ios - 'NSInvalidArgumentException '-[NSTaggedPointerString count]: unrecognized selector sent to instance 0xa006162696a614e6'

Question

我是一名新手，正在设计一个天气预报应用程序作为我在 iOS 中的第一个项目。

我设计了一个UISearchBar来搜索城市名称。搜索栏的工作方式如下:

1) 第 1 步: 输入“Naji”。它在表格 View 中显示两个城市:纳吉纳吉

2) 第 2 步: 输入“纳吉”。它没有显示任何内容，因为结果为空。

(错误部分)

3)第 3 步:每当我输入“Najib”时，应用程序都会终止并出现以下错误:

*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[NSTaggedPointerString count]: unrecognized selector sent to instance 0xa006162696a614e6'

---

代码是:

    -(void)searchBar:(UISearchBar *)asearchBar textDidChange:(NSString *)searchText{

    if ([searchText length] <=                return;
    if ([searchText length]==0) {
            [discityname removeAllObjects];
            [discityname addObjectsFromArray:cityname];
    }else{
            [discityname removeAllObjects];
            NSURLSession *session=[NSURLSession sharedSession];
            NSString *complete_url=[NSString stringWithFormat:@"https://query.yahooapis.com/v1/public/yql?q=select%%20*%%20from%%20geo.places%%20where%%20text%%3D%%22%@%%25%%22&format=json&diagnostics=true&callback=",searchText];
            NSURLSessionDataTask *dataTask=[session dataTaskWithURL:[NSURL URLWithString:complete_url]completionHandler:^(NSData *data, NSURLResponse *response, NSError *error){
            NSDictionary *json=[NSJSONSerialization JSONObjectWithData:data options:0 error:nil];
            NSLog(@"%@",json);
            NSMutableDictionary *currentDict = [[NSMutableDictionary alloc]initWithDictionary:json[@"query"]];
            int a=[[currentDict valueForKey:@"count"] intValue];
            if (a!=0) {
                cityname=[[NSMutableArray alloc]init];
           // if ([currentDict valueForKey: @"results"] != [NSNull null])
          //  {
                NSMutableDictionary *conDict = [[NSMutableDictionary alloc]initWithDictionary:currentDict[@"results"]];
                NSMutableArray *places=[conDict valueForKey:@"place"];
                NSMutableArray *country=[places valueForKey:@"country"];
                NSMutableArray *countryname=[country valueForKey:@"content"];
                NSMutableArray *placeName= [places  valueForKey:@"name"];
                NSMutableArray *centroid=[places valueForKey:@"centroid"];
                NSMutableArray *latitude=[centroid valueForKey:@"latitude"];
                NSMutableArray *longitude=[centroid valueForKey:@"longitude"];
                //if ([placeName isKindOfClass:[NSMutableArray class]]) {
                //}
                //cityname=[[NSMutableArray alloc]initWithArray:placeName];
                NSLog(@"%lu",placeName.count);
                for (int i=0; i<placeName.count; i++) {
                    [cityname addObject:[placeName objectAtIndex:i]];
                     [countrynames addObject:[countryname objectAtIndex:i]];
                }
                dispatch_async(dispatch_get_main_queue(), ^{
                    discityname=[[NSMutableArray alloc]initWithArray:cityname];
                    discountryname=[[NSMutableArray alloc]initWithArray:countrynames];
                    latitudenames=[[NSMutableArray alloc]initWithArray:latitude];
                    longitudenames=[[NSMutableArray alloc]initWithArray:longitude];
                    [tableView reloadData];
                });
            }
           // for(int i=0;i<[placeName count];i++)
          //i  NSLog(@"The place is: %@ , %@, %@, %@",placeName[i],countryname[i],latitude[i],longitude[i]);
        }];
        [dataTask resume];
        for (NSString *string in discityname) {
            NSRange r=[string rangeOfString:searchText options:NSCaseInsensitiveSearch];
            if (r.location!=NSNotFound) {
                [discityname addObject:string];
            }
        }
    }
    [tableView reloadData];
}

答案:它应该显示以 json 形式返回的城市名称“Najib”。

Answer 1

你可以在Xcode中添加异常断点，它会告诉你哪一行崩溃了在 Xcode 中，打开左侧面板，转到显示断点导航器，在左下角单击加号图标并添加异常断点。