重構iOS代碼：減少代碼行數

Question

第一次我在這裏發佈一個問題，其中包含實際可用的代碼！但是，我相信有一種方法可以減少代碼中的行數。我希望上師能指引我。

這裏是故事板窗口以供參考：

我有一個主視圖控制器與containerView。 ContainerView擁有自己的導航控制器。主視圖控制器左邊的每個按鈕（B1-B5）塞格的相應場景編號。即B2將場景2推到堆棧上。 B4將場景4推到堆疊上。如果visibleViewContoller是場景5，並且用戶按下B1，它將彈出所有viewController，直到我們到達場景1爲止。等等等等。

再次跌破代碼工作正常，我只是希望縮小代碼的大小爲B1和B2：

- (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]; 
} 
} 

現在想象寫相同的代碼三次爲B3Pressed，B4Pressed和B5Pressed。我認爲這是太多的代碼，我幾乎肯定有一個更好的方法來做到這一點。

謝謝。

Answer 1

1.）您不得使用==來比較對象。使用isEqual:。

2.）循環。

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

我覺得這回答您的問題：

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