ios - iOS-异步创建的数组仅返回一个对象

Question

我真的很难弄清楚这一点。我正在将JSON异步解析为数组。在异步函数上运行NSLog可以打印出包含多个对象的数组，这正是我想要的。但是，当我在ViewController中返回的数组上运行NSLog时，它只会打印出数组的最后一个对象。然后，我对其进行计数，实际上，数组中只有1个对象。为什么只从具有多个对象的数组中返回只有一个对象的数组？下面是我的代码。感谢您的输入。

函数异步执行

- (NSArray *)locationsFromJSONFile:(NSURL *)url {
NSURLRequest *request = [NSURLRequest requestWithURL:url
                                             cachePolicy:NSURLRequestReloadIgnoringLocalAndRemoteCacheData
                                     timeoutInterval:30.0];


NSURLResponse *response;
NSData *data = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:nil];

NSError *error;
NSMutableDictionary *allTeams = [NSJSONSerialization
                                   JSONObjectWithData:data
                                   options:NSJSONReadingMutableContainers
                                   error:&error];

if( error )
{
    NSLog(@"%@", [error localizedDescription]);
}
else {
    team = allTeams[@"10"];
    for ( NSDictionary *teamArray in team )
    {
        teams = [NSArray arrayWithObjects: teamArray[@"team"], nil];

}
}

return teams;
}

ViewController

- (void)viewDidLoad
{
[super viewDidLoad];


JSONLoader *jsonLoader = [[JSONLoader alloc] init];
NSURL *url = [[NSBundle mainBundle] URLForResource:@"teams" withExtension:@"json"];

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
    teams = [jsonLoader locationsFromJSONFile:url];

    int i = [teams count];

    NSString *string = [NSString stringWithFormat:@"%d", i];

    NSLog(@"%@", string);

});
}

Answer 1

您需要将对象添加到现有数组中，而不是每次都重新初始化它

    teams = [NSArray arrayWithObjects: teamArray[@"team"], nil];

应该

else {
    team = allTeams[@"10"];
    teams = [NSMutableArray array]; 
    for ( NSDictionary *teamArray in team )
    {
        [teams addObject:teamArray[@"team"]];

}

另外，您正在使用此代码发送同步请求

NSData *data = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:nil];

您应该使用 sendAsynchronous方法。同步请求可能会被操作系统本身终止，并可能导致稍后的混乱和错误。