ios - 模态关闭后显示错误消息

Question

所以就像问题所暗示的那样，我试图在用户向服务器发送一些数据后不卡住 UI。就我而言，他们可能会发送大量数据，而服务器端我必须为多个查询执行 foreach 循环。

虽然发生了所有这些，但我不希望用户等待响应，因此我在单击“发送”后关闭模态 VC。数据仍然被插入到数据库中，但如果出现错误怎么办？现在我在模态 VC 被关闭后显示一个 UIAlertView 但我得到一个错误的访问错误。显示错误的最佳方式是什么？

- (void)send:(id)sender{
if ([[Data sharedInstance].someData objectForKey:@"time"] != NULL) {
    [self dismissViewControllerAnimated:YES completion:^(){
        NSMutableDictionary *paramDic = [NSMutableDictionary new];
        [paramDic setValue:[[Data sharedInstance].someData objectForKey:@"oneArray"] forKeyPath:@"oneArray"];
        [paramDic setValue:[NSArray arrayWithObjects:[[[Data sharedInstance].someData objectForKey:@"two"] valueForKey:@"Name"], [[[Data sharedInstance].someData objectForKey:@"two"] valueForKey:@"State"], [[[Data sharedInstance].someData objectForKey:@"two"] valueForKey:@"Country"], nil] forKeyPath:@"twoArray"];
        [paramDic setValue:[[[Data sharedInstance].someData objectForKey:@"three"] stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding] forKeyPath:@"three"];
        [paramDic setValue:[[NSUserDefaults standardUserDefaults] valueForKey:@"username"] forKey:@"username"];
        NSData *jsonData = [NSJSONSerialization dataWithJSONObject:paramDic options:NSJSONWritingPrettyPrinted error:nil];
        NSURL *url = [NSURL URLWithString:@"http://localhost/myapp/handleData.php"];
        NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
        [request setHTTPMethod:@"POST"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
        NSString *length = [NSString stringWithFormat:@"%d", jsonData.length];
        [request setValue:length forHTTPHeaderField:@"Content-Length"];
        [request setValue:@"json" forHTTPHeaderField:@"Data-Type"];
        [request setHTTPBody:jsonData];
        [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *error){
            NSHTTPURLResponse *httpResponse = (NSHTTPURLResponse*)response;
            if (![httpResponse statusCode] == 200 || ![[[NSString alloc]initWithData:data encoding:NSUTF8StringEncoding] isEqualToString:@"success"]) {
                UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Error" message:@"Problem on the server.  Please try again later." delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
                [alert show];
            }
        }];
    }];
}

我现在就是这样做的...什么是更好的方法？

Answer 1

以上回答很有值(value)。但我认为这是导致问题的原因。

改变:

UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Error" message:@"Problem on the server.  Please try again later." delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];

收件人:

UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Error" message:@"Problem on the server.  Please try again later." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];

希望这对您有所帮助。