Question

我在这里遇到了类似的问题，但没有找到我正面临的问题的答案。

我到现在为止能够解析JSON数据并存储在字典中。以下是JSON数据的原始形式：

{"stores":[{"address":"7801 Citrus Park Town Center Mall","city":"Tampa","name":"Macy's","latitude":"28.068052","zipcode":"33625","storeLogoURL":"http://strong-earth-32.heroku.com/images/macys.jpeg","phone":"813-926-7300","longitude":"-82.573301","storeID":"1234","state":"FL"},

{"address":"27001 US Highway 19N","city":"Clearwater","name":"Dillards's","latitude":"27.9898988","zipcode":"33761","storeLogoURL":"http://strong-earth-32.heroku.com/images/Dillards.jpeg","phone":"727-296-2242","longitude":"-82.7294986","storeID":"1235","state":"FL"},

依旧......

如您所见，它是一个字典数组的字典。因此，我首先将原始数据存储在字典中，为value提取key = stores并将其存储在数组中。之后，我提取了每个字段并将其存储在自定义对象tempStore中。这是失败的时候。

- (void)viewDidLoad
{
[super viewDidLoad];
[populatedStoreArray addObject:@"blah"];
NSString *jsonRawData = [[NSString alloc] initWithContentsOfURL:[NSURL URLWithString:@"http://strong-earth-32.heroku.com/stores.aspx"]];

if([jsonRawData length] == 0)
{
    [jsonRawData release];
    return;
}
SBJsonParser * parser = [[SBJsonParser alloc]init];
resultData = [[parser objectWithString:jsonRawData error:nil]copy];
NSArray *storeArray = [[NSArray alloc]init];
storeArray= [resultData objectForKey:@"stores"];
Store *tempStore = [[Store alloc]init];

/*NSLog(@"show me stores: %@", storeArray);*/
for(int i=1;i<[storeArray count];i++)
{
    NSDictionary *tempDictionary = [storeArray objectAtIndex:i];
    if([tempDictionary objectForKey:@"address"]!=nil)
    {
        tempStore.address= [tempDictionary objectForKey:@"address"];
        //NSLog(@"Address: %@",tempStore.address);
    }
    //and so on for other keys
    [populatedStoreArray addObject:tempStore]; 
    NSLog(@"In array: %@",[populatedStoreArray objectAtIndex:i]);
}

这是tempStore对象：

- (id) init 
{
    if (self = [super init])
    {
    self.address = @"address";
    self.city = @"city";
    self.name = @"name";
    self.latitude = @"latitude";
    self.longitude = @"longitude";
    self.state = @"state";
    self.phone = @"phone";
    self.storeid = @"storeID";
    self.url = @"storeLogoURL";
    self.zipcode = @"zipcode";
    }
    return self;
}

现在，我使用populatedStoreArray来填充表格的单元格。我不确定要显示的格式，但我主要担心的是当我尝试打印populatedStoreArray时，即使已经填充了tempStore，其内容仍为null。我在这里错过了什么？此外，populatedStoreArray在.h文件中声明为属性。

@property (nonatomic, strong) NSMutableArray * populatedStoreArray;

提前致谢。

Answer 1

请先分配您的 NSMutableArray ，然后首先合成您的阵列

喜欢populatedStoreArray = [[NSMutableArray alloc] init];

Answer 2

正如Cocoa Matters上面所说，你需要确保你的populatedStoreArray被分配和初始化，当你尝试将对象添加到nil数组时，objective-c没有错误，所以它看起来像你添加它们，但你的不是

此外，我不知道它是否是您的实际代码，但我注意到您只分配并初始化tempStore一次。所以你循环遍历数组并设置tempStore.address并每次添加相同的对象，所以你只会在数组中得到一个tempStore对象，你需要在每次迭代中分配和初始化一个新的tempStore对象。你的循环：

Store *tempStore;

/*NSLog(@"show me stores: %@", storeArray);*/
for(int i=1;i<[storeArray count];i++)
{
    tempStore = = [[Store alloc]init];

    NSDictionary *tempDictionary = [storeArray objectAtIndex:i];
    if([tempDictionary objectForKey:@"address"]!=nil)
    {
        tempStore.address= [tempDictionary objectForKey:@"address"];
        //NSLog(@"Address: %@",tempStore.address);
    }
    //and so on for other keys
    [populatedStoreArray addObject:tempStore]; 
    NSLog(@"In array: %@",[populatedStoreArray objectAtIndex:i]);
}

在iOS（Objective-c）中解析JSON数据并在TableView中显示。获得空值

2 个答案: