iphone - tableViewCell 中的 UITapgestureRecogniser

Question

我在 UITableViewCell 中有一个 UILabel，它显示电话号码。当我单击它时，我应该能够调用该特定号码。因此，为了使其可点击，我向其中添加了一个 UITapGestureRecognizer 。但我不知道如何引用单击了哪个水龙头，我的意思是单击了哪个数字。

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
    lblphone = [[UILabel alloc] initWithFrame:CGSizeMake(20,30,50,100)];
    lblphone.tag = 116;
    lblphone.backgroundColor = [UIColor clearColor];
    lblphone.text=[NSString stringwithFormat:@"%@",phone];
    [lblphone setFont:[UIFont fontWithName:@"Helvetica" size:12]];
    [lblphone setLineBreakMode:UILineBreakModeWordWrap];
    [lblphone setUserInteractionEnabled:YES];

    UITapGestureRecognizer *tapGestureRecognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(labelButton:)];
    [tapGestureRecognizer setNumberOfTapsRequired:1];
    [lblphone addGestureRecognizer:tapGestureRecognizer];
    [tapGestureRecognizer release];
    [cell addSubview:lblphone];
}

-(IBAction)labelButton:(UITapGestureRecognizer*)tapGestureRecognizer
{
    NSIndexPath *indexPath;
    UITableViewCell *cell = [self.tableView cellForRowAtIndexPath:indexPath];
    UILabel *phoneno = (UILabel*)[cell viewWithTag:116];
    NSString *phoneNumber = [@"telprompt://" stringByAppendingString:phoneno.text];
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
}

但每次我都能看到 NSString *phoneNumber; 中最后一个单元格的电话号码；

如何引用 UITapGestureRecognizer 中单击的标签？

Answer 1

在cellForRowAtIndexPath中使用lblphone.tag = indexPath.row + 1;而不是lblphone.tag = 116;

您也可以在不使用标签的情况下实现此目的。

参见下面的示例

-(void)labelButton:(UITapGestureRecognizer*)tapGestureRecognizer
{
   UILabel *phoneno = (UILabel *) tapGestureRecognizer.view;
}