iphone - 将 JSON 提取到 iOS 应用程序中

Question

http://xtremeinspection.com/new2you4kids/app/android/get_category_brand.php

上面是 JSON 的 URL。

我想在“PARENT_CATEGORY_ID”=“0”的 xcode 上显示数据。

我使用了以下编码：

-(void) loadJSON
{
loadJSONData = [NSURL URLWithString:mainURL];

    NSLog(@"Fetch JSON URL : %@",loadJSONData);

    dispatch_async(kBgQueue, ^{
        NSData* data = [NSData dataWithContentsOfURL:
                        loadJSONData];
        [self performSelectorOnMainThread:@selector(fetchedData:)
                               withObject:data waitUntilDone:YES];
    });
}

- (void)fetchedData:(NSData *)responseData
{
    NSError* error;
    json = [NSJSONSerialization
            JSONObjectWithData:responseData
            options:kNilOptions
            error:&error];
    adultJSON = [json objectForKey:@"category"];
    NSLog(@"Data: %@", adultJSON);

    NSDictionary *typeData = [json objectForKey:@"category"];
    categTypeArray = [[NSMutableArray alloc] initWithCapacity:0];
    for (NSDictionary *types in typeData)
    {
        if([[NSString stringWithFormat:@"%@",[adultJSON valueForKey:@"PARENT_CATEGORY_ID"]] isEqualToString:@"0"])
        {
            [categTypeArray addObject:[types valueForKey:@"PARENT_CATEGORY_ID"]];
        }
    }
    NSLog(@"Data: %@", adultJSON);
    NSLog(@"Category Data: %@", categTypeArray);
}

但是categTypeArray 返回null。

我应该使用哪种方法？

Answer 1

错误在于以下代码：

if([[NSString stringWithFormat:@"%@",[adultJSON valueForKey:@"PARENT_CATEGORY_ID"]] isEqualToString:@"0"])

你写adultJSON而不是types.

将该代码更改为：

if([[NSString stringWithFormat:@"%@",[types valueForKey:@"PARENT_CATEGORY_ID"]] isEqualToString:@"0"])