Question

我希望NSMuttableArray中的每个对象都显示4秒。然后以这种方式迭代数组中的所有项目。相反，我得到的结果是所有项目都出现，并且在这段时间后消失在一起。

-(void)showTreasures{


for (int i = 0; i < _treasures.count; i++)
{
   SKSpriteNode *obj = [_treasures objectAtIndex:i];

SKAction *show = [SKAction runBlock:^{
    obj.hidden = NO;
}];

SKAction *wait = [SKAction waitForDuration:4];


SKAction *hide = [SKAction runBlock:^{
    obj.hidden = YES;
}];

SKAction *sequence = [SKAction sequence:@[show, wait, hide]];
[obj runAction:sequence completion:^{

    NSLog(@"Item %d", i);
}];

}

}

Answer 1

我认为递归方法，调用自身的方法很好。您可以创建所有对象的数组（SKSpriteNode）并将其传递给获取第一个（或最后一个）对象并运行适当操作的方法，删除该对象并再次调用该方法：

NSMutableArray *arrOfObject = //arrat with all of the sprites you want to show
[self runShowAction:arrOfObject];

-(void)runShowAction:(NSMutableArray*)array {

    //if no object in the array return
    if(array.count <= 0) return;

    SKSpriteNode *obj = [array firstObject];
    //Run your code here
    //...


    //On completion remove object from array and run this method again
    [obj runAction:sequence completion:^{

        NSLog(@"Item %d", i);

        [array removeObject:obj];
        [self runShowAction:array];
    }];
}

Answer 2

为什么不使用数组来建立所有操作的列表，然后运行所有这些操作的序列？

-(void)showTreasures
{
    NSMutableArray *actions = [NSMutableArray array];

    for (int i = 0; i < _treasures.count; i++)
    {
        SKSpriteNode *obj = [_treasures objectAtIndex:i];

        SKAction *show = [SKAction runBlock:^{
            obj.hidden = NO;
        }];

        SKAction *wait = [SKAction waitForDuration:4];

        SKAction *hide = [SKAction runBlock:^{
            obj.hidden = YES;
        }];

        SKAction *finish = [SKAction runBlock:^{
            NSLog(@"Item %d", i);
        }];

        [actions addObjectsFromArray:@[show, wait, hide, finish]];
    }

    SKAction *sequence = [SKAction sequence:actions];
    [obj runAction:sequence completion:^{
        NSLog(@"Finished all items");
    }];
}

目标c循环具有时间延迟

2 个答案: