我正在做一个小型游戏。我的仆从的实例名称是sideMinion。香蕉正在下降,一切正常,但我的removeChild()无法正常工作。我得到的错误是
错误#2025:提供的DisplayObject必须是调用方的子级。 removeChild()或hitTestObject无法正常工作。
这就是我在香蕉中所拥有的。类:
package {
import flash.display.MovieClip;
import flash.utils.Timer;
import flash.events.TimerEvent;
import flash.events.MouseEvent;
import flash.events.Event;
import flash.events.Event;
public class banana extends MovieClip {
var velX: Number = 0;
var velY: Number = 0;
var falling: Boolean = false;
var gravity: Number = 0;
public function banana() {
var timing: Timer = new Timer(25, 0);
timing.addEventListener(TimerEvent.TIMER, moveMe);
timing.start();
}
private function moveMe(event: TimerEvent)
{
x += velX;
y += velY;
if (falling) velY += gravity;
/* trace("[BANANA] position:", x, y, "speed:", velX, velY);*/
}
public function setSpeed(dx,dy) {
velX = dx;
velY = dy;
}
public function setSpot(atX,atY){
this.x=atX;
this.y=atY;
}
public function makeItFall (){
falling=true;
}
}
}
在我的主程序中:
import flash.display.MovieClip;
import flash.events.KeyboardEvent;
import flash.events.Event;
var leftKey:Boolean;
var rightKey:Boolean;
var upKey:Boolean;
var downKey:Boolean;
var jump:Boolean = false;
var xvelocity:int = 9;
var yvelocity:int = 3;
var gravity:Number = 7;
stage.addEventListener(Event.ENTER_FRAME, changeVelocity);
stage.addEventListener(KeyboardEvent.KEY_UP, checkKeyUp);
stage.addEventListener(KeyboardEvent.KEY_DOWN, checkKeyDown);
function changeVelocity(evt:Event){
moveMinion();
yvelocity += gravity;
}
function moveMinion(){
if (leftKey == true){
sideMinion.x -= xvelocity;
sideMinion.left();
}
if (rightKey == true){
sideMinion.x += xvelocity;
sideMinion.right();
}
}
function checkKeyDown(e:KeyboardEvent){
if (e.keyCode == Keyboard.LEFT){
leftKey = true;
}
else if (e.keyCode == Keyboard.RIGHT){
rightKey = true;
}
}
function checkKeyUp(e:KeyboardEvent){
if (e.keyCode == Keyboard.LEFT){
leftKey = false;
}
else if (e.keyCode == Keyboard.RIGHT){
rightKey = false;
}
}
btnStart.addEventListener(MouseEvent.CLICK, makeItFall);
function makeItFall(e:MouseEvent){
var numBananas = 6;
var theBananas: Array = new Array();
theBananas = [];
for (var i = 0; i < numBananas; i++) {
var aBanana: banana = new banana();
theBananas.push(aBanana);
btnStart.visible=false;
aBanana.y=30;
theBananas[i].setSpot(Math.random()*450+50,Math.random()*200+20);
theBananas[i].setSpeed((Math.random()), 1);
stage.addChild(aBanana);
}
var health: uint= 1;
addEventListener(Event.ENTER_FRAME, pickUpBananas);
function pickUpBananas(event:Event){
for( var i= 0; i<theBananas.length; ++i){
if (sideMinion.hitTestObject(theBananas[i])){
removeChild(theBananas[i]);
health=health+1;
trace(health);
}
}
}
}
stop();
编辑:格式代码
答案 0 :(得分:1)
当您将儿童添加到舞台时,您还必须从舞台上删除它:
stage.removeChild(theBananas[i]);
将来,在某些情况下,如果您不了解实际的父母,也可以使用父母财产:
theBananas[i].parent.removeChild(theBananas[i]);
在你的游戏中,我假设当你从舞台上移走香蕉时你也想从香蕉阵中取出香蕉,这样你的阵列就不会最终已经删除了香蕉。所以,这里有几个修改:
for(var i:int = theBananas.length-1; i>-1; i--){ //inverted loop
if (sideMinion.hitTestObject(theBananas[i])){
stage.removeChild(theBananas[i]);
theBananas.splice(i,1); //removing it from the array
health=health+1;
trace(health);
}
}
反转循环明显地从最后一个元素一直循环到第一个元素,因为如果你从数组中删除第一个元素,那么第二个元素就会跳过&#39;进入它的位置并循环将跳过它。
我希望我们所有人都能很快看到你的游戏! :)