ios json not working when using include() - php

Question

I have an app that receives a JSON file generated by my jason.php script and displays the data in a table view.
It works fine until I try to use 'include(db_connect.php)' in my jason.php file to pass the database log in details to it.
Running my php script, with 'include(db_connect.php)', does work in a browser (returns the JSON file formatted correctly) but it doesn’t work on my phone.
However..
It does work on my phone if I just paste the contents of db_connect.php into the jason.php file...and it returns exactly the same JSON file in a browser.
Both ways return exactly the same JSON text in browser.
All the app does is expect to receive a JSON file from a specified URL, it does’t pass anything to it. Just visits the URL and stores whats returned in an NSData object.
If anyone knows why this is happening I would be grateful to know!
Thanks
jason.php: this returns a the JSON script perfectly in my browser
<?php
require("db_connect.php");
//Check to see if we can connect to the server
if(!$connection)
{
die("Database server connection failed.");
}
else
{
//Attempt to select the database
$dbconnect = mysql_select_db($db, $connection);
//Check to see if we could select the database
if(!$dbconnect)
{
die("Unable to connect to the specified database!");
}
else
{
$query = "SELECT * FROM cities";
$resultset = mysql_query($query, $connection);
$records = array();
//Loop through all our records and add them to our array
while($r = mysql_fetch_assoc($resultset))
{
$records[] = $r;
}
//Output the data as JSON
echo json_encode($records);
}
}
?>
db_connect.php the log in details
<?php
$host = "xxxxx"; //Your database host server
$db = "xxxxx"; //Your database name
$user = "xxxxx"; //Your database user
$pass = "xxxxx"; //Your password
$connection = mysql_connect($host, $user, $pass);
?>
jason_pasted.php this is exactly the same as jason.php but the contents of db_connect.php are just pasted in - produces exactly the same result in browser, and also works when used in my app.
<?php
$host = "xxxxx"; //Your database host server
$db = "xxxxxx"; //Your database name
$user = "xxxxx"; //Your database user
$pass = "xxxxxx"; //Your password
$connection = mysql_connect($host, $user, $pass);
//Check to see if we can connect to the server
if(!$connection)
{
die("Database server connection failed.");
}
else
{
//Attempt to select the database
$dbconnect = mysql_select_db($db, $connection);
//Check to see if we could select the database
if(!$dbconnect)
{
die("Unable to connect to the specified database!");
}
else
{
$query = "SELECT * FROM cities";
$resultset = mysql_query($query, $connection);
$records = array();
//Loop through all our records and add them to our array
while($r = mysql_fetch_assoc($resultset))
{
$records[] = $r;
}
//Output the data as JSON
echo json_encode($records);
}
}
?>
ViewController.m extract from the app code
-(void) retrieveData
{
NSURL *url = [NSURL URLWithString:jsonURL];
NSData *data = [NSData dataWithContentsOfURL:url];
json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
//set up cities array
citiesArray = [[NSMutableArray alloc]init];
for (int i=0;i<[json count]; i++)
{
//create city object
NSString *cID = [[json objectAtIndex:i] objectForKey:#"id"];
NSString *cName = [[json objectAtIndex:i] objectForKey:#"cityName"];
NSString *cState = [[json objectAtIndex:i] objectForKey:#"cityState"];
NSString *cPopulation = [[json objectAtIndex:i] objectForKey:#"cityPopulation"];
NSString *cCountry = [[json objectAtIndex:i] objectForKey:#"country"];
City *myCity = [[City alloc] initWithCityID:cID
andCityName:cName
andCityState:cState
andCityPopulation:cPopulation
andCityCountry:cCountry];
//add city oject to city array
[citiesArray addObject:myCity];
}
[davesTableView reloadData];
}
TL;DR the app works perfectly with jason_pasted.php but not jason.php.
jason.php and jason_pasted.php return exactly the same JSON script when opened in a browser.
String returned from jason.php and jason_pasted.php
(
{
cityName = London;
cityPopulation = 8173194;
cityState = London;
country = "United Kingdom";
id = 1;
},
{
cityName = Bombay;
cityPopulation = 12478447;
cityState = Maharashtra;
country = India;
id = 2;
},
{
cityName = "Kuala Lumpur";
cityPopulation = 1627172;
cityState = "Federal Territory";
country = Malaysia;
id = 3;
},
{
cityName = "New York";
cityPopulation = 8336697;
cityState = "New York";
country = "United States";
id = 4;
},
{
cityName = Berlin;
cityPopulation = 3538652;
cityState = Berlin;
country = Deutschland;
id = 5;
}
)
error returned only when NSUrl points to jason.php
2014-02-13 11:43:34.760 JSONios[4655:60b] JSON error: Error Domain=NSCocoaErrorDomain Code=3840
"The operation couldn’t be completed. (Cocoa error 3840.)" (JSON text did not start with array or
object and option to allow fragments not set.)
UserInfo=0x14659c40 {NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}

Answer 1

This is placed in an answer for formatting:
Do not ignore errors!
Incorrect:
json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
Note: The docs do not specify that nil may be passed for the error parameter.
Correct:
// There is an assumption that the JSON head is a dictionary.
NSError *error;
NSDictionary *jsonAsDict = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
if (jsonAsDict == nil) {
NSLog(#"JSON error: %#", error);
}
else { // For debug only
NSLog(#"JSON: %#", jsonAsDict);
}
Now, what happens with this code?
Also please provide the JSON string if possible.
Oh, I personally to not care how the php creates the JSON, all I need to see is the JSON.
Still having trouble: NSLog the data as a string:
NSLog(#"data: %#", [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding]);
If no data add the error parameter to
dataWithContentsOfURL:options:error:
in place of dataWithContentsOfURL: