0

无法使用 AFNetworking 在块内同步接收 JSON。我检查了这个解决方案。在方法结束时它总是为零。

这是我的方法：

- (BOOL)whois:(NSString *)domain withZone: (NSString*) zone
{        
    __block NSString *resultCode;

    NSURL *url = [[NSURL alloc] initWithString:@"myurl"];
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];

    AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {     
        resultCode = [JSON valueForKeyPath:[NSString stringWithFormat:@"%@.%@", domain,zone]];  //checked with NSLog, works well              
    } failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
        NSLog(@"Request Failed with Error: %@, %@", error, error.userInfo);
    }];

    NSOperationQueue *queue = [[NSOperationQueue alloc] init];
    [queue addOperation: operation];
    [operation waitUntilFinished];

    if(resultCode == @"available") //nil here
    {
        return YES;
    }
    return NO; 
}
4

2 回答 2

0

而不是创建一个，NSOperationQueue开始你然后调用它会阻塞主线程直到它完成。那么你的 resultCode 不应该为零。AFJSONRequestOperation[operation start][operation waitUntilFinished]

正如@mattt在您链接的帖子中所说，强烈建议不要像这样冻结线程。考虑找出另一种方法来执行此操作，例如从成功块调用您希望继续的新方法，以及从失败块调用不同的失败方法。

于 2012-11-13T01:27:34.813 回答
0

您的方法无法在其当前设计下运行。

- (BOOL)whois:(NSString *)domain withZone: (NSString*) zone
{        
    __block NSString *resultCode;

    NSURL *url = [[NSURL alloc] initWithString:@"myurl"];
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];

    // *** Runs 1st
    AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {     

        // *** runs 3rd
        resultCode = [JSON valueForKeyPath:[NSString stringWithFormat:@"%@.%@", domain,zone]];  //checked with NSLog, works well              
    } failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
        NSLog(@"Request Failed with Error: %@, %@", error, error.userInfo);
    }];

    // *** Runs 2nd
    NSOperationQueue *queue = [[NSOperationQueue alloc] init];
    [queue addOperation: operation];
    [operation waitUntilFinished];

    if(resultCode == @"available") //nil here
    {
        return YES;
    }
    return NO; 
}

因为块中的材料是第三次异步运行，所以您将无法以当前设计的方式将该值返回给更大的方法。也许使用这样的东西：

- (void)whois:(NSString *)domain withZone: (NSString*) zone
{        

    NSURL *url = [[NSURL alloc] initWithString:@"myurl"];
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];

    __weak id weakSelf = self;
    // Runs 1st
    AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {     

        NSString *resultCode = [JSON valueForKeyPath:[NSString stringWithFormat:@"%@.%@", domain,zone]];  //checked with NSLog, works well       
        [weakSelf receivedResultCode:resultCode];       
    } failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
        NSLog(@"Request Failed with Error: %@, %@", error, error.userInfo);
    }];
}

- (void) receivedResultCode:(NSString *)resultCode {
    NSOperationQueue *queue = [[NSOperationQueue alloc] init];
    [queue addOperation: operation];
    [operation waitUntilFinished];

    if(resultCode == @"available") //nil here
    {
        // do @YES stuff
    }
    else {
        // do @NO stuff
    }
}

显然，您必须更改调用它的人的设计，因为它不会以您指定的方式返回值。也许有更好的解决方案，但我认为这是它工作所需的设计类型。

于 2014-04-07T20:12:00.937 回答