SBJson int 解析

Question

我有一个 json 字符串：

{"1":{"homeTeam":"Home01","awayTeam":"Away01","homeScore":0,"awayScore":0,"gameID":1},"2":{"homeTeam":"Home11","awayTeam":"Away11","homeScore":0,"awayScore":0,"gameID":2},"3":{"homeTeam":"Home21","awayTeam":"Away21","homeScore":0,"awayScore":0,"gameID":3}}

我想将其转换为 Objective-C 类：

#import <Foundation/Foundation.h>


@interface ScoreKeeperGame : NSObject {

}

@property (nonatomic, retain) NSString *homeTeam;
@property (nonatomic, retain) NSString *awayTeam;
@property (nonatomic, assign) int homeScore;
@property (nonatomic, assign) int awayScore;
@property (nonatomic, assign) int gameID;

- (id) init: (NSString *) homeTeam 
   awayTeam: (NSString *) awayTeam;

- (id) init: (NSDictionary *) game;

@end

我通过 NSDictionary“game” 传入 json（json 字符串代表哈希图，因此第一个游戏是）：

{"homeTeam":"Home01","awayTeam":"Away01","homeScore":0,"awayScore":0,"gameID":1}

当我尝试使用此时类：

#import "ScoreKeeperGame.h"


@implementation ScoreKeeperGame
@synthesize homeTeam=_homeTeam, awayTeam = _awayTeam, homeScore = _homeScore, awayScore     = _awayScore, gameID = _gameID;

- (id) init: (NSString *) homeTeam 
   awayTeam: (NSString *) awayTeam
{
    self = [super init];
    self.homeTeam = homeTeam;
    self.awayTeam = awayTeam;

NSLog(@"away: %@", awayTeam);
return self;
}

- (id) init: (NSDictionary *) game
{
    self = [super init];
    if(self)
    {
    self.homeTeam = [game objectForKey:@"homeTeam"];
    self.awayTeam = [game objectForKey:@"awayTeam"];
    self.awayScore = (int) [game objectForKey:@"awayScore"];
    self.gameID = [game objectForKey:@"gameID"];

    NSLog(@"game awayScore: %d", self.awayScore);
    NSLog(@"game gameID: %@", [NSNumber numberWithInt:self.gameID]);
    }
return self;
}

@end

awayScore 和gameId 被打印为大数字（也许是指针）？

我已经尝试过

self.awayScore = [NSNumber numberWithInt: [game objectForKey:@"awayScore"]];

但这似乎也不起作用。

如何从游戏对象中获取 int 值？

po game 产生：

{
awayScore = 0;
awayTeam = Away01;
gameID = 1;
homeScore = 0;
homeTeam = Home01;
}

谢谢！

Answer 1

如果您想从 JSON 响应中获取整数值，您可以尝试此操作，

self.awayScore = [NSNumber numberWithInt: [[game objectForKey:@"awayScore"] intValue]];

因为您的响应将在 nsstring 中。

Answer 2

iOs 6.0，现在只是 [game[@"awayScore"] intValue]

Answer 3

您必须首先将其作为 NSNumber 取出。

NSNumber *myNum = [game objectForKey:@"awayScore"];

self.awayScore = [myNum intValue];

哎哟！