如何在ios中使用快速枚举从scrollview的所有子视图中获取固定数量的子视图

时间:2014-10-09 10:25:48

标签: ios scrollview subview nsfastenumeration

如何在ios中使用快速枚举从scrollview的所有子视图中选择子视图的修订号?

1 个答案:

答案 0 :(得分:0)

执行以下操作: 1.在创建按钮时给你的按钮添加标签。例如:

UIButton *button = [UIButton buttonWithType:UIButtonTypeRoundedRect];
button.tag = 1;
[button addTarget:self
       action:@selector(aMethod:)
forControlEvents:UIControlEventTouchUpInside]; 
[button setTitle:@"Show View" forState:UIControlStateNormal];
button.frame = CGRectMake(80.0, 210.0, 160.0, 40.0);
[view addSubview:button];

2.不明白为什么要使用快速枚举,你可以通过以下方式简单地获取按钮:

UIButton *btn1 = (UIButton *)[self.yourScrollView viewWithTag:1];
UIButton *btn2 = (UIButton *)[self.yourScrollView viewWithTag:2];
UIButton *btn3 = (UIButton *)[self.yourScrollView viewWithTag:3];
UIButton *btn4 = (UIButton *)[self.yourScrollView viewWithTag:4];
UIButton *btn5 = (UIButton *)[self.yourScrollView viewWithTag:5];

2.如果您的要求是使用快速枚举,请使用NSMutableArray来保存按钮:

NSMutableArray *arrayOfButtons = [[NSMutableArray alloc] initWithCapacity:5];
for (id btn in self.yourScrollView.subviews)
{
    @autoreleasepool {
        if (btn.tag > 0 || btn.tag<=5)
        {
            [arrayOfButtons addObject:btn];
        }
    }
}
希望它会对你有所帮助。