如何以编程方式调用具有ID发送者的IBAction方法？

Question

我的代码中包含以下IBAction方法。

-(IBAction)handleSingleTap:(id)sender
{
    // need to recognize the called object from here (sender)
}


UIView *viewRow = [[UIView alloc] initWithFrame:CGRectMake(20, y, 270, 60)];
// Add action event to viewRow
UITapGestureRecognizer *singleFingerTap = 
[[UITapGestureRecognizer alloc] initWithTarget:self 
                                        action:@selector(handleSingleTap:)];
[self.view addGestureRecognizer:singleFingerTap];
[singleFingerTap release];
//
UILabel *infoLabel = [[UILabel alloc] initWithFrame:CGRectMake(5,30, 100, 20)];
infoLabel.text = @"AAAANNNNVVVVVVGGGGGG";
//[viewRow addSubview:infoLabel];
viewRow.backgroundColor = [UIColor whiteColor];

// display the seperator line 
UILabel *seperatorLablel = [[UILabel alloc] initWithFrame:CGRectMake(0,45, 270, 20)];
seperatorLablel.text = @" ___________________________";
[viewRow addSubview:seperatorLablel];
[scrollview addSubview:viewRow];

如何在允许它接收该方法的调用者对象的同时调用IBAction方法？

Answer 1

方法签名是手势识别器和UIControl所共有的。 两者都可以正常工作而不会发出警告或错误。 要确定发件人，请首先确定类型...

- (IBAction)handleSingleTap:(id)sender
{
// need to recognize the called object from here (sender)
    if ([sender isKindOfClass:[UIGestureRecognizer self]]) {
        // it's a gesture recognizer.  we can cast it and use it like this
        UITapGestureRecognizer *tapGR = (UITapGestureRecognizer *)sender;
        NSLog(@"the sending view is %@", tapGR.view);
    } else if ([sender isKindOfClass:[UIButton self]]) {
        // it's a button
        UIButton *button = (UIButton *)sender;
        button.selected = YES;
    }
    // and so on ...
}

要调用它，直接调用它，让它连接的UIControl调用它，或者让手势识别器调用它。 他们都会工作。

Answer 2

您不必调用它，因为您将方法用作UITapGestureRecognizer Selector ，因此只要轻按该应用程序，它将自动调用。 另外，如果您可以在action:@selector(handleSingleTap:)中的方法名称后识别冒号，则意味着将UITapGestureRecognizer类型的对象发送给该方法。 如果您不想发送任何对象，则只需从方法中删除冒号和(id)sender 。

Answer 3

如您所愿：

[self handleSingleTap:self.view];

sender可以使用任何您喜欢的id类型。 您还可以发送带有标签的UIButton实例。