Question

weather =     (
            {
        description = mist;
        icon = 50n;
        id = 701;
        main = Mist;
    },
            {
        description = "heavy intensity drizzle";
        icon = 09n;
        id = 302;
        main = Drizzle;
    }
)

我想要获取＆＃34; main＆＃34;由此，我正在使用此

这样做

NSDictionary *dic1 = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
NSDictionary *dic4 = [dic1 objectForKey:@"weather"];

NSString * weatherStr =  
    [NSString stringWithFormat:@"Weather type:%@",[dic4 valueForKey:@"main"]];
weathertypeLabel.text =  
    [NSString stringWithFormat:@"Weather type:%@",weatherStr];

但我的o / p是

Weather type:(
    Mist,
    Drizzle
)

Answer 1

当你这么说时，你只是在开玩笑：

sudo service networking restart

尽管您声明了，NSDictionary *dic4 = [dic1 objectForKey:@"weather"]; 不是字典。它是一个字典数组（NSArray），包含两个字典。因此，您一次从中提取dic4键的值。

如果那不是你想要的，那么你需要决定你想要的词典，"main"或dic4[0]。解压缩，然后获取其dict4[1]。

Answer 2

NSString * weatherStr =  
[NSString stringWithFormat:@"Weather type:%@",[dic4 valueForKey:@"main"]];

使用此

 NSString * weatherStr =  
[NSString stringWithFormat:@"Weather type:%@",[[dic4 objectForKey:@"weather"]objectForKey :@"main"]];

如何在objective-c中传递此JSON的值？

2 个答案: