Converting Json Object to String - Objective C

Question

Imagine I make an request to my api that return an mysql_query("Select * from user where id=1").

After the request is done, it returns the user info in json. I do some debuging, NSlog(@"JSON : %@",json); and it gives me this:

JSON : (
    {
    aboutme = "";
    active = 0;
    birthday = "1992-10-14";
    "city_id" = 0;
    email = "[email protected]";
    fbid = "";
    firstname = test;
    gender = 1;
    id = 162;
    lastname = test;
    password = "$2a$12$8iy.sGr.4V/Ea3GfHZe0m.SLDrvoSj3/wYRlWsNce1yyCMeCbDrMC";
    "phone_number" = "";
    "recovery_date" = "0000-00-00 00:00:00";
    "register_date" = "2013-06-06 02:44:20";
    salt = "8iy.sGr.4V/Ea3GfHZe0m";
    "user_type_id" = 1;
    username = test;
}
)

Now I parse it with AFJONDecode and when I get the [json valueForKey:@"username"]; and debug it ( NSlog(@"username = %@",[json valueForKey@"username"]); ) and I get this:

username = (
    testeteste739
)

It gives me an object ( because in the Json, the username = test and not username = "test").

So, how can i convert this object to string?

** UPDATE **

I resolve it by the following way:

NSArray *username = [JSON valueForKey:@"username"];
username = [username objectAtIndex:0];

Is there any better way to overpass this? Thanks

The value of username is an NSString with a value of @"test". Just because it doesn't log with quotes doesn't mean it isn't a string. — rmaddy
– rmaddy, Commented Jun 6, 2013 at 1:57
I don't think so, because if i do NSlog(@"username = --%@--", username) i get: username = --( testeteste1009 )-- — Tiago Almeida
– Tiago Almeida, Commented Jun 6, 2013 at 2:04
First off, what you have above is not a JSON string. It is a dump of NSDictionary and NSArray objects, representing the interalized form of the JSON. In other words, it is already "parsed". — Hot Licks
– Hot Licks, Commented Jun 6, 2013 at 2:14
What you have there is an NSArray (apparently named "json") which contains a single array element which is an NSDictionary. To get, eg, username you'd first extract element zero of json (NSDictionary* dict = [json objectAtIndex:0]) and then extract the appropriate array element (NSString* username = [dict objectForKey:@"username"];). — Hot Licks
– Hot Licks, Commented Jun 6, 2013 at 2:18
And note that since the above is an Objective-C object dump, and not JSON, string values that have no blanks or special characters are shown without enclosing quotes. — Hot Licks
– Hot Licks, Commented Jun 6, 2013 at 2:20

Prince Kumar Sharma · Accepted Answer · 2013-06-06 05:27:03Z

27

As JSON is a dictionary object, so you can get your json data to JSONDic NSDictionary variable and parse it to string as follows:-

NSDictionary *JSONDic=[[NSDictionary alloc] init];
NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:JSONDic
                                                   options:NSJSONWritingPrettyPrinted 
                                                     error:&error];
NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];

answered Jun 6, 2013 at 5:27

Prince Kumar Sharma

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4 Comments

Hot Licks Over a year ago

That is, of course, the way to create a JSON string, given an internal representation in NSArray/NSDictionary objects. (Hopefully your JSONDict object has contents when you actually do this.) I think the OP is still struggling to get values OUT of the JSON.

Prince Kumar Sharma Over a year ago

As question title specifies the he want to convert JSON to string I did this , he didn't specify in the title that he want values out of JSON.

Hot Licks Over a year ago

He doesn't know what he wants.

Tiago Almeida Over a year ago

All i wanted was to get the value "username". I already resolve it, so thanks everyone and sorry for all my confusion!

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