将 NSString 传递到 UIAlertview 的消息内容中

Question

我想显示这样的东西:SampleUser poked you.在UIAlertView的message，但实际上我得到错误。我知道如何用一个简单的字符串，我不知道如何用一个包含另一个字符串的字符串。

UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Poke" message:@"%@ poked you.", self.senderString delegate:self cancelButtonTitle:@"Yes" otherButtonTitles:@"No", nil];
[alertView show];

Answer 1

您应该首先创建您的composati_ NSString，然后在您的UIAlertView中调用它:

NSString *message = [NSString stringWithFormat:@"%@ poked you.", userName];
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Poke" message:message, self.senderString delegate:self cancelButtonTitle:@"Yes" otherButtonTitles:@"No", nil];
[alertView show]; 

Answer 2

这里的问题是，对于message:参数，您试图发送这个:

@"%@ poked you.", userName

这没有任何意义。

相反，您需要发送一个NSString对象作为参数。

NSString *message = [NSString stringWithFormat:@"%@ poked you.", self.senderString];

现在我们已经创建了一个NSString对象，我们可以使用这个对象作为消息参数。

您可以在调用中嵌入这个对象来创建警报视图，但为了可读性和调试，最好这样做。

NSString *message = [NSString stringWithFormat:@"%@ poked you.", self.senderString];
UIAlertView *pokeAlert = [[UIAlertView alloc] initWithTitle:@"Poke" 
                                                    message:message 
                                                   delegate:self
                                          cancelButtonTitle:@"Yes" 
                                          otherButtonTitles:@"No", nil];
[pokeAlert show];

Answer 3

UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Poke" message:[NSString stringWithFormat:@"%@ poked you.", self.senderString] delegate:self cancelButtonTitle:@"Yes" otherButtonTitles:@"No", nil];
[alertView show];