如何從url ini解析json數據iOS無響應輸出數據

我在解析iOS中的json數組時遇到問題，錯誤在於Objective-C源代碼中json的結構，我附加了objective-c和JSON ，這是我的源代碼ViewController.m：如何從url ini解析json數據iOS無響應輸出數據

- (void)connectionDidFinishLoading:(NSURLConnection *)connection 
{ 
NSDictionary *allDataDictionary = [NSJSONSerialization JSONObjectWithData:webData options:0 error:nil]; 
NSDictionary *JSON = [allDataDictionary objectForKey:@"JSON"]; 
NSArray *arrayOfJSON = [JSON objectForKey:@"genres"]; 

for (NSDictionary *diction in arrayOfJSON) { 
    NSDictionary *JSON = [diction objectForKey:@"genres"]; 
    NSString *id = [feed objectForKey:@"id"]; 
    NSString *name = [feed objectForKey:@"name"]; 

    [array addObject:id]; 
    [array addObject:name]; 

} 
[[self myTableView ]reloadData]; 

} 
- (IBAction)getData:(id)sender { 



NSURL *url = [NSURL URLWithString:@"http://api.themoviedb.org/3/genre/list?api_key=65b56d5d3fa43ad24d10b2786b3d0b96"]; 


NSURLRequest *request = [NSURLRequest requestWithURL:url]; 

Connection = [NSURLConnection connectionWithRequest:request delegate:self]; 
if(Connection) 
{ 
    webData = [[NSMutableData alloc]init]; 
} 
}

這是我的源代碼ViewController.h：

#import <UIKit/UIKit.h> 

@interface ViewController : UIViewController<UITableViewDataSource, UITableViewDelegate, NSURLConnectionDataDelegate> 
@property (weak, nonatomic) IBOutlet UITableView *myTableView; 

- (IBAction)getData:(id)sender; 
@end

這是我的JSON數據：

{"genres":[{"id":28,"name":"Action"},{"id":12,"name":"Adventure"},  {"id":16,"name":"Animation"},{"id":35,"name":"Comedy"},{"id":80,"name":"Crime"},{"id":105,"name":"Disaster"},{"id":99,"name":"Documentary"},{"id":18,"name":"Drama"},{"id":82,"name":"Eastern"},{"id":2916,"name":"Erotic"},{"id":10751,"name":"Family"},{"id":10750,"name":"Fan Film"},{"id":14,"name":"Fantasy"},{"id":10753,"name":"Film Noir"},{"id":10769,"name":"Foreign"},{"id":36,"name":"History"},{"id":10595,"name":"Holiday"},{"id":27,"name":"Horror"},{"id":10756,"name":"Indie"},{"id":10402,"name":"Music"},{"id":22,"name":"Musical"},{"id":9648,"name":"Mystery"},{"id":10754,"name":"Neo-noir"},{"id":1115,"name":"Road Movie"},{"id":10749,"name":"Romance"},{"id":878,"name":"Science Fiction"},{"id":10755,"name":"Short"},{"id":9805,"name":"Sport"},{"id":10758,"name":"Sporting Event"},{"id":10757,"name":"Sports Film"},{"id":10748,"name":"Suspense"},{"id":10770,"name":"TV movie"},{"id":53,"name":"Thriller"},{"id":10752,"name":"War"},{"id":37,"name":"Western"}]}

對不起，它不工作。這裏我的更新代碼

- (NSArray *) genreList 
{ 
    if(!gendreList) 
{ 

    NSURL *url = [NSURL URLWithString:@"http://api.themoviedb.org/3/genre/list?api_key=65b56d5d3fa43ad24d10b2786b3d0b96"]; 
    NSURLRequest *request = [NSURLRequest requestWithURL:url]; 
    [NSURLConnection sendAsynchronousRequest:request 
             queue:[NSOperationQueue mainQueue] 
          completionHandler:^(NSURLResponse *response, 
               NSData *data, NSError *connectionError) 
    { 
     if (data.length > 0 && connectionError == nil) 
     { 
      NSDictionary *jsonDict = [NSJSONSerialization JSONObjectWithData:data options:0 error:NULL]; 

      gendreList = [[jsonDict valueForKey:@"genres"] valueForKey:@"name"]; 
      self.idGenreList = [[jsonDict valueForKey:@"genres"] valueForKey:@"id"]; 

          } 
     else if (connectionError != nil) 
     { 
      NSLog(@"No Internet!"); 
     } 
    }]; 
} 
    self.gendreList = gendreList; 
NSLog(@"%@",self.genreList); 
return gendreList; 
}

但它仍然沒有輸出響應。你能告訴我那個代碼有什麼問題嗎？對不起，我新手在iOS編程的傢伙..

來源

2014-03-05 user3213595

那麼什麼是錯誤？ – Zen

'NSDictionary * JSON = [allDataDictionary objectForKey：@「JSON」]的用途是什麼;'有沒有與我們分享的JSON級別？ –

NSDictionary * JSON = [allDataDictionary objectForKey：@「JSON」];不需要...你可以直接調用這個方法。NSArray * arrayOfJSON = [allDataDictionary objectForKey：@「genres」]; –

下面是代碼

- (void)connectionDidFinishLoading:(NSURLConnection *)connection 
{ 
    NSDictionary *allDataDictionary = [NSJSONSerialization JSONObjectWithData:webData options:0 error:nil]; 
    NSArray *arrayOfJSON = [allDataDictionary objectForKey:@"genres"]; 

    for (NSDictionary *diction in arrayOfJSON) { 
      NSString *id = [diction objectForKey:@"id"]; 
      NSString *name = [diction objectForKey:@"name"]; 

      [array addObject:id]; 
      [array addObject:name]; 

    } 
    [[self myTableView ]reloadData]; 

}

你解析數據後，你會直接獲取數據..

來源

2014-03-05 05:54:17

ü得到的迴應，現在ü希望解析數據。完美

ü做了Web服務，現在笏ü需要做的就是它解析爲一個數組首先下載SBJSON文件中的鏈接

https://github.com/stig/json-framework/

然後，將它們複製到你的工作區。然後，在的viewController添加此

#import "SBJson.h"

你的JSON數據包含在字典的形式

SO價值觀，解析他們

SBJsonParser * parser=[SBJsonParser new]; 
NSDictionary * jsonData=(NSDictionary *)[parser objectWithString:outputData]; 
NSDictionary * dict=(NSDictionary *)[NSDictionary objectForKey:@"genres"]; 
NsDictionary * id=(NSDictionary*)[dict objectForKey:@"id"]; 
NsDictionary * name=(NSDictionary*)[dict objectForKey:@"name"];

我認爲這將有助於

來源

2014-03-05 05:56:53 Chandru

如何從url ini解析json數據iOS無響應輸出數據

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