How to handle PHP errors in Swift - php

Question

I'm using PHP & JSON to extract some data from a database.
This is my PHP file
<?php
error_reporting(0);
ini_set('error_reporting', E_ALL);
ini_set('display_errors','Off');
$mysqli = new mysqli("localhost", "root", $_REQUEST['password'], "");
if ($mysqli->connect_errno) {
echo "Failed to connect to DB.";
die();
} else {
$dbs = array();
$res = $mysqli->query("SHOW DATABASES");
$res->data_seek(0);
if ($res->num_rows > 0) {
while($row = $res->fetch_assoc()) {
$dbs[] = $row;
}
echo json_encode($dbs);
} else {
echo "Failed to get list of databases from server.";
die();
}} ?>
If the password is wrong, then the system outputs "Failed to connect to DB"
In my program, I have things to handle errors, but I am stuck at one part.
let urlString = "http://\(hostTextField.text):\(portTextField.text)/dblist.php? &password=\(passTextField.text)"
let url: NSURL = NSURL(string: urlString)!
let urlSession = NSURLSession.sharedSession()
println(url)
println(urlSession)
//2
let jsonQuery = urlSession.dataTaskWithURL(url, completionHandler: { data, response, error -> Void in
println(response)
println(data)
if (error != nil) {
println("Can't connect using credentials")
dispatch_async(dispatch_get_main_queue(), {
HUDController.sharedController.hide(afterDelay: 0.1)
})
sleep(1)
var refreshAlert = UIAlertController(title: "Camaleon Reports", message: "Can't connect to the database", preferredStyle: UIAlertControllerStyle.Alert)
refreshAlert.addAction(UIAlertAction(title: "Retry", style: .Default, handler: { (action: UIAlertAction!) in
println("Yes Logic")
}))
self.presentViewController(refreshAlert, animated: true, completion: nil)
return }
var err: NSError?
var jsonResult: [Dictionary<String, String>] = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: &err) as [Dictionary<String, String>]
// 3
if (err != nil) {
println("Still cant connect....")
println("JSON Error \(err!.localizedDescription)")
}
var jsonDB : [Dictionary<String, String>] = jsonResult
for currentDictionary in jsonDB{
var currentEntry = currentDictionary["Database"] as String!
My program crashes if I don't have the right password, but have the right IP address and Port/User for the MYSQL Database.
It crashes with this:
fatal error: unexpectedly found nil while unwrapping an Optional value
and points towards jsonResult. It makes sense, cause I don't retrieve two strings.
My problem is that if my password is off, then my PHP file echoes a string. How can I search for that string so that I can use an if statement and stop my application from crashing?

Answer 1

Your problem is likely in this line (wrapped for clarity):
var jsonResult: [Dictionary<String, String>] =
NSJSONSerialization.JSONObjectWithData(data,
options: NSJSONReadingOptions.MutableContainers,
error: &err) as [Dictionary<String, String>]
When your PHP script reports the error by just returning the string, it has returned invalid JSON. When you use NSJSONSerialization.JSONObjectWithData to parse it, that method will return nil if the JSON is invalid, as yours is.
You then take that value and assign it to a Swift variable that you've declared is not an optional. Trying to assign nil to a variable not declared with either ? or ! is a runtime error in Swift. (You don't get an error at compile time because you're using as to cast the value.)
One way to fix this would be to change your PHP so the error is proper JSON:
echo "{ \"error\": \"Failed to connect to DB.\" }"; # or something, my PHP is rusty
But that still leaves your Swift program in a fragile state; getting anything but proper JSON back from the server will make it crash.
Better is to declare the jsonResult variable as being an optional:
var jsonResult: [Dictionary<String, String>]? =
NSJSONSerialization.JSONObjectWithData(data,
options: NSJSONReadingOptions.MutableContainers,
error: &err) as [Dictionary<String, String>]?
Then in your code you can explicitly check whether jsonResult is nil, and if it is, you know an error has occurred, and can go back and look at the data object to see what it was.
Even that, though, can leave you in trouble. The root of a JSON document doesn't have to be a dictionary; it could be an array. And even if it is a dictionary, the values may not all be strings; they could be numbers, booleans, or nested arrays or dictionaries!
Objective-C's relatively lax type checking makes this easy to deal with, but Swift is stricter. Best might be to use one of the Swift-specific JSON libraries. That'll make your code far more robust.
Good luck!

Answer 2

There are two issues. One is the PHP and one is the Swift.
Your PHP really should never just report an error message. I'd suggest that it always return JSON. This will make it easier for your client code to detect and handle errors appropriately.
<?php
header("Content-Type: application/json");
$response = array();
error_reporting(0);
ini_set('error_reporting', E_ALL);
ini_set('display_errors','Off');
if (!isset($_REQUEST['password'])) {
$response["success"] = false;
$response["error_code"] = 1;
$response["error_message"] = "No password provided";
echo json_encode($response);
exit();
}
$mysqli = new mysqli("localhost", "root", $_REQUEST['password'], "");
if ($mysqli->connect_errno) {
$response["success"] = false;
$response["error_code"] = 2;
$response["mysql_error_code"] = $mysqli->connect_errno;
$response["error_message"] = $mysqli->connect_error;
echo json_encode($response);
exit();
}
if ($res = $mysqli->query("SHOW DATABASES")) {
$dbs = array();
$res->data_seek(0);
if ($res->num_rows > 0) {
while($row = $res->fetch_assoc()) {
$dbs[] = $row;
}
$response["success"] = true;
$response["results"] = $dbs;
} else {
$response["success"] = false;
$response["error_code"] = 3;
$response["error_message"] = "Failed to get list of databases from server.";
}
$res->close();
} else {
$response["success"] = false;
$response["error_code"] = 4;
$response["mysql_error_code"] = $mysqli->errno;
$response["error_message"] = $mysqli->error;
}
$mysqli->close();
echo json_encode($response);
?>
Note, this:
Specifies application/json header for Content-Type;
Always returns a dictionary, containing
a "success" key, which is either true or false;
if an error, an error_code indicating the type of error (1 = no password provided; 2 = connect failed; 3 = no databases found; 4 = some SQL error);
if an error, an error_msg string indicating the error message string; and
if a success, a results array (much like you used to return at the root level).
On the Swift side, you need to :
Change it to look for these various server app-level errors (note, I make the top level structure a dictionary, and your original array of dictionaries a particular value;
You might want to proactively check the statusCode of the response object, to make sure the server gave you a 200 return code (e.g. 404 means that the page was not found, etc.);
You might also want to check for JSON parsing errors (in case some bug in the server prevented well-formed JSON from being returned); and
You really should be percent-escaping the password (because if it included + or & characters, it wouldn't get transmitted successfully otherwise).
Thus, you might have something like:
let encodedPassword = password.stringByAddingPercentEncodingForURLQueryValue()!
let body = "password=\(encodedPassword)"
let request = NSMutableURLRequest(URL: URL!)
request.HTTPBody = body.dataUsingEncoding(NSUTF8StringEncoding)!
request.HTTPMethod = "POST"
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) { data, response, error in
// detect fundamental network error
guard error == nil && data != nil else {
print("network error: \(error)")
return
}
// detect fundamental server errors
if let httpResponse = response as? NSHTTPURLResponse where httpResponse.statusCode != 200 {
// some server error
print("status code was \(httpResponse.statusCode); not 200")
return
}
// detect parsing errors
guard let responseObject = try? NSJSONSerialization.JSONObjectWithData(data!, options: []) as? [String : AnyObject] else {
// some problem parsing the JSON response
print(String(data: data!, encoding: NSUTF8StringEncoding))
return
}
// now parse app-level response to make sure `status` was `true`
guard let success = responseObject!["success"] as? NSNumber else {
print("Problem extracting the `success` value") // we should never get here
return
}
if !success.boolValue {
print("server reported error")
if let errorCode = responseObject!["error_code"] as? NSNumber {
switch (errorCode.integerValue) {
case 1:
print("No password provided")
case 2:
print("Connection failed; probably bad password")
case 3:
print("No databases found")
case 4:
print("Some SQL error")
default:
print("Unknown error code: \(errorCode)") // should never get here
}
}
if let errorMessage = responseObject!["error_message"] as? String {
print(" message=\(errorMessage)")
}
return
}
if let databases = responseObject!["results"] as? [[String : AnyObject]] {
print("databases = \(databases)")
}
}
task.resume()
The percent-escaping code is in a String category:
extension String {
// see RFC 3986
func stringByAddingPercentEncodingForURLQueryValue() -> String? {
let characterSet = NSCharacterSet(charactersInString:"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-._~")
return self.stringByAddingPercentEncodingWithAllowedCharacters(characterSet)
}
}
A couple of other ancillary observations:
Never send a passwords in the clear. Put them in the body of a POST request (not the URL), and then use a https URL.
I'd personally not use the MySQL password part of the app-level authentication. I'd keep MySQL authentication logic encoded on the server-side, and then use your own user authentication table.