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我目前有一个UISwitch打开和关闭时分别递增和递减计数器。

当计数器为 0 时，计数器不会递减。从功能上讲，这很有效，但是我注意到了一个错误，想知道是否有人遇到过这种情况。

本质上，如果您非常快速地双击 UISwitch 在其远位置（完全打开或关闭），计数器将增加两倍，因为我想象 UISwitch 没有完全达到关闭状态，因此只是简单地添加到计数器没有首先减少它。

这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];
}

然后是减法评分：

// subtract from rating

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

最后，如果滑块处于某个位置会发生什么：

// 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开启状态变为关闭状态，或者有更好的方法？

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1 回答 1

3

检测是否确实发生了变化的一种解决方案是保留一个额外的 BOOL 变量来跟踪最后的开关状态。

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改变时执行。

于 2012-10-18T01:05:51.927 回答