0

JSON 响应像 - {"response":{"Success":"Y","items":[{"userid":"255"}]}}

我试图这样解析:

-(void)connectionDidFinishLoading:(NSURLConnection *)connection
{

    NSString *jsonStr = [[NSString alloc] initWithData:mutaebleData encoding:NSUTF8StringEncoding];
    NSLog(@"JSonSTr : %@", jsonStr);
    
    SBJSON *json = [[SBJSON alloc]init];
    
    NSDictionary *dic = (NSDictionary *) [json objectWithString:jsonStr];
    NSDictionary *dic1 = (NSDictionary *) [dic objectForKey:@"response"];
    NSDictionary *dic2 = (NSDictionary *) [dic1 objectForKey:@"Success"];
    NSDictionary *dic3 = (NSDictionary *) [dic1 objectForKey:@"items"];

    NSDictionary *dic4 = (NSDictionary *) [dic3 objectForKey:@"userid"]; // App crash in this line
}

如何获取用户标识值?

4

4 回答 4

2

这就是问题:

 NSDictionary *dic4 = (NSDictionary *) [dic3 objectForKey:@"userid"];

你应该使用:

NSDictionary *dic4 = (NSDictionary *) [[dic3 objectAtIndex:0] objectForKey:@"userid"];
于 2013-01-28T07:27:15.573 回答
0
-(void)connectionDidFinishLoading:(NSURLConnection *)connection
{

NSString *jsonStr = [[NSString alloc] initWithData:mutaebleData encoding:NSUTF8StringEncoding];
NSLog(@"JSonSTr : %@", jsonStr);

SBJSON *json = [[SBJSON alloc]init];

NSDictionary *dic = (NSDictionary *) [json objectWithString:jsonStr];
NSDictionary *dic1 = (NSDictionary *) [dic objectForKey:@"response"];
NSDictionary *dic2 = (NSDictionary *) [dic1 objectForKey:@"Success"];
NSArray *arr3 = (NSArray *) [dic1 objectForKey:@"items"];

NSString *str = [[arr objectAtIndex:0]objectForKey:@"userid"];
NSlog(@"userid ==%@",str);
// NSDictionary *dic4 = (NSDictionary *) [dic3 objectForKey:@"userid"]; // App crash in this line
 }
于 2013-01-28T07:30:40.110 回答
0

您可以添加 SBJSON 框架文件。并通过解析

NSDictionary *responseDict = [response JSONValue];
just parse it as normal dictionary.

NSString *userIDValue=[NSString stringWithFormat:@"%@",[[[[responseDict valueForKey:@"response"]valueForKey:@"items"]objectAtIndex:0]valueForKey:@"userid"]];
于 2013-01-28T07:48:22.920 回答
-1
NSDictionary *dic = (NSDictionary *) [json objectWithString:jsonStr];
 NSDictionary *dic1 = (NSDictionary *) [dic objectForKey:@"response"];
 NSDictionary *dic2 = (NSDictionary *) [dic1 objectForKey:@"Success"];
 NSDictionary *dic3 = (NSDictionary *) [dic1 objectForKey:@"items"];
    
 for(NSDictionary *str in dic3)
 {
    NSLog(@"str:%@",[str valueForKey:@"userid"]);
 }
于 2013-01-28T08:00:45.017 回答