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Andrew Ho
Andrew Ho

Reputation: 638

SBJson displays null

I am trying to parse some json data with SBJson to show the current temperature. The example code from this tutorial works perfect: Tutorial: Fetch and parse JSON

When I change the code to my json feed i get a null. I am kind of new to JSON but followed every tutorial and documentation I found. The json source i used: JSON Source

My code with sbjson:

NSString *responseString = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
self.responseData = nil;

NSArray* currentw = [(NSDictionary*)[responseString JSONValue] objectForKey:@"current_weather"];

//choose a random loan
NSDictionary* weathernow = [currentw objectAtIndex:0];

//fetch the data
NSNumber* tempc = [weathernow objectForKey:@"temp_C"];
NSNumber* weatherCode = [weathernow objectForKey:@"weatherCode"];


NSLog(@"%@ %@", tempc, weatherCode);

and of course I have already implemented the other sbjson code.

Upvotes: 0

Views: 560

Answers (3)

rid
rid

Reputation: 63442

There is no current_weather key in the JSON data you posted. The structure is:

{ "data": { "current_condition": [ { ..., "temp_C": "7", ... } ], ... } }

Here's a visual representation:

JSON visual representation

Therefore, to get to temp_C, you'd need to first obtain the top-level data property:

NSDictionary* json = (NSDictionary*)[responseString JSONValue];
NSDictionary* data = [json objectForKey:@"data"];

then, from that, obtain the current_location property:

NSArray* current_condition = [data objectForKey:@"current_condition"];

and finally, from the current_location array, get the element you're interested in:

NSDictionary* weathernow = [current_condition objectAtIndex:0];

Also note that temp_C and weatherCode are strings, not numbers. To transform them to numbers, instead of:

NSNumber* tempc = [weathernow objectForKey:@"temp_C"];
NSNumber* weatherCode = [weathernow objectForKey:@"weatherCode"];

you could use something like:

int tempc = [[weathernow objectForKey:@"temp_C"] intValue];
int weatherCode = [[weathernow objectForKey:@"weatherCode"] intValue];

(or floatValue / doubleValue if the value is not supposed to be an int, but rather a float or a double)

You would then use %d (or %f for float / double) as a format string:

NSLog(@"%d %d", tempc, weatherCode);

Upvotes: 2

Prasad G
Prasad G

Reputation: 6718

Use NSJSONSerialization instead of JSONValue.

NSData* data = [responseString dataUsingEncoding:NSUTF8StringEncoding];
            NSDictionary* jsonDict = [NSJSONSerialization
                                      JSONObjectWithData:data
                                      options:kNilOptions
                                      error:&error];
 NSLog(@"jsonDict:%@",jsonDict);

In your link, there is no current_weather key.

NSString* tempc = [[[[jsonDict objectForKey:@"data"] objectForKey:@"current_condition"] objectAtIndex:0] objectForKey:@"temp_C"];

Upvotes: 0

Ivan Alek
Ivan Alek

Reputation: 1919

Provided link returns json without current_weather parameter. There is only current_condition parameter, please review this.

Upvotes: 0

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