Question

我试图创建一个表格视图，单元格包含姓名，电话和邮件。如果用户触摸电话，则应用程序呼叫电话（单击以呼叫）。如果用户触摸邮件，我需要使用新电子邮件打开邮件应用程序，然后点击“＃34;触摸的地址”。

首先，我尝试开发包含电子邮件的可触摸标签，但我收到错误＆＃34; 选择器发送到实例＆＃34;。

我需要识别tapgesture，然后确定它是电子邮件还是电话，获取文本和电话或打开邮件应用程序。

当前代码：

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
    ...
    cell.lblEmail.userInteractionEnabled = YES;
    UITapGestureRecognizer *tapGesture =
    [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(labelTap)];
    [cell.lblEmail addGestureRecognizer:tapGesture];
    return cell;
}

- (void)labelTap:(UIGestureRecognizer *)sender {
    UILabel *labelSender = (UILabel*) sender;
    NSLog(@"%text: %@", labelSender.text);
}

错误：

2015-02-03 12:28:29.726 MyApp[80080:4019203] -[MyAppContactsViewController labelTap]: unrecognized selector sent to instance 0x7ff6d8cb4110
2015-02-03 12:28:29.743 MyApp[80080:4019203] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[MyAppContactsViewController labelTap]: unrecognized selector sent to instance 0x7ff6d8cb4110'
*** First throw call stack:
(
    0   CoreFoundation                      0x000000010709cf35 __exceptionPreprocess + 165
    1   libobjc.A.dylib                     0x0000000106d35bb7 objc_exception_throw + 45
    2   CoreFoundation                      0x00000001070a404d -[NSObject(NSObject) doesNotRecognizeSelector:] + 205
    3   CoreFoundation                      0x0000000106ffc27c ___forwarding___ + 988
    4   CoreFoundation                      0x0000000106ffbe18 _CF_forwarding_prep_0 + 120
    5   UIKit                               0x00000001059912e6 _UIGestureRecognizerSendActions + 262
    6   UIKit                               0x000000010598ff89 -[UIGestureRecognizer _updateGestureWithEvent:buttonEvent:] + 532
    7   UIKit                               0x0000000105994ba6 ___UIGestureRecognizerUpdate_block_invoke662 + 51
    8   UIKit                               0x0000000105994aa2 _UIGestureRecognizerRemoveObjectsFromArrayAndApplyBlocks + 254
    9   UIKit                               0x000000010598ab1d _UIGestureRecognizerUpdate + 2796
    10  UIKit                               0x0000000105624ff6 -[UIWindow _sendGesturesForEvent:] + 1041
    11  UIKit                               0x0000000105625c23 -[UIWindow sendEvent:] + 667
    12  UIKit                               0x00000001055f29b1 -[UIApplication sendEvent:] + 246
    13  UIKit                               0x00000001055ffa7d _UIApplicationHandleEventFromQueueEvent + 17370
    14  UIKit                               0x00000001055db103 _UIApplicationHandleEventQueue + 1961
    15  CoreFoundation                      0x0000000106fd2551 __CFRUNLOOP_IS_CALLING_OUT_TO_A_SOURCE0_PERFORM_FUNCTION__ + 17
    16  CoreFoundation                      0x0000000106fc841d __CFRunLoopDoSources0 + 269
    17  CoreFoundation                      0x0000000106fc7a54 __CFRunLoopRun + 868
    18  CoreFoundation                      0x0000000106fc7486 CFRunLoopRunSpecific + 470
    19  GraphicsServices                    0x0000000109fa59f0 GSEventRunModal + 161
    20  UIKit                               0x00000001055de420 UIApplicationMain + 1282
    21  MyApp                               0x00000001040769a3 main + 115
    22  libdyld.dylib                       0x0000000107a9c145 start + 1
    23  ???                                 0x0000000000000001 0x0 + 1
)
libc++abi.dylib: terminating with uncaught exception of type NSException

Answer 1

更改此行：

[[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(labelTap)];

到

[[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(labelTap:)];

这是必需的，因为labelTap selector与labelTap:不同。前者是一个没有参数的选择器（它不存在，因此是例外），而后者是带有1个参数的选择器，你确实已经定义了它。

Answer 2

维诺德，克劳斯是对的。我换了UIButton，工作得很好。我已选择使用此代码识别女巫邮件：Detecting which UIButton was pressed in a UITableView

带有可触摸标签的UITableViewCell

2 个答案: