Question

我想在SampleUser poked you. UIAlertView中显示如下内容：message，但实际上我遇到了错误。我知道怎么用一个简单的字符串来做，我不知道怎么用包含另一个字符串的字符串来做。

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

Answer 1

您应首先创建您的作品 - NSString，然后在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，您尝试发送此信息：

@"%@ 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。

您可以创建嵌入在调用中的此对象来创建警报视图，但为了便于阅读和调试，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];

将NSString传递给UIAlertview的消息内容

3 个答案: