重构iOS代码：减少代码行数

Question

我第一次在这里发布一个实际有效的代码！但是，我相信有一种方法可以减少代码中的行数。我希望大师可以告诉我的方式。

以下是故事板窗口供参考：

我有一个带有containerView的主ViewController。 ContainerView拥有自己的导航控制器。主视图控制器左侧的每个按钮（B1-B5）都会显示相应的场景编号。即B2将场景2推入堆栈。 B4将场景4推到堆栈。如果visibleViewContoller是场景5，并且用户按下B1，它将弹出所有viewController，直到我们到达场景1.依此类推。

以下代码再次运行正常，我只是想缩小B1和B2的代码大小：

- (IBAction)B1Pressed:(id)sender {

UINavigationController *navController = [self.childViewControllers objectAtIndex:0];
NSMutableArray *VCs = [navController.viewControllers mutableCopy];
UIViewController *visibleViewController = [navController visibleViewController];

if (visibleViewController == [VCs objectAtIndex:0])
{
    return;
}
else if (visibleViewController ==[VCs objectAtIndex:1])
{
     [navController popViewControllerAnimated:YES];
}

else if (visibleViewController ==[VCs objectAtIndex:2])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];
}

else if (visibleViewController ==[VCs objectAtIndex:3])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];

}
else if (visibleViewController ==[VCs objectAtIndex:4])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];

}
else if (visibleViewController ==[VCs objectAtIndex:5])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];

    }
}

- (IBAction)B2Pressed:(id)sender {

UINavigationController *navController = [self.childViewControllers objectAtIndex:0];
NSMutableArray *VCs = [navController.viewControllers mutableCopy];
UIViewController *visibleViewController = [navController visibleViewController];

if (visibleViewController == [VCs objectAtIndex:0])
{
    STLMEatDrinkViewController *stlmEDVC = [self.storyboard instantiateViewControllerWithIdentifier:@"B2"];
    [navController pushViewController:stlmEDVC animated:YES];
}
else if (visibleViewController ==[VCs objectAtIndex:1])
{
    return;
}

else if (visibleViewController ==[VCs objectAtIndex:2])
{
    [navController popViewControllerAnimated:YES];
}

else if (visibleViewController ==[VCs objectAtIndex:3])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];

}
else if (visibleViewController ==[VCs objectAtIndex:4])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];

}
else if (visibleViewController ==[VCs objectAtIndex:5])
{
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:NO];
    [navController popViewControllerAnimated:YES];   
 }
}

现在想象一下，为B3Pressed，B4Pressed和B5Pressed再写三次相同的代码。我认为代码太多了，我几乎肯定有更好的方法。

谢谢。

Answer 1

1。）您不得使用==比较对象。使用isEqual:。

2。）循环。

int idx = [VCs indexOfObject:visibleViewController];

if (idx == 0) {
    STLMEatDrinkViewController *stlmEDVC = [self.storyboard instantiateViewControllerWithIdentifier:@"B2"];
    [navController pushViewController:stlmEDVC animated:YES];
} else if (idx == 1) {
    return;
} else {
    int i;
    for (i = 2; i < idx; i++) {
        [navController popViewControllerAnimated:NO];
    }

    [navController popViewControllerAnimated:YES];
}

Answer 2

我认为这可以解决您的问题：

- (IBAction)B1Pressed:(id)sender
{
    UINavigationController *navController = [self.childViewControllers objectAtIndex:0];
    UIViewController *B1ViewController = [navController.viewControllers objectAtIndex:0];
    [navController popToViewController:B1ViewController animated:YES];
}