Is there a quick way to POST an NSDictionary to a Python/Django server? - django

Question

I'm looking to send an NSDictionary up to a server running Django, and I would prefer if I had to do little or no work writing an encoder/parser.
Is there an easy way to accomplish this task?

Answer 1

iOS 5 has support for it in the framework. Look at NSJSONSerialization. Here is example code for post. Request object creation has been omitted in the code below.
NSDictionary *postDict = [NSDictionary dictionaryWithObjectsAndKeys:[NSDictionary dictionaryWithObjectsAndKeys:API_KEY, #"apiKey", userName, #"loginUserName", hashPassword, #"hashPassword", nil], #"loginReq", nil];
[request setHTTPMethod:#"POST"];
[request setValue:#"application/json" forHTTPHeaderField:#"content-type"];
NSError *error = nil;
[request setHTTPBody:[NSJSONSerialization dataWithJSONObject:postDict options:0 error:&error]];

Answer 2

iOS doesn't have support to do this as a one-liner, but you could do this:
#interface NSString (URLEncoding)
- (NSString *)urlEncodedUTF8String;
#end
#interface NSURLRequest (DictionaryPost)
+ (NSURLRequest *)postRequestWithURL:(NSURL *)url
parameters:(NSDictionary *)parameters;
#end
#implementation NSString (URLEncoding)
- (NSString *)urlEncodedUTF8String {
return (id)CFURLCreateStringByAddingPercentEscapes(0, (CFStringRef)self, 0,
(CFStringRef)#";/?:#&=$+{}<>,", kCFStringEncodingUTF8);
}
#end
#implementation NSURLRequest (DictionaryPost)
+ (NSURLRequest *)postRequestWithURL:(NSURL *)url
parameters:(NSDictionary *)parameters {
NSMutableString *body = [NSMutableString string];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
[request setHTTPMethod:#"POST"];
[request addValue:#"application/x-www-form-urlencoded"
forHTTPHeaderField:#"Content-Type"];
for (NSString *key in parameters) {
NSString *val = [parameters objectForKey:key];
if ([body length])
[body appendString:#"&"];
[body appendFormat:#"%#=%#", [[key description] urlEncodedUTF8String],
[[val description] urlEncodedUTF8String]];
}
[request setHTTPBody:[body dataUsingEncoding:NSUTF8StringEncoding]];
return request;
}
#end
Then it's as simple as:
NSURL *url = [NSURL URLWithString:#"http://posttestserver.com/post.php"];
NSDictionary *params = [NSDictionary dictionaryWithObjectsAndKeys:
[NSNumber numberWithInt:42], #"number",
#"apple", #"brand", nil];
NSURLRequest *request = [NSURLRequest postRequestWithURL:url parameters:params];
[NSURLConnection sendAsynchronousRequest:request queue:nil completionHandler:nil];
Please note that in this example we're not caring about the response. If you care about it, supply a block so you can do something with it.