有3个视图view1,view2,view3
From view1 there is a function which calls view2
{
view2* view2Controller =[[view2 alloc]initWithNibName:@"view2" bundle:nil];
[ view2Controller SessionMaintained];
// [self.navigationController pushViewController:view2Controller animated:YES];
}
now in view 2 call is made to view3
{
view3* view3Controller =[[view3 alloc]initWithNibName:@"view3" bundle:nil];
[ view3Controller SessionMaintained];
[self.navigationController pushViewController:view3Controller animated:YES];
}
现在我想直接推送view3而不推送view2
答案 0 :(得分:0)
直接创建View3对象并推入View1是什么问题?
答案 1 :(得分:0)
所以我自己回答你的问题
从view1有一个调用view2的函数
{
view2* view2Controller =[[view2 alloc]initWithNibName:@"view2" bundle:nil];
[ view2Controller SessionMaintained];
//push view2 w/o animation
[self.navigationController pushViewController:view2Controller animated:NO];
//call some method (say showView3) in view2
[view2Controller showView3];
[view2Controller release];
}
现在在视图2(showView3方法)中,调用view3
{
view3* view3Controller =[[view3 alloc]initWithNibName:@"view3" bundle:nil];
[ view3Controller SessionMaintained];
[self.navigationController pushViewController:view3Controller animated:YES];
}