Getting error when parsing JSON iOS

Question

After get string from web service, i need to parse them. But something is going wrong.

Here is my code;

NSString *responseString = [request responseString];

    NSLog(@"response String = %@",responseString);

    NSData *tempData = [responseString dataUsingEncoding:NSUTF8StringEncoding];

    NSError *error = nil;

    NSString *innerJson = [NSJSONSerialization JSONObjectWithData:tempData
                                                          options:NSJSONReadingAllowFragments error:&error];      

    NSLog(@"innerJson = %@",innerJson);
    NSArray *entries = [NSJSONSerialization JSONObjectWithData:[innerJson dataUsingEncoding:NSUTF8StringEncoding]
                                                       options:0 error:&error];

    NSLog(@"entries = %@",entries);

    for (NSDictionary *entry in entries) {
        NSLog(@"entry = %@",entry);
        NSString *message = [entry objectForKey:@"message"];
        NSLog(@"message = %@",message );

        NSString* result = [entry objectForKey:@"result"];
        NSLog(@"result = %@", result);

    }

Here is my output;

innerJson = {"result": false,"message":"message!"}//I need parse this string.
entries = {
message = "message!";
result = 0;
}
entry = message

I am taking error in for loop. What am i doing wrong? Thanks for interest and advice.

Error is this :-[__NSCFString objectForKey:]: unrecognized selector sent to instance 0x971d640 2014-01-23 23:29:41.028 postData[10311:70b] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[__NSCFString objectForKey:]: unrecognized selector sent to instance 0x971d640' — Kirdok
– Kirdok, Commented Jan 23, 2014 at 21:36
That pretty obvious. You have an NSString, not an NSDictionary. — rmaddy
– rmaddy, Commented Jan 23, 2014 at 21:38
So you just used the first NSJSONSerialization pass to "unescape" the data. — Hot Licks
– Hot Licks, Commented Jan 23, 2014 at 22:08

Wojtek Surowka · Accepted Answer · 2014-01-23 21:52:41Z

1

The problem is that in the line

NSArray *entries = [NSJSONSerialization

you assign the result to NSArray, and your JSON object is a dictionary, so you should assign to NSDictionary. Then the for loop is unnecessary - you can do:

NSDictionary *entries = [NSJSONSerialization JSONObjectWithData:[innerJson dataUsingEncoding:NSUTF8StringEncoding] options:0 error:&error];

NSLog(@"entries = %@",entries);
NSString *message = [entries objectForKey:@"message"];
NSLog(@"message = %@",message );

NSString* result = [entries objectForKey:@"result"];
NSLog(@"result = %@", result);

Read more details in NSJSONSerialization documentation.

edited Jan 23, 2014 at 21:52

answered Jan 23, 2014 at 21:38

Wojtek Surowka

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8 Comments

Wojtek Surowka Over a year ago

I think it does. The loop for (NSDictionary *entry in entries) really loops through NSDictionary, so the key (which is a string) is assigned to variable of NSDictionary type, and then wrong selector is sent to it.

rmaddy Over a year ago

No, the problem is that entries contains different types of objects but the code assumes it only contains dictionaries.

Hot Licks Over a year ago

The problem is that the user assumes an array when it's a dictionary, and then proceeds to loop when a loop is unnecessary to fetch those two values.

Wojtek Surowka Over a year ago

@maddy: NSJSONSerialization documentation clearly says: the top level object is an NSArray or NSDictionary. In this case it is NSDictionary. And the result of JSONObjectWithData is assigned to NSArray. The rest is outcome of this problem.

Wojtek Surowka Over a year ago

I guess in your example you missed [ ] characters. Anyway, in for JSON of such structure your original code will work, since in this case you will get NSArray consisted of two NSDictionary objects.

|

Artur Kucaj · Accepted Answer · 2014-01-23 22:01:34Z

0

I'm not sure - but I guess this code should work for you:

NSString *responseString = [request responseString];
NSLog(@"response String = %@",responseString);
NSData *tempData = [responseString dataUsingEncoding:NSUTF8StringEncoding];
NSError *error = nil;
/* assuming it is a dictionary */
NSDictionary *inner = [NSJSONSerialization JSONObjectWithData:tempData
                                                      options:NSJSONReadingAllowFragments error:&error];
NSLog(@"inner = %@",inner);
for(id key in inner) {
    id value = [inner objectForKey:key];
    NSLog(@"key=%@, value=%@", key, value);
}

answered Jan 23, 2014 at 22:01

Artur Kucaj

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