Question

我正在研究iPad的匹配程序，当用户选择按钮时，图像被“揭开”，然后当用户选择第二个按钮时，另一个图像被揭开。然后我以编程方式检查匹配，如果没有，则将两个按钮图像恢复到其初始状态。

这个工作正常，除非没有进行匹配，切换发生得太快，以至于你没有时间看你有什么“未被发现”。我试图让它睡觉，但图像不会交换到未覆盖的状态......思考？

此代码如下：

//Take action on the tap of one of the buttons
if(isFirstSelection)
{
    firstSelection = [(UIButton *)sender tag];
    tempImageItem = [tileArray objectAtIndex:firstSelection];
    tempImage = [tempImageItem tileImage];
    firstSelectionName = [[NSString alloc] initWithString:[tempImageItem tileName]];
    [(UIButton *)sender setImage:tempImage forState:UIControlStateNormal];
    tempButton = sender;
    isFirstSelection = NO;       
}else{
    secondSelection = [(UIButton *)sender tag];
    tempImageItem = [tileArray objectAtIndex:secondSelection];
    tempImage = [tempImageItem tileImage];
    secondSelectionName = [[NSString alloc] initWithString:[tempImageItem tileName]];
    [(UIButton *)sender setImage:tempImage forState:UIControlStateNormal];

    //Two game pieces have been removed so check to see if they are a match
    if([firstSelectionName isEqualToString:secondSelectionName])
    {
        //Match found
        //do something

    }else{
        **//NO MATCH FOUND 
        [NSThread sleepForTimeInterval:3]; 
        //Display the checker board pieces again 
        [(UIButton *)sender setImage:[UIImage imageNamed:@"originalImage"] forState:UIControlStateNormal];** 

    }
    //Reset isFirstSelection Flag to YES for next selection
    isFirstSelection = YES;           
}

Answer 1

根据我的理解，您希望在3秒后放置按钮的原始图像，以便用户有时间查看发生的情况。您应该查看NSTimer类以触发该代码

[(UIButton *)sender setImage:[UIImage imageNamed:@"originalImage"] forState:UIControlStateNormal];

从现在起3秒内。

Answer 2

对于“在X秒内运行此代码”，我更喜欢：

[self performSelector:@selector(spinWheel:) withObject:[NSNumber numberWithUnsignedInt:0] afterDelay:delay];

无需使用计时器对象。

强制按钮交换图像

2 个答案: