如何限制节点在触摸时的外观

Question

我有一个名为“ sprite”的节点，当我触摸屏幕时，它在特定的x轴上显示1秒钟，在该“ sprite”节点消失后，我想使屏幕上只能同时是两个“ sprite”节点并且它们不能在同一地点。.因为现在我可以添加“ sprite”节点，就像我的手指可以在屏幕上轻击一样。 这是我现在得到的：

-(void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {

for (UITouch *touch in touches) {

    CGPoint touchlocation = [touch locationInNode:self];
    CGPoint location = CGPointMake(touchlocation.x, 60);

    SKSpriteNode *sprite = [SKSpriteNode spriteNodeWithImageNamed:@"board_ipad@2x"];

    sprite.xScale = 0.4;
    sprite.yScale = 0.6;
    sprite.position = location;
    sprite.physicsBody = [SKPhysicsBody bodyWithRectangleOfSize:sprite.frame.size];
    sprite.physicsBody.dynamic = NO;
    sprite.name = boardCategoryName;

    SKAction *delay = [SKAction waitForDuration:1];
    SKAction *remove = [SKAction removeFromParent];
    SKAction *actionSequence = [SKAction sequence:@[delay,remove]];
    [sprite runAction:actionSequence];

    [self addChild:sprite];

}
}

已编辑

@implementation GameScene
SKNode *groupSprite;

-(void)didMoveToView:(SKView *)view{
    groupSprite  = [[SKNode alloc]init];
    [self addChild:groupSprite];
}

-(void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {

for (UITouch *touch in touches) {


    if (groupSprite.children.count < 2) {



        CGPoint touchlocation = [touch locationInNode:self];
        CGPoint location = CGPointMake(touchlocation.x, 75);

        SKSpriteNode *sprite = [SKSpriteNode spriteNodeWithImageNamed:@"board_ipad@2x"];

            sprite.xScale = 0.4;
            sprite.yScale = 0.6;

        sprite.position = location;
        sprite.physicsBody = [SKPhysicsBody bodyWithRectangleOfSize:sprite.frame.size];
        sprite.physicsBody.dynamic = NO;
        sprite.physicsBody.categoryBitMask = boardCategory;
        sprite.physicsBody.contactTestBitMask = ballCategory;
        sprite.name = boardCategoryName;


        SKAction *delay = [SKAction waitForDuration:1];
        SKAction *remove = [SKAction removeFromParent];
        SKAction *actionSequence = [SKAction sequence:@[delay,remove]];
        [sprite runAction:actionSequence];

        [groupSprite addChild:sprite];


    };

Answer 1

您可以像@kaizoku所指出的那样添加一个数组，但这只是一个浮在其中的变量，使代码可能难以管理。 更好的方法是使用SKNode创建组，然后检查组数。

（我的目标c有点生锈，所以请多多包涵）

...

SKNode *groupSprite = [[SKNode alloc] init];

...

-(void)didMoveToView:(SKView *)view 
{
    self.addChild(groupSprite);
}

...

-(void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {

    for (UITouch *touch in touches) 
    {
        if(groupSprite.children.count < 2)
        {
            CGPoint touchlocation = [touch locationInNode:self];
            CGPoint location = CGPointMake(touchlocation.x, 60);
            SKSpriteNode *sprite = [SKSpriteNode spriteNodeWithImageNamed:@"board_ipad@2x"];

            sprite.xScale = 0.4;
            sprite.yScale = 0.6;
            sprite.position = location;
            sprite.physicsBody = [SKPhysicsBody bodyWithRectangleOfSize:sprite.frame.size];
            sprite.physicsBody.dynamic = NO;
            sprite.name = boardCategoryName;

            SKAction *delay = [SKAction waitForDuration:1];
            SKAction *remove = [SKAction removeFromParent];
            SKAction *actionSequence = [SKAction sequence:@[delay,remove]];
            [sprite runAction:actionSequence];

            [groupSprite addChild:sprite];

        }
    }
}

之所以要这样做，是因为它使您的代码更具可读性。 您有一个包含一组Sprite的场景。 现在，如果出于某种原因，您想从屏幕上删除所有精灵。 您只需将其从组中删除，就不必担心将其从场景中删除并从阵列中删除它。

Answer 2

1. 创建一个SKSpriteNode类型的数组。

   var yourArray = [SKSpriteNode]

2. 触摸时，在此阵列中添加新节点。

   yourArray.append（sprire）

3. 当执行触摸功能时，请检查数组大小是否大于两个。 如果是，则什么也不做。

   如果yourArray.count <2 {//您的代码}