Question

Like the title says..

My NSDictionary initWithObjectsAndKeys for JSON, with the POST method, to do MySQL query via PHP code, by using the json_encoding and json_decoding creates empty data in my SQL database, just the ID int is auto increasing every time I post!

My Xcode code:

-(IBAction)setJsonFromData:(id)sender
{
    NSDictionary *jsonDict = [[NSDictionary alloc] initWithObjectsAndKeys: @"Welcome", @"title", @"Hello", @"article", @"123456789", @"timestamp", nil];

    if([NSJSONSerialization isValidJSONObject:jsonDict])
        {
           NSError *error = nil;
           NSData *result = [NSJSONSerialization dataWithJSONObject:jsonDict     options:NSJSONWritingPrettyPrinted error:&error];
           if (error == nil && result != nil) {
           [self postJSONtoURL:result];
        }
    }
}

-(id)postJSONtoURL:(NSData *)requestJSONdata
{
    NSURL *url = [NSURL URLWithString:@"http://test.com/json.php"];
    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];

    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
    [request setValue:[NSString stringWithFormat:@"%d", [requestJSONdata length]] forHTTPHeaderField:@"Content-Length"];
    [request setHTTPBody: requestJSONdata];

    NSURLResponse *response = nil;
    NSError *error = nil;

    NSData *result = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

    NSLog(@"RESULT: %@", requestJSONdata);

    if (error == nil)
        return result;
    return nil;
}

My PHP code:

if (isset($_REQUEST))
{
    $json = $_REQUEST;
    $data = json_decode($json);

    $title = $_REQUEST['title'];
    $article = $_REQUEST['article'];
    $timestamp = $_REQUEST['timestamp'];

    mysql_query("INSERT INTO news (title, article, timestamp) VALUES ('$title->title','$article->article','$timestamp->timestamp')");
}

mysql_close();  

Answer 1

You’re receiving the json payload in the http body since you do a POST request.
So, you’ll need to decode that:

$http_body = file_get_contents('php://input');
$data = json_decode($http_body);

After that you should be able to access data like:

$data->title

PHP does not automatically decode incoming json data.

see the docs regarding that php://input thing.