Understanding xcode json web request - mysql

Question

this is my first time connecting an app to the web, I just want to make sure i am clear on how this works.
I want to download data from a mysql db hosted online, it also seems to have php attached (hostgator) is the host.
So steps are:
1. In php (on the server module), set up a method to transcribe mysql information into JSON using the attached PHP module. Return an array (json i think is a mutilayered dictionary array object)
In xcode, use apple's json framework to create a url request, and download data into a json object (or array or dictionary?
Go through the data and create objects and save to coredata.
Please let me know if im following the logic correctly.
Also does JSON return 1 object or all the objects on the mysql db. So if i need to input 10,000 objects on coredata i have to make 10,000 requests or one request and parse 10,000 objects worth of info?
Also is this the best way to do what i need to do? I have heard of http request but it seems complicated, and I have no clue what it is.
Sorry for such a noob question.
Thanks for the help.

Answer 1

There are many ways to do it, some are "more correct" than others :-) but you are on the right way.
I explain what I would do in a similar situation:
1- for the PHP engine, you should create an API to access your data. Which data?
The simpliest and probably the first thing you can do TO TEST (only for testing purpose!!!!) is to create a page that receive a query via POST from your ios APP and answer with an encoded JSON string. With JSON you can transfer an entire hierarchy of objects: variables, arrays, dictionary...is up to you to decide how link the objects together.
If the data you want to encode is a table, you can do something similar to this:
// first connect to the database and execute the query, then...
$data = array();
while(!$RS->EOF){
for ($i = 0; $i < $cols; $i++){
$rowName = utf8_encode($RS->Fields[$i]->name);
$rowValue = utf8_encode($RS->Fields[$i]->value);
$rowData[$rowName] = $rowValue;
}
array_push($data, $rowData);
$RS->MoveNext();
}
$response['data'] = $data;
echo json_encode($response);
the result is a JSON object with a first dictionary with a key named "data".
Inside the "data" key there is an array of dictionaries. Each dictionary has the column name for the key, and the column value for the data:
data = (
{
attivita = "this is the first row in the table";
id = 2548;
},
{
attivita = "this is the second row in the table";
id = 2547;
};
}
You can simply use json_encode to create your json string.
On the iPhone side, I suggest you to download and use AFNetworking. It's a very good and complete open source framework, with a lot of builtin objects and methods for http/https requests, XML, JSON ecc...
Using AFNetworking you can make the request in a similar way:
AFHTTPClient *httpClient = [[AFHTTPClient alloc] initWithBaseURL:#"http://url.for.the.php.page"];
NSMutableURLRequest *request = [httpClient requestWithMethod:#"POST" path:mainPath parameters:params];
AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
// your success code here
// remember, the first object is a dictionary with a key called data
NSArray *arrayOfData = [JSON objectForKey:#"data"];
NSMutableArray *arrayOfActivities = [NSMutableArray array];
for (NSDictionary *objectDictionary in arrayOfData) {
// objectDictionary is your dictionary containing
// a key with the column name and a value with the column content
// here you can init your custom objects
}
} failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
// your failure code here
}];
[operation start];
As you can see, the JSONRequestOperationWithRequest: method has a completion handler returning you the decoded JSON, so you can access your dictionaries/arrays directly using objectForKey or objectAtIndex ;-)
Last suggestions: in a production and secure environment, you should avoid sending query over post requests. The code I pasted here is for a private application (used only by me for testing purpose). It's better to use a different API for each kind of request and a secure authentication method (look at oAuth).
I suggest you to give a look to the Instagram or Twitter API (Instagram is simpler), trying to use it. They will give you some ideas on how to create your own API.
Good luck

Answer 2

Your first step of creating a method to return JSON-ified MySQL data is correct. What exactly is returned however, is completely up to you! You will find that JSON is an extremely flexible format that allows you to put nearly any format of data in your response. It may make sense, depending on what you want to do, to have several methods in your PHP module that return different things. For example, if you have a database of books, maybe you want to create a method that allows you to specify an author, and will return a JSON response with all the books in your database written by that author.
You could use HttpRequest to actually perform the call to your module, but I found the Obj-C NSURLRequest quite easy to use:
NSURLRequest *request = [NSURLRequest requestWithURL:
[NSURL URLWithString:#"http://yoururl/books/38917"]];
[[NSURLConnection alloc] initWithRequest:request delegate:self];
Don't think that you have to manually parse the JSON response into your own Objective-C objects; this is a very common task and there exist many great libraries to do so. Check out this tutorial on the SBJSON framework to view one of the most popular solutions.
As an example, here is how you could parse the JSON response into an NSDictionary for easy traversal:
// Create SBJSON object to parse JSON
SBJSON *parser = [[SBJSON alloc] init];
// parse the JSON string into an object - assuming json_string is a NSString of JSON data
NSDictionary *object = [parser objectWithString:json_string error:nil];