我一直在为简单的iOS用户系统开发Web服务。 PHP查询完美,但有一个问题 - iOS工作不太好。数据未收到。我如何继续?这是我的连接类:iOS连接不返回数据
import Foundation
class Connection: NSObject {
var data: NSMutableData = NSMutableData()
func login(username: String, password: String) {
var url: NSURL = NSURL(string: "localhost/getusers.php?username=" + username + "&password=" + password)!
var request: NSURLRequest = NSURLRequest(URL: url)
var conn: NSURLConnection = NSURLConnection(request: request, delegate: self, startImmediately: true)!
}
func connection(didReceiveResponse: NSURLConnection!, didReceiveResponse response: NSURLResponse!) {
println("didReceiveResponse")
}
func connection(connection: NSURLConnection!, didReceiveData conData: NSData!) {
self.data.appendData(conData)
}
func connectionDidFinishLoading(connection: NSURLConnection!) {
println(self.data)
}
deinit {
println("deiniting")
}
}
这里它被称为地方。
@IBAction func attemptLogin(sender: UIButton) {
if(usernameTextField.text == "" || passwordTextField.text == "") {
var alert = UIAlertController(title: "Error", message: "Invalid Credentials", preferredStyle: UIAlertControllerStyle.Alert)
alert.addAction(UIAlertAction(title: "Close", style: UIAlertActionStyle.Default, handler: nil))
self.presentViewController(alert, animated: true, completion: nil)
var connection: Connection = Connection()
connection.login(usernameTextField.text, password: passwordTextField.text)
}
}
最后,这里的PHP服务:
error_reporting(E_ALL);
ini_set('display errors', 1);
$username = 'root';
$password = 'root';
try {
$DBH = new PDO('mysql:host=localhost; dbname=login_test', $username, $password);
$DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$recievedUsername = $_GET['username'];
$recievedPassword = $_GET['password'];
$data = array($recievedUsername, $recievedPassword);
$STH = $DBH->prepare("SELECT id, first_name, last_name FROM `users` WHERE `username` = ? AND `password` = ?");
$STH->execute($data);
$row = $STH->fetch(PDO::FETCH_ASSOC);
echo json_encode($row);
} catch(PDOException $e) {
echo $e->getMessage();
}
'NSURLConnectionDelegate'是一种罕见的鸭子类型的代表 - 参数仅仅是类型为'AnyObject',所以你不一定需要*声明*符合您的委托方法被调用(虽然这是一个很好的做法)。 – 2014-12-06 20:02:08
经过测试,它们都运行。 – Jeffrey 2014-12-06 21:47:38