Question

这是Raywenderlich（http://www.raywenderlich.com/5492/working-with-json-in-ios-5）的JSON编辑样本。我需要进一步整合POST数据，例如用户名，以便从URL获取响应。我该怎么办？感谢。

#define kBgQueue dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0) //1
#define kLatestKivaLoansURL [NSURL URLWithString: @"http://www.url.org/json.php"] //2

#import "ViewController.h"

@interface NSDictionary(JSONCategories)
+(NSDictionary*)dictionaryWithContentsOfJSONURLString:(NSString*)urlAddress;
-(NSData*)toJSON;
@end

@implementation NSDictionary(JSONCategories)
+(NSDictionary*)dictionaryWithContentsOfJSONURLString:(NSString*)urlAddress
{
    NSData* data = [NSData dataWithContentsOfURL: [NSURL URLWithString: urlAddress] ];
    __autoreleasing NSError* error = nil;
    id result = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:&error];
    if (error != nil) return nil;
    return result;
}

-(NSData*)toJSON
{
    NSError* error = nil;
    id result = [NSJSONSerialization dataWithJSONObject:self options:kNilOptions error:&error];
    if (error != nil) return nil;
    return result;    
}
@end

@implementation ViewController

- (void)viewDidLoad
{
    [super viewDidLoad];

    dispatch_async(kBgQueue, ^{
        NSData* data = [NSData dataWithContentsOfURL: kLatestKivaLoansURL];
        [self performSelectorOnMainThread:@selector(fetchedData:) withObject:data waitUntilDone:YES];
    });
}

- (void)fetchedData:(NSData *)responseData {
    //parse out the json data
    NSError* error;
    NSDictionary* json = [NSJSONSerialization JSONObjectWithData:responseData //1
                                                         options:kNilOptions 
                                                           error:&error];

    NSNumber* idd = [json objectForKey:@"id"];
    NSString* name = [json objectForKey:@"name"];
    NSString* email = [json objectForKey:@"email"];

    // 3) Set the label appropriately
    humanReadble.text = [NSString stringWithFormat:@"ID: %@ \nName: %@ \nEmail: %@", idd, name, email];

}

@end

Answer 1

你必须做这样的事情来发布数据形式json。

-(void) retrieveData{

   NSMutableURLRequest *request=[NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"yoururl"]];


  [request setHTTPMethod:@"GET"];


  [request setValue:@"application/json;charset=UTF-8" forHTTPHeaderField:@"content-type"];


   NSError *err;

   NSURLResponse *response;

   NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];

   NSArray *jsonArray = [NSJSONSerialization JSONObjectWithData:responseData options: NSJSONReadingMutableContainers error: &err];


   name.text=[[jsonArray objectAtIndex:0]objectForKey:@"title"];

   til.text=[[jsonArray objectAtIndex:0]objectForKey:@"place"];

  }

Answer 2

您无法使用 dataWithContentsOfURL 发布数据。为此，您必须使用 NSMutableURLRequest 。请参阅下面的精彩教程：

http://www.mysamplecode.com/2013/04/ios-http-request-and-json-parsing.html

iOS发布数据以获取JSON响应

2 个答案: