JSON Parsing in iOS 7

Ajeet picture Ajeet · Oct 16, 2013 · Viewed 83.3k times · Source

I am creating an app for as existing website. They currently has the JSON in the following format :

[

   {
       "id": "value",
       "array": "[{\"id\" : \"value\"} , {\"id\" : \"value\"}]"
   },
   {
       "id": "value",
       "array": "[{\"id\" : \"value\"},{\"id\" : \"value\"}]"
   } 
]

which they parse after escaping the \ character using Javascript.

My problem is when i parse it in iOS using the following command :

NSArray *result = [NSJSONSerialization JSONObjectWithData:jsonData options:kNilOptions error:&localError];

and do this :

NSArray *Array = [result valueForKey:@"array"];

Instead of an Array I got NSMutableString object.

  • The website is already in production so I just cant ask them to change their existing structure to return a proper JSON object. It would be a lot of work for them.

  • So, until they change the underlying stucture, is there any way i can make it work in iOS like they do with javascript on their website?

Any help/suggestion would be very helpful to me.

Answer

Rob picture Rob · Oct 16, 2013

The correct JSON should presumably look something like:

[
    {
        "id": "value",
        "array": [{"id": "value"},{"id": "value"}]
    },
    {
        "id": "value",
        "array": [{"id": "value"},{"id": "value"}]
    }
]

But, if you're stuck this the format provided in your question, you need to make the dictionary mutable with NSJSONReadingMutableContainers and then call NSJSONSerialization again for each of those array entries:

NSMutableArray *array = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingMutableContainers error:&error];
if (error)
    NSLog(@"JSONObjectWithData error: %@", error);

for (NSMutableDictionary *dictionary in array)
{
    NSString *arrayString = dictionary[@"array"];
    if (arrayString)
    {
        NSData *data = [arrayString dataUsingEncoding:NSUTF8StringEncoding];
        NSError *error = nil;
        dictionary[@"array"] = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
        if (error)
            NSLog(@"JSONObjectWithData for array error: %@", error);
    }
}