NSJSONSerialization JSONObjectWithData returns NSString

Question

I have a NSJSONSerialization JSONObjectWithData where I pass a NSData object. As a result I get a NSString instead of an NSDictionary. The complete line I have is:

NSDictionary* jsonDec = [NSJSONSerialization JSONObjectWithData:jData options:NSJSONReadingAllowFragments error:&error];

And the return is:

enter image description here

Why don't I get a NSDictionary back? My backend is PHP on the server...

Thanks in advance!

What does the original JSON look like? Your server definitely hasn't accidentally encoded twice and returned a string object? `NSJSONSerialization` should normally return a dictionary or an array, and if you have `NSJSONReadingAllowFragments` set as you do, it will also happily return one of the primitive types if that's all that it finds. — Tommy, Dec 13 '13 at 19:28
that is the string representation of the dictionary. you could do: NSLog(@"%@",jsonDec); and would see the dict as string in your console log. — thorb65, Dec 13 '13 at 19:31
reason: '-[__NSCFString objectForKey:]: unrecognized selector sent to instance. This is the error I get if I try to do the following: [[jsonDec objectForKey:@"result"] valueForKey:@"UserID"] — CyberK, Dec 13 '13 at 19:33
@CyberK: It seems to me that your server is sending "nested JSON" (for whatever reason). Compare http://stackoverflow.com/questions/16948427/how-can-you-deserialize-an-escaped-json-string-with-nsjsonserialization for a similar issue, perhaps you can use that. — Martin R, Dec 13 '13 at 19:44
The problem is on the server, which is sending twice-encoded JSON; try posting the PHP code you're using there. — Jesse Rusak, Dec 13 '13 at 19:45
Server side this is happening: json_encode(array('replyCode'=>'success','result'=>array('UserID' =>'xxx'))); — CyberK, Dec 13 '13 at 19:52
You should return the result object. Maybe set a response header to indicate Success. — LJ Wilson, Dec 13 '13 at 19:53
I just echo the value in PHP... Strange thing is that everything looks OK. JSON looks valid etc, only I get a NSString back instead of a NSDictionary... — CyberK, Dec 13 '13 at 19:57
The issue seems to be on the server side of things. Web requests as JSON should be dictionaries an/or arrays. — LJ Wilson, Dec 13 '13 at 20:01
echo json_encode(array('replyCode'=>'success','result'=>array('UserID' => mysql_insert_id()))); — CyberK, Dec 13 '13 at 20:02
And what does that look like when you echo that out (raw data)? — LJ Wilson, Dec 13 '13 at 20:04
@CyberK Clearly, that's not the whole story on the PHP side, since your code in the comments below shows base64 decoding, etc. Somewhere else, you're re-encoding it as JSON again. — Jesse Rusak, Dec 14 '13 at 14:58
Yes I know, but what I do is: json_encode, then base64 that json, encode it with AES256, and then base64 that again. That is returned. On iOS I do the same process, but then decoding. And there I can see the result as a string in JSON, but if I then put that in the NSJSONSerializer I get the string back instead of the dictionary... If I put that again in an NSJSONSerializer, I get an array back. Problem is that the request sends multiple possibilities back. So one time that works, other time putting it 2 times in the serializer it fails... — CyberK, Dec 14 '13 at 20:29

user2828120 · Answer 1 · 2013-12-13T19:36:46.437

1

NSDictionary *jsonDec;
NSError* error = nil;
jsonDec = [NSJSONSerialization JSONObjectWithData: jData
                                          options: NSJSONReadingMutableContainers
                                            error: &error];

if (error)
{
     NSLog(@"Error: %@",error);
}

Apple doc: NSJSONReadingMutableContainers - Specifies that arrays and dictionaries are created as mutable objects.

edited Dec 13 '13 at 19:36

answered Dec 13 '13 at 19:30

user2828120

233
4
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Error Domain=NSCocoaErrorDomain Code=3840 "The operation couldn’t be completed. (Cocoa error 3840.)" (JSON text did not start with array or object and option to allow fragments not set.) UserInfo=0x10d25c950 {NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.} – CyberK Dec 13 '13 at 19:32
your variable that you want to serialize – user2828120 Dec 13 '13 at 19:41
NSString *decodedString = [[NSString alloc] initWithData:decMutData encoding:NSUTF8StringEncoding]; NSData *jData = [NSData dataFromBase64String:decodedString]; – CyberK Dec 13 '13 at 19:43
you mess up encodings. – user2828120 Dec 13 '13 at 19:44
why is that? What do I mess up? – CyberK Dec 13 '13 at 19:45
What does mutability have to do with his problem? – Jesse Rusak Dec 13 '13 at 19:46
check this code: NSData* jData = [NSData dataWithContentsOfURL:(NSURL *)] – user2828120 Dec 13 '13 at 19:46
That's not possible, the return string is decrypted before I get to this point so I cannot simply call dataWithContentsOfURL. I use AFNetworking – CyberK Dec 13 '13 at 19:49
Or check this code: NSData *jData = [[NSData alloc] initWithData:decMutData]; – user2828120 Dec 13 '13 at 19:49
if not help - check the coding on the server – user2828120 Dec 13 '13 at 19:56
The [NSData dataFromBase64String is an extension on NSData. – CyberK Dec 13 '13 at 19:56
I do not understand for what you use this method... If you use AFNetworking: success block: (void (^)(NSURLRequest *request, NSHTTPURLResponse *response, id JSON))success. jData = JSON – user2828120 Dec 13 '13 at 20:02
because my return value is a AES256 encrypted string, it first gets decrypted into a base64 string etc... But we are going off topic here I think ;-) – CyberK Dec 13 '13 at 20:10

score 0 · Answer 2 · answered Dec 13 '13 at 19:42

0

I use this when consuming JSON responses (mine usually are arrays of dictionaries). I prefer to use object type id and just use introspection to determine whether it is an NSArray or NSDictionary. BTW - my web requests are either done using AFNetworking (v2) or using the Azure SDK.

id jsonDec = [NSJSONSerialization JSONObjectWithData:jData options:0 error:&error];

answered Dec 13 '13 at 19:42

LJ Wilson

14,445
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What is your answer differs from mine? – user2828120 Dec 13 '13 at 19:43
1

Well you assign the serialized object into an `NSDictionary` which most of the time is a mistake. Unless you know that all your responses will be a JSON dictionary, you should use id and then find out what you have. Most of my requests end up as an `NSArray` of dictionaries. You also use the option to create mutable objects which is not as efficient as using the immutable counterparts. – LJ Wilson Dec 13 '13 at 19:50
And BTW - I DID NOT down-vote your answer. I really wish that when people down-vote an answer, the person also leaves a comment explaining why. – LJ Wilson Dec 13 '13 at 19:51
id is giving me the exact same result as if I wrote NSDictionary – CyberK Dec 13 '13 at 20:00
1

While this is a reasonable practice, I don't think it has anything to do with the problem the OP is having. – Jesse Rusak Dec 14 '13 at 15:00

NSJSONSerialization JSONObjectWithData returns NSString

2 Answers2