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In my iPhone app, I am trying to use the JSON library (YAJL) to create a JSON string that looks like the following format:

{"user": 
  {"name":"Jon", "username":"jon22", "password":"passw@rd", "email":"jon22@example.com"}
}

But I can't figure out the YAJL methods to create this.

I have tried the following:

 NSArray *params = [NSArray arrayWithObjects:@"Jon", @"jon22", @"passw@rd", @"jon22@gmail.com", nil];
 NSArray *keys = [NSArray arrayWithObjects:@"name", @"username", @"password", @"email", nil];

 NSDictionary *userDictionary = [NSDictionary dictionaryWithObjects:params forKeys:keys];
 NSString *JSONString = [userDictionary yajl_JSONString];

However, the returned string is not wrapped in the outside "user". How can I use YAJL to create this json string? Does anyone have experience with this???

Many thanks, Brett

Brett
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2 Answers2

1

I don't use YAJL, but SBJSON, which is the one Apple uses in iOS too...

Anyway, the library is behaving correctly: you're not creating an "user" dictionary!

You need to do something like:

NSDictionary *serialize = [NSDictionary dictionaryWithObject:userDictionary forKey:@"user"];

And then "JSON-ize" this one.

This is because when you call -yajl_JSONString on userDictionary, you are using userDictionary as your "root object". If you want to wrap this inside another dictionary, you need to explicitly do so.

ItalyPaleAle
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    OK, right. This is what I need to know how to do though. Any ideas how to explicitly do this? – Brett Sep 06 '11 at 21:20
  • I told you :) Create a NSDictionary object like the "serialize" i wrote, then call [serialize yajl_JSONString], instead of [userDictionary yajl_JSONString]... Just that! – ItalyPaleAle Sep 06 '11 at 21:22
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    Ah, I misunderstood that. OK, makes perfect sense now! Many thanks! – Brett Sep 06 '11 at 21:24
0

I highly recommend that you check out JSONKit https://github.com/johnezang/JSONKit I have used numerous JSON parsers and have found it to be the easiest.

Also the documentation is awesome.

Shaun
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