我搜索过完整的谷歌搜索。但没有找到答案。 我在解析json结果中的对象数组时遇到了问题。 我有这样的json结果:
[{“created_at”:“Sat Apr 28 13:36:24 +0000 2012”, “text”:“@ LisaPrejean爱你的工作!干得好!”, “source”:“web”,“user”:{“id”:478983313,“name”:“3rd Dimension Media”, “created_at”:“Mon Jan 30 21:51:20 +0000 2012”,“favourites_count”:0}, “retweet_count”:0}]
&安培;从结果我想只提取“created_at”& “文本”
(void)connectionDidFinishLoading:(NSURLConnection *)连接 {
NSString * jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding]; NSLog(@“%@”,jsonString); self.jsonData = nil;NSArray *tweetArray = [jsonString JSONValue]; tweetTextArray = [NSMutableArray array]; for (NSDictionary *tweet in tweetArray)
{ [tweetTextArray addObject:[tweet objectForKey:@“text”]]; [tweetCreated_atArray addObject:[tweet objectForKey:@“created_at”]]; } }
但在线获取EXC_bad_access错误: NSArray * tweetArray = [jsonString JSONValue];
答案 0 :(得分:-1)
试试这个
NSString *responseString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
SBJSON *parser = [[SBJSON alloc] init];
NSDictionary *data = (NSDictionary *) [parser objectWithString:responseString error:nil];
NSString *text = (NSString *) [data objectForKey:@"text"];
NSLog(@"textvalue= %@",text);
NSString * created_at = (NSString *) [data objectForKey:@"created_at"];
NSLog(@"created_at= %@",created_at);