1

我是菜鸟,我想要的是将 NSDictionary 对象转换为 NSInteger.im 解析 json 对象并将其存储在 NSMutableArray 中,然后将其存储在名为 boy 的 NSDictionary 对象中。你们能帮帮我吗。下面是代码。

        NSURLRequest *request = [NSURLRequest requestWithURL:jsonurl cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];
connection1=[[[NSURLConnection alloc] initWithRequest:request delegate:self] autorelease];


NSLog(@"jsonurl:%@",jsonurl);
self.jsonData=[NSData dataWithContentsOfURL:jsonurl];

NSDictionary *items=[NSJSONSerialization JSONObjectWithData:self.jsonData options:NSJSONReadingMutableLeaves error:nil];
NSLog(@"blah:%@",jsonArray);
items1 = [items objectForKey:@"ThingsTodo"];

story = [[NSMutableArray array]retain];
media1 = [[NSMutableArray array]retain];

url=[[NSMutableArray array]retain];
media2=[[NSMutableArray array]retain];
descriptiondesc=[[NSMutableArray array]retain];
for (NSDictionary *item in items1)
{
    [self.story addObject:[item objectForKey:@"Name"]];
    [self.media1 addObject:[item objectForKey:@"Status"]];
    [self.media2 addObject:[item objectForKey:@"Image"]];

}
    NSDictionary *boy=[self.media1 objectAtIndex:indexPath.row];
4

2 回答 2

10

把你的nsdictionary价值放进去nsstring,然后做一个NSInteger喜欢:

NSInteger n = [strVal intValue];
于 2012-02-02T05:16:44.947 回答
2

更具体地说,“intValue”给出“int”,而“integerValue”返回“NSInteger”,所以

NSInteger n = [strVal integerValue];
于 2014-05-25T13:20:37.903 回答