如何从JSON响应中获取字段值

Question

我可以从API检索整个JSON响应并将其存储在NSArray中。

然后我遍历整个响应并将每个结果组存储在NSDictionary和NSLogit中。

不过，我不是我怎么能得到字段值从每个组中像

STNAME, 
CTYNAME, 
DENSITY, 
POP, 
DATE, 
state, 
county 


- (void)fetchedData:(NSData *)responseData { 
    //parse out the json data 
    NSError* error; 
    NSArray* json = [NSJSONSerialization JSONObjectWithData:responseData //1 
                 options:kNilOptions 
                  error:&error]; 

for (int i=0; i<[json count]; i++) { 
    NSDictionary *avatars = [json objectAtIndex:i]; 
     NSLog(@"value at %d, ::: %@", i,avatars); 
    } 

什么，我试图做的就是像

//  NSString *name = avatar[@"STNAME"]; // for any specific state or county name..like objectbyKey 

让我能从sep组中获取每个值。

这是我的JSON数据看起来像上的NSLog

2014-06-23 12:04:06.893 KivaJSONDemo[2693:90b] value at 0, ::: (
    STNAME, 
    CTYNAME, 
    DENSITY, 
    POP, 
    DATE, 
    state, 
    county 
) 
2014-06-23 12:04:06.894 KivaJSONDemo[2693:90b] value at 1, ::: (
    Florida, 
    "Alachua County", 
    "282.65234809", 
    247336, 
    1, 
    12, 
    001 
) 
2014-06-23 12:04:06.894 KivaJSONDemo[2693:90b] value at 2, ::: (
    Florida, 
    "Alachua County", 
    "282.65234809", 
    247336, 
    2, 
    12, 
    001 
) 
2014-06-23 12:04:06.895 KivaJSONDemo[2693:90b] value at 3, ::: (
    Florida, 
    "Alachua County", 
    "283.0454668", 
    247680, 
    3, 
    12, 
    001 
) 
2014-06-23 12:04:06.895 KivaJSONDemo[2693:90b] value at 4, ::: (
    Florida, 
    "Alachua County", 
    "285.30018541", 
    249653, 
    4, 
    12, 
    001 
) 
. 
. 
. 
. 
. 
. 
. 
. 
. 

工作：

我得到了它的工作，但它绝对不是这样做的最有效方式。

NSString *value = @"CTYNAME"; 
NSUInteger idx = [json[0] indexOfObject:value]; 

NSString *value2=@"POP"; 
NSUInteger idx1 = [json[0] indexOfObject:value2]; 

NSString *value3=@"DENSITY"; 
NSUInteger idx2 = [json[0] indexOfObject:value2]; 


if (idx != NSNotFound) 
{ 
    for (NSUInteger i = 1; i < [json count]; i=i+6) 
    { 
     if (idx < [json[i] count]) 
     { 
      NSLog(@"County:%@    Population:%@   Density:%@", json[i][idx],json[i][idx1],json[i][idx2]); 


      [CountyNames addObject:json[i][idx]]; 
      [CountyPopulation addObject:json[i][idx1]]; 
      [CountyDensity addObject:json[i][idx2]]; 

     } 
     else 
     { 
      NSLog(@"Value %@ unavailable in %@", value, json[i]); 
     } 
    } 
} 
else 
{ 
    NSLog(@"Value %@ not found.", value); 
} 

Answer 1

看起来你从服务器回来的数据来自数组，而不是字典数组。你正在做这样的事情。

NSArray *avatars = @[@[@"STNAME", 
         @"CTYNAME", 
         @"DENSITY", 
         @"POP", 
         @"DATE", 
         @"state", 
         @"county"], 
        @[@"Florida", 
         @"Alachua County", 
         @"282.65234809", 
         @247336, 
         @1, 
         @12, 
         @001], 
        @[@"Florida", 
         @"Alachua County", 
         @"282.65234809", 
         @247336, 
         @2, 
         @12, 
         @001], 
        @[@"Florida", 
         @"Alachua County", 
         @"283.0454668", 
         @247680, 
         @3, 
         @12, 
         @001], 
        @[@"Florida", 
         @"Alachua County", 
         @"285.30018541", 
         @249653, 
         @4, 
         @12, 
         @001]]; 

NSString *value = @"DENSITY"; 
NSUInteger idx = [avatars[0] indexOfObject:value]; 
if (idx != NSNotFound) 
{ 
    for (NSUInteger i = 1; i < [avatars count]; ++i) 
    { 
     if (idx < [avatars[i] count]) 
     { 
      NSLog(@"%@", avatars[i][idx]); 
     } 
     else 
     { 
      NSLog(@"Value %@ unavailable in %@", value, avatars[i]); 
     } 
    } 
} 
else 
{ 
    NSLog(@"Value %@ not found.", value); 
} 

Answer 2

这是一个常见的方法，而不是重复的键为每个对象，其中，第一行包含密钥（列名称）和随后的行表示对象（web服务将回答几个行，其中的值数组映射到列）。生产什么你输入的期望，考虑以下因素：

NSMutableArray *resultDictionaries = [NSMutableArray array]; // this will be our result 
NSArray *keys = json[0]; 

for (int i=1; i<json.count; i++) { 
    NSArray *row = json[i]; 
    NSDictionary *dictionary = [NSDictionary dictionaryWithObjects:row forKeys:keys]; 
    [resultDictionaries addObject:dictionary]; 
}