creating an array of format array[][@"string"];

Question

I am trying to create an array similar to the one that is created like this:

NSData *jsonData = [NSData dataWithContentsOfURL:[NSURL URLWithString:@"www.example.com/j.php"]];
NSError *error = nil;
if (jsonData) {
     result = [NSJSONSerialization JSONObjectWithData:jsonData
                                              options:NSJSONReadingMutableContainers 
                                                error:&error];
}

where www.example.com/j.php Is JSON that looks like this:

[{"name":"matt","genre":"Photography","info":"foo","date":"2014-06-12"},{"name":"jamie","genre":"Art","info":"Bar","date":"2014-06-13"}]

This array is then used like

result[indexPath.row][@"name"];

How can I build a new NSArray with my own name, genre and info that is in the same format as the one above??

I have tried to print the array using logs but it comes up as (null)

and I know I do not create it like this:

[NSArray arrayWithObjects: @"1", @"2", @"3", @"4", @"5", @"6", @"7", @"8", @"9", nil]

Go to json.org and learn the JSON syntax. Then understand that a JSON "array" maps to an NSArray while a JSON "object" maps to an NSDictionary. — Hot Licks
– Hot Licks, Commented Jul 7, 2014 at 12:07

Yogesh Suthar · Accepted Answer · 2014-07-07 12:16:50Z

5

Its NSArray of NSDictionary.

NSArray *tmp = @[@{@"name": @"matt", @"genre" : @"Photography"}, @{@"name": @"jamie", @"genre" : @"Art"}];

You have to create it using the given example.

edited Jul 7, 2014 at 12:16

answered Jul 7, 2014 at 12:02

Yogesh Suthar

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12 Comments

Hot Licks Over a year ago

You are incorrect on that last update. You can access an NSDictionary using [] notation, in later versions of iOS.

Hot Licks Over a year ago

That error message is coming from the NSLog you're surrounding it all with.

maxisme Over a year ago

Thank you @YogeshSuthar how would I then add more results to the array? Like this? [tmp addObject: @[@{@"name": @"matt", @"genre" : @"Photography"}]]; ?

Hot Licks Over a year ago

(Hint: Do NSLog(@"%@", something);, not NSLog(something);.)

Yogesh Suthar Over a year ago

@HotLicks Ohhh yes yes I was trying it in NSLog. Thanks for correction. :)

|

said · Accepted Answer · 2014-07-07 13:07:36Z

0

use this code i hope help you :

 for (NSDictionary *data in jsonObjects) {


            NSString *name_data           =  [data objectForKey:@"name"];
            NSString *secteur_data      =  [data objectForKey:@"genre"];
            NSString *société_data      =  [data objectForKey:@"info"];

     DictionaryOfname = [NSDictionary dictionaryWithObjectsAndKeys:
                               [data objectForKey:@"name"],@"name",
                                  [data objectForKey:@"genre"],@"genre",
                                 [data objectForKey:@"info"],@"info",
                                   nil];

            [Listeofname addObject:DictionaryOfname];
    }

edited Jul 7, 2014 at 13:07

answered Jul 7, 2014 at 12:11

said

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3 Comments

trojanfoe Over a year ago

I can't see that even compiling.

trojanfoe Over a year ago

What does [ objectForKey:@"name"]; do then?

Hot Licks Over a year ago

And leading caps should not be used for variable names in Objective-C.

Jasmin · Accepted Answer · 2014-07-07 13:12:55Z

0

You can as @Yogesh Suthar suggested and Can access using valueForKeyPath or just valueForKey Method.

NSArray * tmp = @[
                      @{@"name": @"matt", @"genre" : @"Photography"},
                      @{@"name": @"jamie", @"genre" : @"Art"}
                 ];


    NSLog(@"name is : %@", tmp[0][@"name"]);
    NSLog(@"name is : %@", [tmp valueForKeyPath:@"name"][0]);

answered Jul 7, 2014 at 13:12

Jasmin

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