UISwitch双击

Question

我目前有一个UISwitch，当打开和关闭时，分别递增和递减一个计数器。

当计数器为0时，计数器不会递减。在功能上，这是完美的工作，但我注意到一个bug，并想知道是否有人经历过这一点。

基本上，如果你非常快速地双击UISwitch的远端位置(完全打开或关闭)，计数器将递增两次，因为我想象UISwitch没有完全进入关闭状态，因此只是简单地再次添加到计数器，而不是第一次递减它。

下面是我用来检查交换机的代码：

// Sliders modified

- (IBAction)personalityChanged:(id)sender {
    if ([personality isOn] ){
        [[[GlobalData sharedGlobalData]personalitySliderValue] replaceObjectAtIndex:currentRecord-1 withObject:@"1"];
        rating ++;
        NSLog(@"The value of personality slider is %@", [[[GlobalData sharedGlobalData]personalitySliderValue] objectAtIndex:currentRecord-1]);
        [personality set]
    }
    else {
        [[[GlobalData sharedGlobalData]personalitySliderValue] replaceObjectAtIndex:currentRecord-1 withObject:@"0"];
        [self subtractFromRating:nil];
        NSLog(@"The value of personality slider is %@", [[[GlobalData sharedGlobalData]personalitySliderValue] objectAtIndex:currentRecord-1]);
    }
    [self checkRating:nil];
}

然后减去评级：

// subtract from rating

-(void)subtractFromRating:(id)sender{
    if (rating == 0) {
        // do nothing
    }
    else
    {
        rating --;
    }
}

最后，如果滑块位于某个位置，会发生什么情况：

// check rating

-(void)checkRating:(id)sender{
    switch (rating) {
        case 0:
            [matchRating setText:@""];
            [ratingGraphic setImage:[UIImage imageNamed:@""]];
            NSLog(@"rating is 0");
            break;
        case 1:
            [matchRating setText:@"Single Match"];
            [ratingGraphic setImage:[UIImage imageNamed:@"ratinggraphic1.png"]];
            NSLog(@"rating is 1");
            break;
        case 2:
            [matchRating setText:@"Potential Match"];
            [ratingGraphic setImage:[UIImage imageNamed:@"ratinggraphic2.png"]];
            NSLog(@"rating is 2");
            break;
        case 3:
            [matchRating setText:@"Great Match"];
            [ratingGraphic setImage:[UIImage imageNamed:@"ratinggraphic3.png"]];
            NSLog(@"rating is 3");
            break;
        case 4:
            [matchRating setText:@"Hot Match"];
            [ratingGraphic setImage:[UIImage imageNamed:@"ratinggraphic4.png"]];
            NSLog(@"rating is 4");
            break;
        default:
            break;
    }
}

有没有一种方法可以确保开关在返回之前完全从on状态转换到off状态，或者是更好的方法？

Answer 1

检测是否确实发生了变化的一个解决方案是保留一个额外的BOOL变量，用于跟踪最后一个开关状态。

BOOL lastValue = NO; // initial switch state
- (IBAction)personalityChanged:(id)sender {
    if (personality.isOn != lastValue) {
        lastValue = personality.isOn;
        if ([personality isOn] ){
            [[[GlobalData sharedGlobalData]personalitySliderValue] replaceObjectAtIndex:currentRecord-1 withObject:@"1"];
            rating ++;
            NSLog(@"The value of personality slider is %@", [[[GlobalData sharedGlobalData]personalitySliderValue] objectAtIndex:currentRecord-1]);
            [personality set]
        }
        else {
            [[[GlobalData sharedGlobalData]personalitySliderValue] replaceObjectAtIndex:currentRecord-1 withObject:@"0"];
            [self subtractFromRating:nil];
            NSLog(@"The value of personality slider is %@", [[[GlobalData sharedGlobalData]personalitySliderValue] objectAtIndex:currentRecord-1]);
        }
        [self checkRating:nil];
    }
}

然后，这将仅在开关状态实际发生变化时执行。