ios - 如何以编程方式调用带有 id sender 的 IBAction 方法？

Question

我的代码中有以下 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];

如何调用 IBAction 方法，同时允许它接收该方法的调用者对象？

Answer 1

方法签名对于手势识别器和 UIControls 是通用的。两者都将在没有警告或错误的情况下工作。要确定发送者，首先要确定类型...

- (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，让它连接的一个UIControl来调用它，或者让手势识别器调用它。他们都会工作。