Send JSON data with NSURLConnection in Xcode

Question

I am trying to send JSON data to a web server via the code below. For some reason, the request does not seem to be going out. What does it look like I am missing? Also the result from NSURLConnection (retStr) is always empty?

NSDictionary *data = [NSDictionary dictionaryWithObject:@"test sending ios" forKey:@"value1"];
NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:data options:kNilOptions error:&error];

    NSURL *url = [NSURL URLWithString:@"http://webserveraddress"];
NSMutableURLRequest *req = [NSMutableURLRequest requestWithURL:url cachePolicy:nil timeoutInterval:60];
[req setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[req setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[req setValue:[NSString stringWithFormat:@"%d", [jsonData length]] forHTTPHeaderField:@"Content-Length"];
[req setHTTPMethod:@"POST"];
[req setHTTPBody:jsonData];
NSString *retStr = [[NSString alloc] initWithData:[NSURLConnection sendSynchronousRequest:req returningResponse:nil error:nil] encoding:NSUTF8StringEncoding];

just an opinion you should send POST variable to your server and retrieve json from it. It will be easier especially if you connect through an http connection. easier and faster. And you can simply put in the HTTPBody NSString stringWithFormat:@"var1=%@&var2=%@&var3=%@..." and on the server you have in php $_POST['var1'], $_POST['var2'] etc... — Nicolas Manzini
– Nicolas Manzini, Commented Jun 14, 2012 at 15:19
@NicolasManzini Can you give me some example code for this. I tried rearranging the code above and removing NSJSONSerialization, and I am still not getting any POST on my web server? Would I need to remove the: [[NSString alloc] initWithData:[NSURLConnection sendSynchronousRequest:req returningResponse:nil error:nil] encoding:NSUTF8StringEncoding]? — Nick
– Nick, Commented Jun 14, 2012 at 17:32

WrightsCS · Accepted Answer · 2014-02-03 23:35:41Z

3

To send simple data in post vars to your webserver running php you simply do this in

Example

NSString * key = [NSString stringWithFormat:@"var1=%@&var2=%@&var3=%@",@"var1String" ,@"var2string" ,[NSnumber numberWithBool:YES]];

NSURL * url = [NSURL URLWithString:@"http://webserver.com/yourScriptPHP.php"];

NSMutableURLRequest * request = [NSMutableURLRequest requestWithURL:url];

[request setHTTPMethod:@"POST"];
[request setHTTPBody:[key dataUsingEncoding:NSUTF8StringEncoding]];

[[NSURLConnection alloc] initWithRequest:request delegate:self];
// this is for you to be able to get your server answer. 
// you will need to make your class a delegate of NSURLConnectionDelegate and NSURLConnectionDataDelegate
myClassPointerData = [[NSMutableData data] retain];

Implement

-(void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data {
    [myClassPointerData appendData:data]
}

-(void)connection:(NSURLConnection *)connection DidFinishLoading {
    // do what you want with myClassPointerData the data that your server did send you back here
    // for info on your server php script you just need to do: echo json_encode(array('var1'=> $var1, 'var2'=>$var2...));
    // to get your server sending an answer
}

edited Feb 3, 2014 at 23:35

WrightsCS

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answered Jun 15, 2012 at 8:20

Nicolas Manzini

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1 Comment

Nicolas Manzini Over a year ago

dont forget to release your NSURLCOnnecion and your myClassPointerData

ganesh manoj · Accepted Answer · 2012-06-14 13:15:34Z

0

once check like this

ASIFormDataRequest *request=[ASIFormDataRequest requestWithURL:[NSURL URLWithString:@"ur url"]];

[request setPostValue:appdelegate.userid forKey:@"userid"];
[request setPostValue:self.nameLbl.text forKey:@"username"];
[request setPostValue:self.website.text forKey:@"website"];

NSLog(@"~~~~~~ request~~~~~ %@",request);

[request setTimeOutSeconds:5000];
[request startSynchronous];

answered Jun 14, 2012 at 13:15

ganesh manoj

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Comments

codeburn · Accepted Answer · 2013-04-02 12:58:01Z

0

Instead of sending as JSON, you could simply send it the normal way - by setting the request's HTTPBody with POST method.

Only differenc you need to make if you need to make a call such as 'http://abc.example.com/user_profile.json?user[first_name]=fdffdf&user[last_name]=dffdf'

is, you need to have your dictionary keys have a prefix. That is, 'first_name=fdffdf' needs to be changed to 'user[first_name]=fdffdf'.

Try this bit of code to change your parameter dictionary to the required format for JSON..

for (id key in [self allKeys]) {
NSString *newKey = [NSString stringWithFormat:@"%@[%@]", parent, key];
[result setObject:[self objectForKey:key] forKey:newKey];
}

answered Apr 2, 2013 at 12:58

codeburn

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