Encode JSON data for URL - ios

Question

In iOS, I want to send JSON data in URL to make service call. I tried following code snipped but Encoded URL seems wrong. Because in JSON there is a colon character (:) between key and value and comma character (,) for separation. But, i am not able to encode colon(:) as %3A and comma(,) as %2C
Code Snippet:
- (NSURL *)getEncodedUrl {
// Build dictionnary with parameters
NSString *abc = #"abc";
NSNumber *limitNumber = [NSNumber numberWithInt:2];
NSMutableDictionary *dictionnary = [NSMutableDictionary dictionary];
[dictionnary setObject:limitNumber forKey:#"limit"];
[dictionnary setObject:abc forKey:#"abc"];
NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:dictionnary options:0 error:&error];
if (!jsonData) {
debug("Json error %#",error);
return nil;
} else {
NSString *JSONString = [[NSString alloc] initWithBytes:[jsonData bytes] length:[jsonData length] encoding:NSUTF8StringEncoding];
debug("Json op %#",JSONString);
NSString* params = [JSONString stringByAddingPercentEscapesUsingEncoding:
NSASCIIStringEncoding];
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:#"http://baseUrl.com?param=#",params]];
debug("URL = %#",url);
return url;
}
}
OUTPUT:~
URL = http://baseUrl.com?param=%7B%22abc%22:%22abc%22,%22limit%22:2%7D
(Include colon and comma characters)
But I want following o/p:
http://baseUrl.com?param=%7B%22abc%22%3A%22abc%22%2C%22limit%22%3A2%7D
(No colon and comma characters)
Online Encoding-Decoding Site that I am referring as of now.
http://www.url-encode-decode.com/

Answer 1

you can simply use
NSString *url = #"http://baseUrl.com?param=%7B%22abc%22:%22abc%22,%22limit%22:2%7D";
NSString *encodeImgUrl = [url stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

Answer 2

I'd recommend that you send the JSON as POST data instead of the GET that you're using. It'd be more straightforward to package it as MIME data and any encoding you do would be easier to understand.

Answer 3

So you are trying to generate the query portion of a URL here. Colons are a perfectly legitimate character to include in URL queries. I wrote an article covering the intricacies of escaping URL queries in Cocoa:
http://www.mikeabdullah.net/escaping-url-queries-in-cocoa.html
Since you're keen to perform extra escaping, I suggest taking my sample code and extending it to specially ask for : and ; characters to be escaped too.

Answer 4

I made small mistake in API call that is why I am getting wrong result. There is no need to encode colon(:) as %3A and comma(,) as %2C.
One more thing I would like to share with you. You can use base64 string instead of encoding JSON part.