Send data from a PHP script to iOS

Question

I'm trying to write an iOS (5) app that uses json objects to get data from a server. im calling a php script from the app, that gets the data from the server, encodes it as json, and then outputs it. the line in the php script that does that is:

$json_encoded =  json_encode($normal_array, JSON_FORCE_OBJECT)
echo $json_encoded;

but when I do:

NSURL *url = [NSURL URLWithString:@"192.168.1.100/testing/get.php?number=5"];
NSData *data = [NSData dataWithContentsOfURL:url];

the data variable is of size 0 bytes and contains no data.

Can anyone tell me why the dataWithContentsOfURL gets something thats empty? Is it not enough to echo? Do i need to send a response or something like that? If so, how would I send a response from the php script? If i call the exact same script in a browser, it outputs:

{
    "0": {
        "0":"asdfgh",
        "1":"2012-06-11 16:21:15"
    },
    "1": {
        "0":"jumps over",
        "1":"2012-06-11 16:20:52"
    },
    "2": {
        "0":"the quick brown fox",
        "1":"2012-06-11 16:20:40"
    },
    "3": {
        "0":"the quick brown fox",
        "1":"2012-06-11 16:20:40"
    },
    "4": {
        "0":"the quick brown fox",
        "1":"2012-06-11 16:20:21"
    }
}

which is a valid JSON object, as far as I know.

MetalFrog · Accepted Answer · 2012-06-12 12:48:14Z

1

Make sure you send a header with a content type.

header('Content-type: application/json');
print json_encode($array);
exit();

answered Jun 12, 2012 at 12:48

MetalFrog

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Comments

Luc · Accepted Answer · 2012-06-12 12:50:00Z

0

My guess is that your device doesnt see that local ip address? Either that or maybe it's json parsing is much stricter. Did you output the json header with your php script?

answered Jun 12, 2012 at 12:50

Luc

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Omar Abdelhafith · Accepted Answer · 2012-06-12 12:52:16Z

0

The url outputs string, so use the following

NSURL *url = [NSURL URLWithString:@"192.168.1.100/testing/get.php?number=5"];
NSString *jsonVal = [NSString stringWithContentsOfURL:url];
NSLog(jsonVal);

answered Jun 12, 2012 at 12:52

Omar Abdelhafith

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