从目标C发送变量到url字符串

Question

我在下面的代码尝试在我的iPhone应用程序中创建反馈表单并通过php邮件程序页传递问题时遇到问题，我不知道下面的错误可能是原因。

下面的代码显示了这个问题，特别是

NSString *email = [self.emailTxt text]; 
NSString *message = [self.messageTxt text]; 
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"%@?email=%@&message=", FeedbackURL, email, message]]; 
NSURLRequest *request = [NSURLRequest requestWithURL:url]; 

[NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) { 
    NSString *title = (!error)? @"Message Sent" : @"Error"; 
    NSString *message = (!error)? @"Thank you for your Feedback, we hope it will help us to make this product even better." : @"There was an connection error. Please make sure you have internet connection and try again later."; 
    UIAlertView *alert = [[UIAlertView alloc] initWithTitle:title message:message delegate:nil cancelButtonTitle:@"Close" otherButtonTitles:nil, nil]; 
    [alert show]; 

    NSLog(@"Error: %@", error.localizedDescription); 
}]; 


if (email && (email.length > 0)) { 
    NSUserDefaults *userDefaults = [NSUserDefaults standardUserDefaults]; 
    [userDefaults setObject:email forKey:UserEmailKey]; 
    [userDefaults synchronize]; 
} 

Answer 1

你忘了第三%@指的这个@"%@?email=%@&message="“未使用的格式字符串数据的说法”。

@"%@?email=%@&message=" 

第三个参数未填写，因为您缺少%@。

Answer 2

看起来像你缺少你的格式字符串一个%@：

@"%@?email=%@&message=" 

应该

@"%@?email=%@&message=%@"