问题描述
我创建了一个具有串行视图(A、B、C、D、...)的应用程序，我需要将 D 弹回 B.有人可能会说为什么不使用:
I've created an application with serial views (A,B,C,D,...), and I need to pop back D to B. Someone may say Why not using:
[self.navigationController popToViewController:[[self.navigationController viewControllers] objectAtIndex:1] animated:YES];
但是，这不是一个好的解决方案.因为这个方法需要你获取我们的B"所在的索引.
However, it is not a good solution. Because this method needs you to get the index which our "B" is store in.
问题:如何获取viewControllers中B"的索引?
Question: How to get the index of "B" in the viewControllers?
UIViewControllers 中的格式应为:
"< A: 0x6e70710 >","
format in the UIViewControllers should be:
"< A: 0x6e70710 >",
"< C: 0x6e30370 >",
"< B: 0x6988a70 >",
"< D: 0x6ea8950 >",
"< E: 0x6eaaad0 >"
我尝试过使用 rangeOfString 和 hasPrefix 来获取B"视图的索引，但失败了.
I've tried and failed to use rangeOfString and hasPrefix to get the "B" view's index.
推荐答案
这里我想知道NavigationController通过Concept Of Stack管理ViewController.ie LAST COME FIRST OUT.这里有两种相同的方法.
Here i would like to know you that NavigationController manage the ViewController by the Concept Of Stack.ie LAST COME FIRST OUT.Here are two Approach for doing the Same.
1)在这里您可以进行迭代以获得所需的 ViewController，如下所示
1)Here you can make iteration for getting the desirable ViewController as Below
for (id controller in [self.navigationController viewControllers])
{
if ([controller isKindOfClass:[BViewController class]])
{
[self.navigationController popToViewController:controller animated:YES];
break;
}
}
2) 这里有 A,B,C,D 控制器.意味着 B 将在第三位，所以你能做什么
2) Here you have A,B,C,D Controllers. means B would be on 3rd Position so what can you do
您可以将索引硬连线如下
you can hard wired the Index as Below
[self.navigationController popToViewController:[[self.navigationController viewControllers] objectAtIndex:2] animated:YES];
希望对你有所帮助..
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