我不太确定如何解决这个问题。基本上，我有一个循环100个按钮（10行10）的算法，并根据随机的起点和终点分配一个数值。单击时，按钮应进入游戏的下一部分，并将自己从地图上移开。为此，我使用了以下代码：
public function createMap()
{
var startPointX:int = int(10 * Math.random()); //creates random values that set the x and y values of the start and end points
var startPointY:int = int(10 * Math.random());
var endPointX:int = int(10 * Math.random());
var endPointY:int = int(10 * Math.random());
var spaceDiffX:int = startPointX - endPointX; //the difference in X values
var spaceDiffY:int = startPointY - endPointY; //the difference in Y values
//Creation loop - loops through all of the buttons to create them
for (i = 0; i <= 9; i++)
{
mapY[i] = mapX; //Adds the rows into each column
for (j = 0; j <= 9; j++)
{
mapY[i][j] = new LocationButton(); //Creates a new button
mapY[i][j].x = 6 * 20 * i; //Sets X position
mapY[i][j].y = 6 * 10 * j; //Sets Y position
//places start point
if (i == startPointX && j == startPointY) //checks if this point on the grid is equal to the random start point
{
mapY[i][j].buttonType = 1; //sets it to the start point
}
//places end point
if (i == endPointX && j == endPointY) //checks if this point is equal to the random end point
{
mapY[i][j].buttonType = 2; //sets it to the end point
}
//finds path on x axis
if (i != startPointX && j == startPointY) //checks if this point is on the same Y value as the start point
{
if (spaceDiffX >= 0) //checks to see if the start point is on the left or right of the end point, left is negative, right is positive
{
if (i < startPointX && i >= endPointX) //checks if i is between the start X and end X
{
mapY[i][j].buttonType = 3; //sets it to a "shortest path point"
}
}
else
{
if (i > startPointX && i <= endPointX) //checks to see if i is between the start X and end X
{
mapY[i][j].buttonType = 3; //sets it to a "shortest path point"
}
}
}
if (j != startPointY && i == endPointX) //checks if this point is on the same X value as the end point
{
if (spaceDiffY >= 0) //checks to see if the start point is over or under the end point, above is positive, below is negative
{
if (j < startPointY && j >= endPointY) //checks if j is between the end Y and the start Y
{
mapY[i][j].buttonType = 3; //sets it to a "shortest path point"
}
}
else
{
if (j > startPointY && j <= endPointY)
{
mapY[i][j].buttonType = 3; //sets it to a "shortest path point"
}
}
}
if (mapY[i][j].buttonType != 0) //checks whether or not the button should be added
{
mapY[i][j].addEventListener(MouseEvent.CLICK, createBattleGUI); //adds event listener
stage.addChild(mapY[i][j]); //adds it to stage
}
}
}
}
然而，问题在于eventListener之后的代码是如何起作用的。它调用的函数从这个循环开始：
for (i = 0; i <= 9; i++)
{
for (j = 0; j <= 9; j++)
{
if (mapY[i][j].buttonType != 0)
{
stage.removeChild(mapY[i][j]);
}
}
}
一旦该循环通过，它应该 - 删除所有按钮。但事实并非如此。我开始跟踪一些变量，并注意到在第二个循环中，它实际上循环遍历相同的10个对象10次，而不是循环遍历所有100个对象。 *表示第一个for循环中的trace语句（在事件之前），没有它的是在第二个for循环中跟踪的按钮：
name - buttonType - i - j
*instance33 1 1 6
*instance53 3 2 6
*instance73 3 3 6
*instance93 3 4 6
*instance113 3 5 6
*instance133 3 6 6
*instance153 3 7 6
*instance173 3 8 6
instance181 0 0 0
instance183 0 0 1
instance185 0 0 2
instance187 0 0 3
instance189 0 0 4
instance191 0 0 5
instance193 0 0 6
instance195 0 0 7
instance197 0 0 8
instance199 0 0 9
instance181 0 1 0
instance183 0 1 1
instance185 0 1 2
instance187 0 1 3
instance189 0 1 4
instance191 0 1 5
instance193 0 1 6
instance195 0 1 7
instance197 0 1 8
instance199 0 1 9
我意识到这是一个很长的问题，但我无法想象我的生活。谢谢你的帮助。