ios - 在iOS中解析Json而没有常见的 key

Question

我的JSON字符串是:

[
    {'Local':'webaddress'},
    {'QA':'webaddress1'}
]

我的代码是:

NSMutableDictionary *dataDictonary = [NSJSONSerialization
                                       JSONObjectWithData:responseData
                                       options:0
                                       error:nil];


        NSArray *keys = [dataDictonary allKeys];
        NSArray *values = [dataDictonary allValues];
        int i=0;
        NSLog(@"",[keys count]);
        NSLog(@"",[values count]);
        int i=0;
        for ( NSString *items in keys )
        {
            NSLog(@"----");
            NSLog(@"Name: %@", items);
            NSLog(@"Address: %@", values[i++]);
            NSLog(@"----");
        }

在这里，我得到的是大小，因为NSlog中没有空白，并且无法解析此值，请不要这样做。请帮忙..

Answer 1

您的JSON是一个内部包含对象的数组，因此您需要将其转换为NSArray:

NSString *json_string = @"[{\"Local\": \"webaddress\" }, {\"QA\": \"webaddress1\" }]";
NSError *error;
NSArray *JSON =
[NSJSONSerialization JSONObjectWithData: [json_string dataUsingEncoding:NSUTF8StringEncoding]
                                options: NSJSONReadingMutableContainers
                                  error: &error];

NSLog(@"Local: %@", JSON[0][@"Local"]); // output is: Local: webaddress

更新

// itherate through array
for(NSDictionary *dictionary in JSON)
{
    //now you can iterate throug each dicitonary
    NSEnumerator *enumerator = [dictionary keyEnumerator];
    id key;

    while((key = [enumerator nextObject])){
        NSLog(@"key=%@ value=%@", key, [dictionary objectForKey:key]);
    }
}

日志如下所示:

键=本地值=网址

键=质量检查值=网址1