace(sum_LightFeatherX);

    dark_feather.x += sum_LightFeatherX;
    for (var j = 1; j <= bad_answers; j++) {
        addChild(dark_feather);
        dark_feather.x += 12 + (j*16);
        dark_feather.y = 0;
        trace("Mørk X-verdi: " + dark_feather.x);
    }

    /*
        //Resetter poengsummen
        good_answers = 0;
        bad_answers = 0;
    */
}

Answer 1

你可以只用一个for循环来做你想要的东西，看看：

ei32cu.write_database()

此代码示例将在2个var good_answers:uint = 2; var bad_answers:uint = 4; function avsluttFunc(evt:MouseEvent) { for (var i:int = 0; i < good_answers + bad_answers; i++) { var feather:DisplayObject = i < good_answers ? new LightFeather() : new DarkFeather(); feather.x += 12 + i * (feather.width + 1); feather.y = 0; addChild(feather); } }个实例旁边创建4个DarkFeather个实例。

修改：

如何将对象添加到数组中？

LightFeather

然后将它们从舞台上移除，你可以做到例如：

// feathers array should be accessible for both codes (adding and removing objects)
var feathers:Array = [];

for (var i:int = 0; i < good_answers + bad_answers; i++) {
    var feather:DisplayObject = i < good_answers ? new LightFeather() : new DarkFeather();  
    addChild(feather);
    feathers.push(feather);
}

希望可以提供帮助。

在循环中为每个maxvalue添加X量的addChild

1 个答案: