JSON Parsing in iOS 7

Question

I am creating an app for as existing website. They currently has the JSON in the following format :

[

   {
       \"id\": \"value\",
       \"array\": \"[{\         


        

Answer 1

//-------------- get data url--------

NSURLRequest *request=[NSURLRequest requestWithURL:[NSURL URLWithString:@"http://echo.jsontest.com/key/value"]];

NSData *data=[NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
NSLog(@"response==%@",response);
NSLog(@"error==%@",Error);
NSError *error;

id jsonobject=[NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingMutableContainers error:&error];

if ([jsonobject isKindOfClass:[NSDictionary class]]) {
    NSDictionary *dict=(NSDictionary *)jsonobject;
    NSLog(@"dict==%@",dict);
}
else
{
    NSArray *array=(NSArray *)jsonobject;
    NSLog(@"array==%@",array);
}

Answer 2

• You may always unescape the jsonData before deliver it to NSJSONSerialization. Or you may use the string got to construct another json object to get the array.

• NSJSONSerialization is doing right, the value in your example should be a string.

Answer 3

As another answer has said, that value is a string.

You can get around it by turning that string into data, as it seems to be a valid json string and then parse that json data object back into an array which you can add to your dictionary as the value for the key.

Answer 4

#define FAVORITE_BIKE @"user_id=%@&bike_id=%@"
@define FAVORITE_BIKE @"{\"user_id\":\"%@\",\"bike_id\":\"%@\"}"
NSString *urlString = [NSString stringWithFormat:@"url here"];
NSString *jsonString = [NSString stringWithFormat:FAVORITE_BIKE,user_id,_idStr];
NSData *myJSONData =[jsonString dataUsingEncoding:NSUTF8StringEncoding];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:[NSURL URLWithString:urlString]];
[request setHTTPMethod:@"POST"];
NSMutableData *body = [NSMutableData data];
[body appendData:[NSData dataWithData:myJSONData]];
[request setHTTPBody:body];
NSError *error;
NSURLResponse *response;
NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSString *str = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
if(str.length > 0)
{
    NSData* data = [str dataUsingEncoding:NSUTF8StringEncoding];
    NSMutableDictionary *resDict =[NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
}

Answer 5

JSON default method :

+ (NSDictionary *)stringWithUrl:(NSURL *)url postData:(NSData *)postData httpMethod:(NSString *)method
{
    NSDictionary *returnResponse=[[NSDictionary alloc]init];

    @try
    {
        NSMutableURLRequest *urlRequest = [NSMutableURLRequest requestWithURL:url
                                                                  cachePolicy:NSURLRequestReloadIgnoringCacheData
                                                              timeoutInterval:180];
        [urlRequest setHTTPMethod:method];

        if(postData != nil)
        {
            [urlRequest setHTTPBody:postData];
        }

        [urlRequest setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
        [urlRequest setValue:@"application/json" forHTTPHeaderField:@"Accept"];
        [urlRequest setValue:@"text/html" forHTTPHeaderField:@"Accept"];

        NSData *urlData;
        NSURLResponse *response;
        NSError *error;
        urlData = [NSURLConnection sendSynchronousRequest:urlRequest
                                        returningResponse:&response
                                                    error:&error];
        returnResponse = [NSJSONSerialization
                          JSONObjectWithData:urlData
                          options:kNilOptions
                          error:&error];
    }
    @catch (NSException *exception)
    {
        returnResponse=nil;
    }
    @finally
    {
        return returnResponse;
    }
}

Return method:

+(NSDictionary *)methodName:(NSString*)string{
    NSDictionary *returnResponse;
    NSData *postData = [NSData dataWithBytes:[string UTF8String] length:[string length]];
    NSString *urlString = @"https//:..url....";
    returnResponse=[self stringWithUrl:[NSURL URLWithString:urlString] postData:postData httpMethod:@"POST"];    
    return returnResponse;
}

Answer 6

The correct JSON should presumably look something like:

[
    {
        "id": "value",
        "array": [{"id": "value"},{"id": "value"}]
    },
    {
        "id": "value",
        "array": [{"id": "value"},{"id": "value"}]
    }
]

But, if you're stuck this the format provided in your question, you need to make the dictionary mutable with NSJSONReadingMutableContainers and then call NSJSONSerialization again for each of those array entries:

NSMutableArray *array = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingMutableContainers error:&error];
if (error)
    NSLog(@"JSONObjectWithData error: %@", error);

for (NSMutableDictionary *dictionary in array)
{
    NSString *arrayString = dictionary[@"array"];
    if (arrayString)
    {
        NSData *data = [arrayString dataUsingEncoding:NSUTF8StringEncoding];
        NSError *error = nil;
        dictionary[@"array"] = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
        if (error)
            NSLog(@"JSONObjectWithData for array error: %@", error);
    }
}