重构 iOS 代码：减少代码行数

Question

我第一次在这里发布一个带有实际工作代码的问题！但是，我相信有一种方法可以减少代码行数。希望有大神给我指路。

这是供参考的故事板窗口：

我有一个带有 containerView 的 Main ViewController。 ContainerView 有它自己的导航控制器。主视图控制器左侧的每个按钮 (B1-B5) 与其各自的场景编号相连。即 B2 会将场景 2 推入堆栈。 B4 将场景 4 推入堆栈。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。

下面的代码再次运行良好，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];   
 }
}

现在想象为 B3Pressed、B4Pressed 和 B5Pressed 再编写 3 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来解决这个问题。

谢谢。

Answer 1

1.) 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];
}