Ios 开关双抽头

Ios 开关双抽头,ios,uiswitch,Ios,Uiswitch,我目前有一个ui开关,它在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。从功能上来说,这是完美的,但是我注意到了一个bug,我想知道是否有人经历过这种情况 基本上,如果您非常快速地在远位置双击UISwitch(完全打开或关闭),计数器将增加两倍,因为我想象UISwitch没有完全进入关闭状态,因此只是再次添加到计数器,而不先减小它 下面是我用来检查交换机的代码: // Sliders modified - (IBAction)personalityChanged:(id

我目前有一个
ui开关
,它在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。从功能上来说,这是完美的,但是我注意到了一个bug,我想知道是否有人经历过这种情况

基本上,如果您非常快速地在远位置双击UISwitch(完全打开或关闭),计数器将增加两倍,因为我想象UISwitch没有完全进入关闭状态,因此只是再次添加到计数器,而不先减小它

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 --;
    }
}
最后是滑块处于某个位置时发生的结果:

// 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;
    }
}

有没有办法确保开关在返回之前完全从on状态切换到off状态,或者有更好的方法?检测是否确实发生了变化的解决方案是保留一个额外的BOOL变量,该变量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];
    }
}

这将仅在开关状态实际发生变化时执行。

感谢wsidell,它工作起来就像做梦一样,现在滑块的行为就像它们应该做的那样。。哇!