Question

我真的不知道这是不是重复的帖子，但我真的不知道如何找到我正在搜索的内容。我有一个来自json的数据，这个数据的格式是这样的：

{
Age = (
"15",
"23",
"44"
),
Name = (
"Jack",
"Mark",
"Steve"
),
City = (
"New York",
"Los Angeles",
"Miami"
);
}

我需要有关Dictionary风格的数据，如下所示：

{
Age = 15;
Name = Jack;
City = New York;
},
{
Age = 23;
Name = Mark;
City = Los Angeles;
},
{
Age = 44;
Name = Steve;
City = Miami;
};

我使用此代码从json Data：

创建NSDictionary

NSArray *jsonResponse = [responseObject objectForKey:@"Age"];
NSMutableDictionary *allKeys = [NSMutableDictionary dictionary];

for (int i = 0; i < [jsonResponse count]; ++i)
{
  NSArray *jsonAge = [[responseObject objectForKey:@"Age"]objectAtIndex:i];
  NSArray *jsonName = [[responseObject objectForKey:@"Name"]objectAtIndex:i];
  NSArray *jsonCity = [[responseObject objectForKey:@"City"]objectAtIndex:i];

  [allKeys  setObject:jsonAge forKey:@"Age"];
  [allKeys  setObject:jsonName forKey:@"Name"];
  [allKeys  setObject:jsonCity forKey:@"City"];
}

我按照预期得到了这个结果：

   (
    {
    Age = 15;
    Name = Jack;
    City = New York;
    }
   )

问题是，我需要所有数据连续，这个提议给我分组在一个简单的组中。我怎么能在这样的一个团体中得到整个团体？

{
Age = 15;
Name = Jack;
City = New York;
},
{
Age = 23;
Name = Mark;
City = Los Angeles;
},
{
Age = 44;
Name = Steve;
City = Miami;
};

一些建议非常有用，非常感谢!!!!

Answer 1

更好的解决方案是在解析JSON对象后创建模型类并设置值。模型类将简化您在代码中的其他位置管理/修改数据的方式。

@interface Person : NSObject
@property (nonatomic, strong) NSString *age;
@property (nonatomic, strong) NSString *name;
@property (nonatomic, strong) NSString *city;
@end

然后，您可以解析数据并将它们存储在Person个对象的数组中。

NSArray *ageArray = [responseObject objectForKey:@"Age"];
NSArray *nameArray = [responseObject objectForKey:@"Name"];
NSArray *cityArray = [responseObject objectForKey:@"City"];

NSMutableArray *personArray = [NSMutableArray new];

for (int i = 0; i < ageArray.count; i++) {

  Person *person = [Person new];
  person.age = [ageArray objectAtIndex: i];
  person.name = (i < nameArray.count) ? [nameArray objectAtIndex: i] : nil;
  person.city = (i < cityArray.count) ? [cityArray objectAtIndex: i] : nil;

  [personArray addObject: person];
}

注意：如果可以修改JSON服务器响应以表示多个Person对象而不是单个属性的数组，则应该更新它。

从数组json Xcode Objective-c格式化字典

1 个答案: