Question

是否可以在单个视图中的iOS子视图之间移动而无需返回视图？如果是这样，怎么样？代码示例welcome。

我现在如何做，但这需要返回“主页”视图：在myFile.m

中

// Define the other views

DF_Results                  *df_Results;
CR_FiringRange              *cr_FiringRange;
CR_AssetLoader              *cr_AssetLoader;

DB_GeneralData              *db_GeneralData;
DB_ArmorData                *db_ArmorData;
DB_WeaponsData              *db_WeaponsData;



@synthesize ScreenButton01;
@synthesize ScreenButton02;

@synthesize ScreenButton03;
@synthesize ScreenButton04;

@synthesize ScreenButton05;
@synthesize ScreenButton06;
@synthesize ScreenButton07;
.
.
.

-(IBAction) ScreenButton02Clicked:(id) sender {
    df_Results = [[DF_Results alloc]
                               initWithNibName:@"DF_Results"
                               bundle:nil];

    [UIView beginAnimations:@"flipping view" context:nil];
    [UIView setAnimationDuration:1.0];
    [UIView setAnimationCurve:UIViewAnimationCurveEaseInOut];
    [UIView setAnimationTransition:UIViewAnimationTransitionFlipFromRight forView:self.view cache:YES];
    [self.view addSubview:df_Results.view];
    [UIView commitAnimations];
}

在DF_Results中返回我们点击另一个按钮但我希望能够直接跳到另一个子视图：

-(IBAction) ResultsScreenButton01Clicked:(id) sender {
    [UIView beginAnimations:@"flipping view" context:nil];
    [UIView setAnimationDuration:1.0];
    [UIView setAnimationCurve:UIViewAnimationCurveEaseIn];
    [UIView setAnimationTransition:UIViewAnimationTransitionFlipFromLeft forView:self.view.superview cache:YES];
    [self.view removeFromSuperview];
    [UIView commitAnimations];
}

Answer 1

在iOS 4.0中，UIView有许多基于块的动画过渡任务，根据文档推荐的方法。例如。这个可能很有用：

(void)transitionFromView:(UIView *)fromView toView:(UIView *)toView duration:(NSTimeInterval)duration options:(UIViewAnimationOptions)options completion:(void (^)(BOOL finished))completion

有关详细信息，请参阅UIView文档。

是否可以在单个视图中移动iOS子视图而无需返回视图？

1 个答案: